How to count occurrences of text in a file?





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







18

















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question




















  • 1





    With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    Mar 28 at 21:55






  • 1





    Do you have any database software available to use?

    – SpacePhoenix
    Mar 29 at 8:58






  • 1





    Related

    – Julien Lopez
    Mar 30 at 0:17













  • The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

    – j0h
    Mar 31 at 19:36


















18

















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question




















  • 1





    With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    Mar 28 at 21:55






  • 1





    Do you have any database software available to use?

    – SpacePhoenix
    Mar 29 at 8:58






  • 1





    Related

    – Julien Lopez
    Mar 30 at 0:17













  • The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

    – j0h
    Mar 31 at 19:36














18












18








18


6








I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question


















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"






command-line bash sort uniq






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 28 at 22:25









dessert

25.4k673107




25.4k673107










asked Mar 28 at 21:51









j0hj0h

6,5721657121




6,5721657121








  • 1





    With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    Mar 28 at 21:55






  • 1





    Do you have any database software available to use?

    – SpacePhoenix
    Mar 29 at 8:58






  • 1





    Related

    – Julien Lopez
    Mar 30 at 0:17













  • The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

    – j0h
    Mar 31 at 19:36














  • 1





    With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    Mar 28 at 21:55






  • 1





    Do you have any database software available to use?

    – SpacePhoenix
    Mar 29 at 8:58






  • 1





    Related

    – Julien Lopez
    Mar 30 at 0:17













  • The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

    – j0h
    Mar 31 at 19:36








1




1





With “bash”, do you mean the plain shell or the command line in general?

– dessert
Mar 28 at 21:55





With “bash”, do you mean the plain shell or the command line in general?

– dessert
Mar 28 at 21:55




1




1





Do you have any database software available to use?

– SpacePhoenix
Mar 29 at 8:58





Do you have any database software available to use?

– SpacePhoenix
Mar 29 at 8:58




1




1





Related

– Julien Lopez
Mar 30 at 0:17







Related

– Julien Lopez
Mar 30 at 0:17















The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

– j0h
Mar 31 at 19:36





The log is from an appache2 server, not really a database. bash is what I would prefer, in a general use case. I see the python and perl solutions, if they are good for someone else, that is great. the initial sorting was done with sort -V though I think that wasn't required. I sent the top 10 abusers of the login page to the system admin with recommendations for banning respective subnets. for example, One IP hit the login page over 9000 times. that IP, & its class D subnet is now blacklisted. I'm sure we could automate this, though that is a different question.

– j0h
Mar 31 at 19:36










8 Answers
8






active

oldest

votes


















13














You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done


grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



Example run



$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3





share|improve this answer





















  • 12





    This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

    – David
    Mar 29 at 1:56






  • 1





    @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

    – D. Ben Knoble
    Mar 29 at 3:56






  • 3





    I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

    – David
    Mar 29 at 5:26











  • By the way, why is it written as <log grep ... and not grep ... log?

    – Santiago
    Apr 3 at 17:09











  • @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

    – dessert
    Apr 3 at 17:28



















36














You can use cut and uniq tools:



cut -d ' ' -f1 test.txt  | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181


Explanation :





  • cut -d ' ' -f1 : extract first field (ip address)


  • uniq -c : report repeated lines and display the number of occurences






share|improve this answer





















  • 6





    One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

    – dessert
    Mar 28 at 22:22






  • 2





    This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

    – Guntram Blohm
    Mar 29 at 8:44



















13














If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



If you really need the given output format, a single-pass way to do it in Awk would be



awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



awk '
NR==1 {last=$1}
$1 != last {print last, "count: " c[last]; last = $1}
{c[$1]++}
END {print last, "count: " c[last]}
'


Ex.



$ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3





share|improve this answer


























  • it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

    – Peter A. Schneider
    Mar 29 at 11:12











  • @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

    – steeldriver
    Mar 29 at 12:07













  • Ah, yes, I see.

    – Peter A. Schneider
    Mar 29 at 12:36



















8














Here is one possible solution:





IN_FILE="file.log"
for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
do
echo -en "${IP}tcount: "
grep -c "$IP" "$IN_FILE"
done



  • replace file.log with the actual file name.

  • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

  • then grep -c will count each of these values within the file.




$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5





share|improve this answer





















  • 1





    Prefer printf...

    – D. Ben Knoble
    Mar 29 at 3:58






  • 1





    This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

    – terdon
    Mar 29 at 16:07



















5














Some Perl:



$ perl -lae '$k{$F[0]}++; }{ print "$_ count: $k{$_}" for keys(%k)' log 
13.57.233.99 count: 1
18.206.226.75 count: 2
13.57.220.172 count: 9
5.135.134.16 count: 5
18.213.10.181 count: 3


This is the same idea as Steeldriver's awk approach, but in Perl. The -a causes perl to automatically split each input line into the array @F, whose first element (the IP) is $F[0]. So, $k{$F[0]}++ will create the hash %k, whose keys are the IPs and whose values are the number of times each IP was seen. The }{ is funky perlspeak for "do the rest at the very end, after processing all input". So, at the end, the script will iterate over the keys of the hash and print the current key ($_) along with its value ($k{$_}).



And, just so people don't think that perl forces you to write script that look like cryptic scribblings, this is the same thing in a less condensed form:



perl -e '
while (my $line=<STDIN>){
@fields = split(/ /, $line);
$ip = $fields[0];
$counts{$ip}++;
}
foreach $ip (keys(%counts)){
print "$ip count: $counts{$ip}n"
}' < log





share|improve this answer































    4














    Maybe this is not what the OP want; however, if we know that the IP address length will be limited to 15 characters, a quicker way to display the counts with unique IPs from a huge log file can be achieved using uniq command alone:



    $ uniq -w 15 -c log

    5 5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] ...
    9 13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] ...
    1 13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] ...
    2 18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] ...
    3 18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] ...


    Options:



    -w N compares no more than N characters in lines



    -c will prefix lines by the number of occurrences



    Alternatively, For exact formatted output I prefer awk (should also work for IPV6 addresses), ymmv.



    $ awk 'NF { print $1 }' log | sort -h | uniq -c | awk '{printf "%s count: %dn", $2,$1 }'

    5.135.134.16 count: 5
    13.57.220.172 count: 9
    13.57.233.99 count: 1
    18.206.226.75 count: 2
    18.213.10.181 count: 3


    Note that uniq won't detect repeated lines in the input file if they are not adjacent, so it may be necessary to sort the file.






    share|improve this answer





















    • 1





      Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

      – Martin Thornton
      Mar 29 at 23:17













    • I like it, I didnt know uniq could count!

      – j0h
      Mar 31 at 12:57



















    1














    FWIW, Python 3:



    from collections import Counter

    with open('sample.log') as file:
    counts = Counter(line.split()[0] for line in file)

    for ip_address, count in counts.items():
    print('%-15s count: %d' % (ip_address, count))


    Output:



    13.57.233.99     count: 1
    18.213.10.181 count: 3
    5.135.134.16 count: 5
    18.206.226.75 count: 2
    13.57.220.172 count: 9





    share|improve this answer

































      0














      cut -f1 -d- my.log | sort | uniq -c


      Explanation: Take the first field of my.log splitting on dashes - and sort it. uniq needs sorted input. -c tells it to count occurrences.






      share|improve this answer


























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        8 Answers
        8






        active

        oldest

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        8 Answers
        8






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        13














        You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



        for i in $(<log grep -o '^[^ ]*' | uniq); do
        printf '%s count %dn' "$i" $(<log grep -c "$i")
        done


        grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



        Example run



        $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
        5.135.134.16 count 5
        13.57.220.172 count 9
        13.57.233.99 count 1
        18.206.226.75 count 2
        18.213.10.181 count 3





        share|improve this answer





















        • 12





          This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

          – David
          Mar 29 at 1:56






        • 1





          @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

          – D. Ben Knoble
          Mar 29 at 3:56






        • 3





          I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

          – David
          Mar 29 at 5:26











        • By the way, why is it written as <log grep ... and not grep ... log?

