how to fill dictionary values with the counts of similar keys in another dictionary












0















I have a dictionary (studentPerf) which has all of the students in a school, with tuples as keys. I want to count the number of male students and the number of female students in the school, and use this to update the values in a second dictionary. The second dictionary (dictDemGender) has 2 keys, male and female, and 0s as the values. How can I change the 0s in dictDemGender to reflect the number of males and females in the school?



Could I do this with dictionary comprehension?



I've given the first few entries to studentPerf:



studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
('Able','male','senior'):[0.87,0.79,0.81,0.81],
('Don','male','junior'):[0.82,0.77,0.8,0.8],
('Will','male','senior'):[0.86,0.78,0.77,0.78],
('John','male','junior'):[0.74,0.81,0.87,0.73]}

#Creates a dictionary with genders as keys and 0s as the values to fill later
dictDemGender = {k:0 for k in genders}

dictDemGender = ?


I did ask a similar question but had diagnosed the problem incorrectly. I previously asked for help with finding an average score. What I actually need is a count of the different key possibilities. I need to be able to do so without any outside packages unfortunately.










share|improve this question

























  • Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

    – slider
    Nov 19 '18 at 17:37






  • 1





    What exactly is the expected output? {'male':5, 'female':0}?

    – timgeb
    Nov 19 '18 at 17:37








  • 1





    You asked a similar question yesterday. What have you tried on your own?

    – slider
    Nov 19 '18 at 17:37











  • expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

    – Jacob Myer
    Nov 19 '18 at 17:41











  • @slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

    – Jacob Myer
    Nov 19 '18 at 19:55
















0















I have a dictionary (studentPerf) which has all of the students in a school, with tuples as keys. I want to count the number of male students and the number of female students in the school, and use this to update the values in a second dictionary. The second dictionary (dictDemGender) has 2 keys, male and female, and 0s as the values. How can I change the 0s in dictDemGender to reflect the number of males and females in the school?



Could I do this with dictionary comprehension?



I've given the first few entries to studentPerf:



studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
('Able','male','senior'):[0.87,0.79,0.81,0.81],
('Don','male','junior'):[0.82,0.77,0.8,0.8],
('Will','male','senior'):[0.86,0.78,0.77,0.78],
('John','male','junior'):[0.74,0.81,0.87,0.73]}

#Creates a dictionary with genders as keys and 0s as the values to fill later
dictDemGender = {k:0 for k in genders}

dictDemGender = ?


I did ask a similar question but had diagnosed the problem incorrectly. I previously asked for help with finding an average score. What I actually need is a count of the different key possibilities. I need to be able to do so without any outside packages unfortunately.










share|improve this question

























  • Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

    – slider
    Nov 19 '18 at 17:37






  • 1





    What exactly is the expected output? {'male':5, 'female':0}?

    – timgeb
    Nov 19 '18 at 17:37








  • 1





    You asked a similar question yesterday. What have you tried on your own?

    – slider
    Nov 19 '18 at 17:37











  • expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

    – Jacob Myer
    Nov 19 '18 at 17:41











  • @slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

    – Jacob Myer
    Nov 19 '18 at 19:55














0












0








0








I have a dictionary (studentPerf) which has all of the students in a school, with tuples as keys. I want to count the number of male students and the number of female students in the school, and use this to update the values in a second dictionary. The second dictionary (dictDemGender) has 2 keys, male and female, and 0s as the values. How can I change the 0s in dictDemGender to reflect the number of males and females in the school?



Could I do this with dictionary comprehension?



I've given the first few entries to studentPerf:



studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
('Able','male','senior'):[0.87,0.79,0.81,0.81],
('Don','male','junior'):[0.82,0.77,0.8,0.8],
('Will','male','senior'):[0.86,0.78,0.77,0.78],
('John','male','junior'):[0.74,0.81,0.87,0.73]}

#Creates a dictionary with genders as keys and 0s as the values to fill later
dictDemGender = {k:0 for k in genders}

dictDemGender = ?


I did ask a similar question but had diagnosed the problem incorrectly. I previously asked for help with finding an average score. What I actually need is a count of the different key possibilities. I need to be able to do so without any outside packages unfortunately.










share|improve this question
















I have a dictionary (studentPerf) which has all of the students in a school, with tuples as keys. I want to count the number of male students and the number of female students in the school, and use this to update the values in a second dictionary. The second dictionary (dictDemGender) has 2 keys, male and female, and 0s as the values. How can I change the 0s in dictDemGender to reflect the number of males and females in the school?



Could I do this with dictionary comprehension?



