find a substring in a selected portion of a list elements
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 '18 at 15:37
jpp
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
add a comment |
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 '18 at 15:37
jpp
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
add a comment |
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
edited Nov 22 '18 at 15:37
jpp
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.
I tried the below code but it finds all the elements where str b is found.
list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])
it returns the following:
['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']
However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?
python string list list-comprehension python-3.5
python string list list-comprehension python-3.5
edited Nov 22 '18 at 15:37
jpp
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
edited Nov 22 '18 at 15:37
jpp
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
edited Nov 22 '18 at 15:37
jpp
103k2167117
edited Nov 22 '18 at 15:37
jpp
103k2167117
edited Nov 22 '18 at 15:37
jpp
103k2167117
103k2167117
asked Nov 22 '18 at 15:33
HT121HT121
1168
asked Nov 22 '18 at 15:33
HT121HT121
1168
asked Nov 22 '18 at 15:33
HT121HT121
1168
1168
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53434191%2ffind-a-substring-in-a-selected-portion-of-a-list-elements%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
add a comment |
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
add a comment |
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
You need to split each string by . and extract the first split:
print([s for s in list1 if '1' in s.split('.')[0]])
['1.5']
For a precise match, use ==:
print([s for s in list1 if s.split('.')[0] == '1'])
['1.5']
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
edited Nov 22 '18 at 15:37
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
answered Nov 22 '18 at 15:35
jppjpp
103k2167117
103k2167117
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
add a comment |
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]]should do it.
– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
And do you know how could i find the actual index of matched element in list1?
– HT121
Nov 22 '18 at 15:47
1
1
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.– jpp
Nov 22 '18 at 15:59
[idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.– jpp
Nov 22 '18 at 15:59
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
– HT121
Nov 22 '18 at 16:08
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
@HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
– jpp
Nov 22 '18 at 16:10
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53434191%2ffind-a-substring-in-a-selected-portion-of-a-list-elements%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