          – Santiago
          Apr 3 at 17:09











        • @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

          – dessert
          Apr 3 at 17:28
















        13














        You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



        for i in $(<log grep -o '^[^ ]*' | uniq); do
        printf '%s count %dn' "$i" $(<log grep -c "$i")
        done


        grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



        Example run



        $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
        5.135.134.16 count 5
        13.57.220.172 count 9
        13.57.233.99 count 1
        18.206.226.75 count 2
        18.213.10.181 count 3





        share|improve this answer





















        • 12





          This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

          – David
          Mar 29 at 1:56






        • 1





          @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

          – D. Ben Knoble
          Mar 29 at 3:56






        • 3





          I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

          – David
          Mar 29 at 5:26











        • By the way, why is it written as <log grep ... and not grep ... log?

          – Santiago
          Apr 3 at 17:09











        • @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

          – dessert
          Apr 3 at 17:28














        13












        13








        13







        You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



        for i in $(<log grep -o '^[^ ]*' | uniq); do
        printf '%s count %dn' "$i" $(<log grep -c "$i")
        done


        grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



        Example run



        $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
        5.135.134.16 count 5
        13.57.220.172 count 9
        13.57.233.99 count 1
        18.206.226.75 count 2
        18.213.10.181 count 3





        share|improve this answer















        You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



        for i in $(<log grep -o '^[^ ]*' | uniq); do
        printf '%s count %dn' "$i" $(<log grep -c "$i")
        done


        grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



        Example run



        $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
        5.135.134.16 count 5
        13.57.220.172 count 9
        13.57.233.99 count 1
        18.206.226.75 count 2
        18.213.10.181 count 3






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 28 at 23:11

























        answered Mar 28 at 22:08









        dessertdessert

        25.4k673107




        25.4k673107








        • 12





          This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

          – David
          Mar 29 at 1:56






        • 1





          @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

          – D. Ben Knoble
          Mar 29 at 3:56






        • 3





          I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

          – David
          Mar 29 at 5:26











        • By the way, why is it written as <log grep ... and not grep ... log?

          – Santiago
          Apr 3 at 17:09











        • @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

          – dessert
          Apr 3 at 17:28














        • 12





          This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

          – David
          Mar 29 at 1:56






        • 1





          @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

          – D. Ben Knoble
          Mar 29 at 3:56






        • 3





          I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

          – David
          Mar 29 at 5:26











        • By the way, why is it written as <log grep ... and not grep ... log?

          – Santiago
          Apr 3 at 17:09











        • @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

          – dessert
          Apr 3 at 17:28








        12




        12





        This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        Mar 29 at 1:56





        This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        Mar 29 at 1:56




        1




        1





        @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

        – D. Ben Knoble
        Mar 29 at 3:56





        @David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?

        – D. Ben Knoble
        Mar 29 at 3:56




        3




        3





        I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

        – David
        Mar 29 at 5:26





        I would not call it a premature optimization, given that the more efficient solution is also simpler, but to each their own.

        – David
        Mar 29 at 5:26













        By the way, why is it written as <log grep ... and not grep ... log?

        – Santiago
        Apr 3 at 17:09





        By the way, why is it written as <log grep ... and not grep ... log?

        – Santiago
        Apr 3 at 17:09













        @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

        – dessert
        Apr 3 at 17:28





        @Santiago Because that’s better in many ways, as Stéphane Chazelas explains here on U&L.

        – dessert
        Apr 3 at 17:28













        36














        You can use cut and uniq tools:



        cut -d ' ' -f1 test.txt  | uniq -c
        5 5.135.134.16
        9 13.57.220.172
        1 13.57.233.99
        2 18.206.226.75
        3 18.213.10.181


        Explanation :





        • cut -d ' ' -f1 : extract first field (ip address)


        • uniq -c : report repeated lines and display the number of occurences






        share|improve this answer





















        • 6





          One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

          – dessert
          Mar 28 at 22:22






        • 2





          This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

          – Guntram Blohm
          Mar 29 at 8:44
















        36














        You can use cut and uniq tools:



        cut -d ' ' -f1 test.txt  | uniq -c
        5 5.135.134.16
        9 13.57.220.172
        1 13.57.233.99
        2 18.206.226.75
        3 18.213.10.181


        Explanation :





        • cut -d ' ' -f1 : extract first field (ip address)