I've given the first few entries to studentPerf:



studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
('Able','male','senior'):[0.87,0.79,0.81,0.81],
('Don','male','junior'):[0.82,0.77,0.8,0.8],
('Will','male','senior'):[0.86,0.78,0.77,0.78],
('John','male','junior'):[0.74,0.81,0.87,0.73]}

#Creates a dictionary with genders as keys and 0s as the values to fill later
dictDemGender = {k:0 for k in genders}

dictDemGender = ?


I did ask a similar question but had diagnosed the problem incorrectly. I previously asked for help with finding an average score. What I actually need is a count of the different key possibilities. I need to be able to do so without any outside packages unfortunately.







python dictionary






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 17:49







Jacob Myer

















asked Nov 19 '18 at 17:34









Jacob MyerJacob Myer

496




496













  • Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

    – slider
    Nov 19 '18 at 17:37






  • 1





    What exactly is the expected output? {'male':5, 'female':0}?

    – timgeb
    Nov 19 '18 at 17:37








  • 1





    You asked a similar question yesterday. What have you tried on your own?

    – slider
    Nov 19 '18 at 17:37











  • expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

    – Jacob Myer
    Nov 19 '18 at 17:41











  • @slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

    – Jacob Myer
    Nov 19 '18 at 19:55



















  • Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

    – slider
    Nov 19 '18 at 17:37






  • 1





    What exactly is the expected output? {'male':5, 'female':0}?

    – timgeb
    Nov 19 '18 at 17:37








  • 1





    You asked a similar question yesterday. What have you tried on your own?

    – slider
    Nov 19 '18 at 17:37











  • expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

    – Jacob Myer
    Nov 19 '18 at 17:41











  • @slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

    – Jacob Myer
    Nov 19 '18 at 19:55

















Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

– slider
Nov 19 '18 at 17:37





Possible duplicate of How do I fill my dictionary values with the values from another dictionary where their keys are the same?

– slider
Nov 19 '18 at 17:37




1




1





What exactly is the expected output? {'male':5, 'female':0}?

– timgeb
Nov 19 '18 at 17:37







What exactly is the expected output? {'male':5, 'female':0}?

– timgeb
Nov 19 '18 at 17:37






1




1





You asked a similar question yesterday. What have you tried on your own?

– slider
Nov 19 '18 at 17:37





You asked a similar question yesterday. What have you tried on your own?

– slider
Nov 19 '18 at 17:37













expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

– Jacob Myer
Nov 19 '18 at 17:41





expected output is {'male':5, 'female:'0'} (of course with the full studentPerf dictionary those numbers will be much higher.

– Jacob Myer
Nov 19 '18 at 17:41













@slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

– Jacob Myer
Nov 19 '18 at 19:55





@slider , I'm not sure if you've seen my post edits yet or not. After I posted yesterday I realized I made a mistake and was looking for the wrong output. I was able to come up with a solution on my own that fit the criteria and I posted that solution here

– Jacob Myer
Nov 19 '18 at 19:55












3 Answers
3






active

oldest

votes


















1














Use collections.Counter:



from collections import Counter

studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
('Able','male','senior'):[0.87,0.79,0.81,0.81],
('Don','male','junior'):[0.82,0.77,0.8,0.8],
('Will','male','senior'):[0.86,0.78,0.77,0.78],
('John','male','junior'):[0.74,0.81,0.87,0.73]}

print(Counter(x[1] for x in studentPerf))
# Counter({'male': 5})


Or, if you need empty counts also:



gender = {'male': 0, 'female': 0}
gender.update(Counter(x[1] for x in studentPerf))
# {'male': 5, 'female': 0}


Or, using dict.fromkeys() with Counter:



d = {'male', 'female'}
gender = dict.fromkeys(d, 0)
gender.update(Counter(x[1] for x in studentPerf))
# {'female': 0, 'male': 5}





share|improve this answer

































    0














    Assuming the expected output is {'male':5, 'female':0}, consider using a Counter.