        • uniq -c : report repeated lines and display the number of occurences






        share|improve this answer





















        • 6





          One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

          – dessert
          Mar 28 at 22:22






        • 2





          This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

          – Guntram Blohm
          Mar 29 at 8:44














        36












        36








        36







        You can use cut and uniq tools:



        cut -d ' ' -f1 test.txt  | uniq -c
        5 5.135.134.16
        9 13.57.220.172
        1 13.57.233.99
        2 18.206.226.75
        3 18.213.10.181


        Explanation :





        • cut -d ' ' -f1 : extract first field (ip address)


        • uniq -c : report repeated lines and display the number of occurences






        share|improve this answer















        You can use cut and uniq tools:



        cut -d ' ' -f1 test.txt  | uniq -c
        5 5.135.134.16
        9 13.57.220.172
        1 13.57.233.99
        2 18.206.226.75
        3 18.213.10.181


        Explanation :





        • cut -d ' ' -f1 : extract first field (ip address)


        • uniq -c : report repeated lines and display the number of occurences







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 28 at 22:34

























        answered Mar 28 at 22:04









        Mikael FloraMikael Flora

        441117




        441117








        • 6





          One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

          – dessert
          Mar 28 at 22:22






        • 2





          This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

          – Guntram Blohm
          Mar 29 at 8:44














        • 6





          One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

          – dessert
          Mar 28 at 22:22






        • 2





          This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

          – Guntram Blohm
          Mar 29 at 8:44








        6




        6





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        Mar 28 at 22:22





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        Mar 28 at 22:22




        2




        2





        This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

        – Guntram Blohm
        Mar 29 at 8:44





        This should be the accepted answer, as the one by dessert needs to read the file repeatedly so is much slower. And you can easily use sort file | cut .... in case you're not sure if the file is already sorted.

        – Guntram Blohm
        Mar 29 at 8:44











        13














        If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



        If you really need the given output format, a single-pass way to do it in Awk would be



        awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


        This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



        awk '
        NR==1 {last=$1}
        $1 != last {print last, "count: " c[last]; last = $1}
        {c[$1]++}
        END {print last, "count: " c[last]}
        '


        Ex.



        $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
        5.135.134.16 count: 5
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3





        share|improve this answer


























        • it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

          – Peter A. Schneider
          Mar 29 at 11:12











        • @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

          – steeldriver
          Mar 29 at 12:07













        • Ah, yes, I see.

          – Peter A. Schneider
          Mar 29 at 12:36
















        13














        If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



        If you really need the given output format, a single-pass way to do it in Awk would be



        awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


        This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



        awk '
        NR==1 {last=$1}
        $1 != last {print last, "count: " c[last]; last = $1}
        {c[$1]++}
        END {print last, "count: " c[last]}
        '


        Ex.



        $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
        5.135.134.16 count: 5
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3





        share|improve this answer


























        • it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

          – Peter A. Schneider
          Mar 29 at 11:12











        • @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

          – steeldriver
          Mar 29 at 12:07













        • Ah, yes, I see.

          – Peter A. Schneider
          Mar 29 at 12:36














        13












        13








        13







        If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



        If you really need the given output format, a single-pass way to do it in Awk would be



        awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


        This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



        awk '
        NR==1 {last=$1}
        $1 != last {print last, "count: " c[last]; last = $1}
        {c[$1]++}
        END {print last, "count: " c[last]}
        '


        Ex.



        $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
        5.135.134.16 count: 5
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3





        share|improve this answer















        If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



        If you really need the given output format, a single-pass way to do it in Awk would be



        awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


        This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



        awk '
        NR==1 {last=$1}
        $1 != last {print last, "count: " c[last]; last = $1}
        {c[$1]++}
        END {print last, "count: " c[last]}
        '


        Ex.



        $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
        5.135.134.16 count: 5
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 28 at 22:36

























        answered Mar 28 at 22:12









        steeldriversteeldriver

        70.6k11115187




        70.6k11115187













        • it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

          – Peter A. Schneider
          Mar 29 at 11:12











        • @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

          – steeldriver
          Mar 29 at 12:07













        • Ah, yes, I see.

          – Peter A. Schneider
          Mar 29 at 12:36



















        • it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

          – Peter A. Schneider
          Mar 29 at 11:12











        • @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

          – steeldriver
          Mar 29 at 12:07













        • Ah, yes, I see.

          – Peter A. Schneider
          Mar 29 at 12:36

















        it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

        – Peter A. Schneider
        Mar 29 at 11:12





        it would be easy to change the cut + uniq based answer with sed to appear in the demanded format.

        – Peter A. Schneider
        Mar 29 at 11:12













        @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

        – steeldriver
        Mar 29 at 12:07







        @PeterA.Schneider yes it would - I believe that was already pointed out in comments to that answer

        – steeldriver
        Mar 29 at 12:07















        Ah, yes, I see.

        – Peter A. Schneider
        Mar 29 at 12:36





        Ah, yes, I see.

        – Peter A. Schneider
        Mar 29 at 12:36











        8














        Here is one possible solution:





        IN_FILE="file.log"
        for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
        do
        echo -en "${IP}tcount: "
        grep -c "$IP" "$IN_FILE"
        done



        • replace file.log with the actual file name.

        • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

        • then grep -c will count each of these values within the file.




        $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3
        5.135.134.16 count: 5





        share|improve this answer





















        • 1





          Prefer printf...

          – D. Ben Knoble
          Mar 29 at 3:58






        • 1





          This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

          – terdon
          Mar 29 at 16:07
















        8














        Here is one possible solution:





        IN_FILE="file.log"
        for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
        do
        echo -en "${IP}tcount: "
        grep -c "$IP" "$IN_FILE"
        done



        • replace file.log with the actual file name.

        • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

        • then grep -c will count each of these values within the file.




        $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3
        5.135.134.16 count: 5





        share|improve this answer





















        • 1





          Prefer printf...

          – D. Ben Knoble
          Mar 29 at 3:58






        • 1





          This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

          – terdon
          Mar 29 at 16:07














        8












        8








        8







        Here is one possible solution:





        IN_FILE="file.log"
        for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
        do
        echo -en "${IP}tcount: "
        grep -c "$IP" "$IN_FILE"
        done



        • replace file.log with the actual file name.

        • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

        • then grep -c will count each of these values within the file.




        $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3
        5.135.134.16 count: 5





        share|improve this answer















        Here is one possible solution:





        IN_FILE="file.log"
        for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
        do
        echo -en "${IP}tcount: "
        grep -c "$IP" "$IN_FILE"
        done



        • replace file.log with the actual file name.

        • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

        • then grep -c will count each of these values within the file.




        $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3
        5.135.134.16 count: 5






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 28 at 22:20

























        answered Mar 28 at 22:07









        pa4080pa4080

        14.8k52872




        14.8k52872








        • 1





          Prefer printf...

          – D. Ben Knoble
          Mar 29 at 3:58






        • 1





          This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

          – terdon
          Mar 29 at 16:07














        • 1





          Prefer printf...

          – D. Ben Knoble
          Mar 29 at 3:58






        • 1





          This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

          – terdon
          Mar 29 at 16:07








        1




        1





        Prefer printf...

        – D. Ben Knoble
        Mar 29 at 3:58





        Prefer printf...

        – D. Ben Knoble
        Mar 29 at 3:58




        1




        1





        This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

        – terdon
        Mar 29 at 16:07





        This means you need to process the entire file multiple times. Once to get the list of IPs and then once more for each of the IPs you find.

        – terdon
        Mar 29 at 16:07











        5














        Some Perl:



        $ perl -lae '$k{$F[0]}++; }{ print "$_ count: $k{$_}" for keys(%k)' log 
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        13.57.220.172 count: 9
        5.135.134.16 count: 5
        18.213.10.181 count: 3


        This is the same idea as Steeldriver's awk approach, but in Perl. The -a causes perl to automatically split each input line into the array @F, whose first element (the IP) is $F[0]. So, $k{$F[0]}++ will create the hash %k, whose keys are the IPs and whose values are the number of times each IP was seen. The }{ is funky perlspeak for "do the rest at the very end, after processing all input". So, at the end, the script will iterate over the keys of the hash and print the current key ($_) along with its value ($k{$_}).



        And, just so people don't think that perl forces you to write script that look like cryptic scribblings, this is the same thing in a less condensed form:



        perl -e '
        while (my $line=<STDIN>){
        @fields = split(/ /, $line);
        $ip = $fields[0];
        $counts{$ip}++;
        }
        foreach $ip (keys(%counts)){
        print "$ip count: $counts{$ip}n"
        }' < log





        share|improve this answer




























          5














          Some Perl:



          $ perl -lae '$k{$F[0]}++; }{ print "$_ count: $k{$_}" for keys(%k)' log 
          13.57.233.99 count: 1
          18.206.226.75 count: 2
          13.57.220.172 count: 9
          5.135.134.16 count: 5
          18.213.10.181 count: 3


          This is the same idea as Steeldriver's awk approach, but in Perl. The -a causes perl to automatically split each input line into the array @F, whose first element (the IP) is $F[0]. So, $k{$F[0]}++ will create the hash %k, whose keys are the IPs and whose values are the number of times each IP was seen. The }{ is funky perlspeak for "do the rest at the very end, after processing all input". So, at the end, the script will iterate over the keys of the hash and print the current key ($_) along with its value ($k{$_}).



          And, just so people don't think that perl forces you to write script that look like cryptic scribblings, this is the same thing in a less condensed form:



          perl -e '
          while (my $line=<STDIN>){
          @fields = split(/ /, $line);
          $ip = $fields[0];
          $counts{$ip}++;
          }
          foreach $ip (keys(%counts)){
          print "$ip count: $counts{$ip}n"
          }' < log





          share|improve this answer


























            5












            5








            5







            Some Perl:



            $ perl -lae '$k{$F[0]}++; }{ print "$_ count: $k{$_}" for keys(%k)' log 
            13.57.233.99 count: 1
            18.206.226.75 count: 2
            13.57.220.172 count: 9
            5.135.134.16 count: 5
            18.213.10.181 count: 3


            This is the same idea as Steeldriver's awk approach, but in Perl. The -a causes perl to automatically split each input line into the array @F, whose first element (the IP) is $F[0]. So, $k{$F[0]}++ will create the hash %k, whose keys are the IPs and whose values are the number of times each IP was seen. The }{ is funky perlspeak for "do the rest at the very end, after processing all input". So, at the end, the script will iterate over the keys of the hash and print the current key ($_) along with its value ($k{$_}).



            And, just so people don't think that perl forces you to write script that look like cryptic scribblings, this is the same thing in a less condensed form:



            perl -e '
            while (my $line=<STDIN>){
            @fields = split(/ /, $line);
            $ip = $fields[0];
            $counts{$ip}++;
            }
            foreach $ip (keys(%counts)){
            print "$ip count: $counts{$ip}n"
            }' < log





            share|improve this answer













            Some Perl:



            $ perl -lae '$k{$F[0]}++; }{ print "$_ count: $k{$_}" for keys(%k)' log 
            13.57.233.99 count: 1
            18.206.226.75 count: 2
            13.57.220.172 count: 9
            5.135.134.16 count: 5
            18.213.10.181 count: 3


            This is the same idea as Steeldriver's awk approach, but in Perl. The -a causes perl to automatically split each input line into the array @F, whose first element (the IP) is $F[0]. So, $k{$F[0]}++ will create the hash %k, whose keys are the IPs and whose values are the number of times each IP was seen. The }{ is funky perlspeak for "do the rest at the very end, after processing all input". So, at the end, the script will iterate over the keys of the hash and print the current key ($_) along with its value ($k{$_}).



            And, just so people don't think that perl forces you to write script that look like cryptic scribblings, this is the same thing in a less condensed form:



            perl -e '
            while (my $line=<STDIN>){
            @fields = split(/ /, $line);
            $ip = $fields[0];
            $counts{$ip}++;
            }
            foreach $ip (keys(%counts)){
            print "$ip count: $counts{$ip}n"
            }' < log






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 29 at 16:14









            terdonterdon

            67.6k13139223




            67.6k13139223























                4














                Maybe this is not what the OP want; however, if we know that the IP address length will be limited to 15 characters, a quicker way to display the counts with unique IPs from a huge log file can be achieved using uniq command alone:



                $ uniq -w 15 -c log

                5 5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] ...
                9 13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] ...
                1 13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] ...
                2 18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] ...
                3 18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] ...


                Options:



                -w N compares no more than N characters in lines



                -c will prefix lines by the number of occurrences



                Alternatively, For exact formatted output I prefer awk (should also work for IPV6 addresses), ymmv.



                $ awk 'NF { print $1 }' log | sort -h | uniq -c | awk '{printf "%s count: %dn", $2,$1 }'

                5.135.134.16 count: 5
                13.57.220.172 count: 9
                13.57.233.99 count: 1
                18.206.226.75 count: 2
                18.213.10.181 count: 3


                Note that uniq won't detect repeated lines in the input file if they are not adjacent, so it may be necessary to sort the file.






                share|improve this answer





















                • 1





                  Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                  – Martin Thornton
                  Mar 29 at 23:17













                • I like it, I didnt know uniq could count!

                  – j0h
                  Mar 31 at 12:57
















                4














                Maybe this is not what the OP want; however, if we know that the IP address length will be limited to 15 characters, a quicker way to display the counts with unique IPs from a huge log file can be achieved using uniq command alone:



                $ uniq -w 15 -c log

                5 5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] ...
                9 13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] ...
                1 13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] ...
                2 18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] ...
                3 18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] ...


                Options:



                -w N compares no more than N characters in lines



                -c will prefix lines by the number of occurrences



                Alternatively, For exact formatted output I prefer awk (should also work for IPV6 addresses), ymmv.



                $ awk 'NF { print $1 }' log | sort -h | uniq -c | awk '{printf "%s count: %dn", $2,$1 }'

                5.135.134.16 count: 5
                13.57.220.172 count: 9
                13.57.233.99 count: 1
                18.206.226.75 count: 2
                18.213.10.181 count: 3


                Note that uniq won't detect repeated lines in the input file if they are not adjacent, so it may be necessary to sort the file.






                share|improve this answer





















                • 1





                  Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                  – Martin Thornton
                  Mar 29 at 23:17













                • I like it, I didnt know uniq could count!

                  – j0h
                  Mar 31 at 12:57














                4












                4








                4







                Maybe this is not what the OP want; however, if we know that the IP address length will be limited to 15 characters, a quicker way to display the counts with unique IPs from a huge log file can be achieved using uniq command alone:



                $ uniq -w 15 -c log

                5 5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] ...
                9 13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] ...
                1 13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] ...
                2 18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] ...
                3 18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] ...


                Options:



                -w N compares no more than N characters in lines



                -c will prefix lines by the number of occurrences



                Alternatively, For exact formatted output I prefer awk (should also work for IPV6 addresses), ymmv.



                $ awk 'NF { print $1 }' log | sort -h | uniq -c | awk '{printf "%s count: %dn", $2,$1 }'

                5.135.134.16 count: 5
                13.57.220.172 count: 9
                13.57.233.99 count: 1
                18.206.226.75 count: 2
                18.213.10.181 count: 3


                Note that uniq won't detect repeated lines in the input file if they are not adjacent, so it may be necessary to sort the file.






                share|improve this answer















                Maybe this is not what the OP want; however, if we know that the IP address length will be limited to 15 characters, a quicker way to display the counts with unique IPs from a huge log file can be achieved using uniq command alone:



                $ uniq -w 15 -c log

                5 5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] ...
                9 13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] ...
                1 13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] ...
                2 18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] ...
                3 18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] ...


                Options:



                -w N compares no more than N characters in lines



                -c will prefix lines by the number of occurrences



                Alternatively, For exact formatted output I prefer awk (should also work for IPV6 addresses), ymmv.



                $ awk 'NF { print $1 }' log | sort -h | uniq -c | awk '{printf "%s count: %dn", $2,$1 }'

                5.135.134.16 count: 5
                13.57.220.172 count: 9
                13.57.233.99 count: 1
                18.206.226.75 count: 2
                18.213.10.181 count: 3


                Note that uniq won't detect repeated lines in the input file if they are not adjacent, so it may be necessary to sort the file.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 31 at 12:13

























                answered Mar 29 at 18:38









                Y. PradhanY. Pradhan

                412




                412








                • 1





                  Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                  – Martin Thornton
                  Mar 29 at 23:17













                • I like it, I didnt know uniq could count!

                  – j0h
                  Mar 31 at 12:57














                • 1





                  Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                  – Martin Thornton
                  Mar 29 at 23:17













                • I like it, I didnt know uniq could count!

                  – j0h
                  Mar 31 at 12:57








                1




                1





                Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                – Martin Thornton
                Mar 29 at 23:17







                Likely good enough in practice, but worth noting the corner cases. Only 6 probably constant characters after the IP ` - - [`. But in theory the address could be up to 8 characters shorter than the maximum so a change of date could split the count for such an IP. And as you hint, this won't work for IPv6.

                – Martin Thornton
                Mar 29 at 23:17















                I like it, I didnt know uniq could count!

                – j0h
                Mar 31 at 12:57





                I like it, I didnt know uniq could count!

                – j0h
                Mar 31 at 12:57











                1














                FWIW, Python 3:



                from collections import Counter

                with open('sample.log') as file:
                counts = Counter(line.split()[0] for line in file)

                for ip_address, count in counts.items():
                print('%-15s count: %d' % (ip_address, count))


                Output:



                13.57.233.99     count: 1
                18.213.10.181 count: 3
                5.135.134.16 count: 5
                18.206.226.75 count: 2
                13.57.220.172 count: 9





                share|improve this answer






























                  1














                  FWIW, Python 3:



                  from collections import Counter

                  with open('sample.log') as file:
                  counts = Counter(line.split()[0] for line in file)

                  for ip_address, count in counts.items():
                  print('%-15s count: %d' % (ip_address, count))


                  Output:



                  13.57.233.99     count: 1
                  18.213.10.181 count: 3
                  5.135.134.16 count: 5
                  18.206.226.75 count: 2
                  13.57.220.172 count: 9





                  share|improve this answer




























                    1












                    1








                    1







                    FWIW, Python 3:



                    from collections import Counter

                    with open('sample.log') as file:
                    counts = Counter(line.split()[0] for line in file)

                    for ip_address, count in counts.items():
                    print('%-15s count: %d' % (ip_address, count))


                    Output:



                    13.57.233.99     count: 1
                    18.213.10.181 count: 3
                    5.135.134.16 count: 5
                    18.206.226.75 count: 2
                    13.57.220.172 count: 9





                    share|improve this answer















                    FWIW, Python 3:



                    from collections import Counter

                    with open('sample.log') as file:
                    counts = Counter(line.split()[0] for line in file)

                    for ip_address, count in counts.items():
                    print('%-15s count: %d' % (ip_address, count))


                    Output:



                    13.57.233.99     count: 1
                    18.213.10.181 count: 3
                    5.135.134.16 count: 5
                    18.206.226.75 count: 2
                    13.57.220.172 count: 9






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Mar 31 at 17:34

























                    answered Mar 31 at 17:25









                    wjandreawjandrea

                    9,53142765




                    9,53142765























                        0














                        cut -f1 -d- my.log | sort | uniq -c


                        Explanation: Take the first field of my.log splitting on dashes - and sort it. uniq needs sorted input. -c tells it to count occurrences.






                        share|improve this answer






























                          0














                          cut -f1 -d- my.log | sort | uniq -c


                          Explanation: Take the first field of my.log splitting on dashes - and sort it. uniq needs sorted input. -c tells it to count occurrences.






                          share|improve this answer




























                            0












                            0








                            0







                            cut -f1 -d- my.log | sort | uniq -c


                            Explanation: Take the first field of my.log splitting on dashes - and sort it. uniq needs sorted input. -c tells it to count occurrences.






                            share|improve this answer















                            cut -f1 -d- my.log | sort | uniq -c


                            Explanation: Take the first field of my.log splitting on dashes - and sort it. uniq needs sorted input. -c tells it to count occurrences.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Mar 31 at 17:04









                            wjandrea

                            9,53142765




                            9,53142765










                            answered Mar 30 at 18:01









                            PhDPhD

                            101




                            101






























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