    >>> from collections import Counter
    >>> c = Counter(male=0, female=0)
    >>> c.update(gen for _, gen, _ in studentPerf)
    >>> c
    Counter({'female': 0, 'male': 5})


    Initializing the two keys with zeros is not really necessary, you could also write



    >>> c = Counter(gen for _, gen, _ in studentPerf)
    >>> c
    Counter({'male': 5})


    because Counter lookup defaults to zero for missing keys:



    >>> c['female']
    0





    share|improve this answer































      0














      As I said, I was looking for a solution that did not require outside packages. I know the way I've gone about this is rather cumbersome but this was for a class and the exercise had these requirements. I found a way to count all of the males and females and input those values into the dictDemGender dictionary.



      genCounts = ([x[1] for x in list(studentPerf.keys())].count('female'), [x[1] for x in list(studentPerf.keys())].count('male'))
      dictDemGender = dict(zip(dictDemGender.keys(), genCounts))





      share|improve this answer























        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
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53379901%2fhow-to-fill-dictionary-values-with-the-counts-of-similar-keys-in-another-diction%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        Use collections.Counter:



        from collections import Counter

        studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
        ('Able','male','senior'):[0.87,0.79,0.81,0.81],
        ('Don','male','junior'):[0.82,0.77,0.8,0.8],
        ('Will','male','senior'):[0.86,0.78,0.77,0.78],
        ('John','male','junior'):[0.74,0.81,0.87,0.73]}

        print(Counter(x[1] for x in studentPerf))
        # Counter({'male': 5})


        Or, if you need empty counts also:



        gender = {'male': 0, 'female': 0}
        gender.update(Counter(x[1] for x in studentPerf))
        # {'male': 5, 'female': 0}


        Or, using dict.fromkeys() with Counter:



        d = {'male', 'female'}
        gender = dict.fromkeys(d, 0)
        gender.update(Counter(x[1] for x in studentPerf))
        # {'female': 0, 'male': 5}





        share|improve this answer






























          1














          Use collections.Counter:



          from collections import Counter

          studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
          ('Able','male','senior'):[0.87,0.79,0.81,0.81],
          ('Don','male','junior'):[0.82,0.77,0.8,0.8],
          ('Will','male','senior'):[0.86,0.78,0.77,0.78],
          ('John','male','junior'):[0.74,0.81,0.87,0.73]}

          print(Counter(x[1] for x in studentPerf))
          # Counter({'male': 5})


          Or, if you need empty counts also:



          gender = {'male': 0, 'female': 0}
          gender.update(Counter(x[1] for x in studentPerf))
          # {'male': 5, 'female': 0}


          Or, using dict.fromkeys() with Counter:



          d = {'male', 'female'}
          gender = dict.fromkeys(d, 0)
          gender.update(Counter(x[1] for x in studentPerf))
          # {'female': 0, 'male': 5}





          share|improve this answer




























            1












            1








            1







            Use collections.Counter:



            from collections import Counter

            studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
            ('Able','male','senior'):[0.87,0.79,0.81,0.81],
            ('Don','male','junior'):[0.82,0.77,0.8,0.8],
            ('Will','male','senior'):[0.86,0.78,0.77,0.78],
            ('John','male','junior'):[0.74,0.81,0.87,0.73]}

            print(Counter(x[1] for x in studentPerf))
            # Counter({'male': 5})


            Or, if you need empty counts also:



            gender = {'male': 0, 'female': 0}
            gender.update(Counter(x[1] for x in studentPerf))
            # {'male': 5, 'female': 0}


            Or, using dict.fromkeys() with Counter:



            d = {'male', 'female'}
            gender = dict.fromkeys(d, 0)
            gender.update(Counter(x[1] for x in studentPerf))
            # {'female': 0, 'male': 5}





            share|improve this answer















            Use collections.Counter:



            from collections import Counter

            studentPerf = {('Jeffery','male','junior'):[0.81,0.75,0.74,0.8],
            ('Able','male','senior'):[0.87,0.79,0.81,0.81],
            ('Don','male','junior'):[0.82,0.77,0.8,0.8],
            ('Will','male','senior'):[0.86,0.78,0.77,0.78],
            ('John','male','junior'):[0.74,0.81,0.87,0.73]}

            print(Counter(x[1] for x in studentPerf))
            # Counter({'male': 5})


            Or, if you need empty counts also:



            gender = {'male': 0, 'female': 0}
            gender.update(Counter(x[1] for x in studentPerf))
            # {'male': 5, 'female': 0}


            Or, using dict.fromkeys() with Counter:



            d = {'male', 'female'}
            gender = dict.fromkeys(d, 0)
            gender.update(Counter(x[1] for x in studentPerf))
            # {'female': 0, 'male': 5}






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 19 '18 at 17:46

























            answered Nov 19 '18 at 17:37









            AustinAustin

            9,8733828




            9,8733828

























                0














                Assuming the expected output is {'male':5, 'female':0}, consider using a Counter.



                >>> from collections import Counter
                >>> c = Counter(male=0, female=0)
                >>> c.update(gen for _, gen, _ in studentPerf)
                >>> c
                Counter({'female': 0, 'male': 5})


                Initializing the two keys with zeros is not really necessary, you could also write



                >>> c = Counter(gen for _, gen, _ in studentPerf)
                >>> c
                Counter({'male': 5})


                because Counter lookup defaults to zero for missing keys:



                >>> c['female']
                0





                share|improve this answer




























                  0














                  Assuming the expected output is {'male':5, 'female':0}, consider using a Counter.



                  >>> from collections import Counter
                  >>> c = Counter(male=0, female=0)
                  >>> c.update(gen for _, gen, _ in studentPerf)
                  >>> c
                  Counter({'female': 0, 'male': 5})


                  Initializing the two keys with zeros is not really necessary, you could also write



                  >>> c = Counter(gen for _, gen, _ in studentPerf)
                  >>> c
                  Counter({'male': 5})


                  because Counter lookup defaults to zero for missing keys:



                  >>> c['female']
                  0





                  share|improve this answer


























                    0












                    0








                    0







                    Assuming the expected output is {'male':5, 'female':0}, consider using a Counter.



                    >>> from collections import Counter
                    >>> c = Counter(male=0, female=0)
                    >>> c.update(gen for _, gen, _ in studentPerf)
                    >>> c
                    Counter({'female': 0, 'male': 5})


                    Initializing the two keys with zeros is not really necessary, you could also write



                    >>> c = Counter(gen for _, gen, _ in studentPerf)
                    >>> c
                    Counter({'male': 5})


                    because Counter lookup defaults to zero for missing keys:



                    >>> c['female']
                    0





                    share|improve this answer













                    Assuming the expected output is {'male':5, 'female':0}, consider using a Counter.



                    >>> from collections import Counter
                    >>> c = Counter(male=0, female=0)
                    >>> c.update(gen for _, gen, _ in studentPerf)
                    >>> c
                    Counter({'female': 0, 'male': 5})


                    Initializing the two keys with zeros is not really necessary, you could also write



                    >>> c = Counter(gen for _, gen, _ in studentPerf)
                    >>> c
                    Counter({'male': 5})


                    because Counter lookup defaults to zero for missing keys:



                    >>> c['female']
                    0






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 19 '18 at 17:42









                    timgebtimgeb

                    50.6k116393




                    50.6k116393























                        0














                        As I said, I was looking for a solution that did not require outside packages. I know the way I've gone about this is rather cumbersome but this was for a class and the exercise had these requirements. I found a way to count all of the males and females and input those values into the dictDemGender dictionary.



                        genCounts = ([x[1] for x in list(studentPerf.keys())].count('female'), [x[1] for x in list(studentPerf.keys())].count('male'))
                        dictDemGender = dict(zip(dictDemGender.keys(), genCounts))





                        share|improve this answer




























                          0














                          As I said, I was looking for a solution that did not require outside packages. I know the way I've gone about this is rather cumbersome but this was for a class and the exercise had these requirements. I found a way to count all of the males and females and input those values into the dictDemGender dictionary.



                          genCounts = ([x[1] for x in list(studentPerf.keys())].count('female'), [x[1] for x in list(studentPerf.keys())].count('male'))
                          dictDemGender = dict(zip(dictDemGender.keys(), genCounts))





                          share|improve this answer


























                            0












                            0








                            0







                            As I said, I was looking for a solution that did not require outside packages. I know the way I've gone about this is rather cumbersome but this was for a class and the exercise had these requirements. I found a way to count all of the males and females and input those values into the dictDemGender dictionary.



                            genCounts = ([x[1] for x in list(studentPerf.keys())].count('female'), [x[1] for x in list(studentPerf.keys())].count('male'))
                            dictDemGender = dict(zip(dictDemGender.keys(), genCounts))





                            share|improve this answer













                            As I said, I was looking for a solution that did not require outside packages. I know the way I've gone about this is rather cumbersome but this was for a class and the exercise had these requirements. I found a way to count all of the males and females and input those values into the dictDemGender dictionary.



                            genCounts = ([x[1] for x in list(studentPerf.keys())].count('female'), [x[1] for x in list(studentPerf.keys())].count('male'))
                            dictDemGender = dict(zip(dictDemGender.keys(), genCounts))






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Nov 19 '18 at 19:52









                            Jacob MyerJacob Myer

                            496




                            496






























                                draft saved

                                draft discarded




















































                                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.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53379901%2fhow-to-fill-dictionary-values-with-the-counts-of-similar-keys-in-another-diction%23new-answer', 'question_page');
                                }
                                );

                                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







                                Popular posts from this blog

                                How to change which sound is reproduced for terminal bell?

                                Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents