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























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        3 Answers
        3






        active

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        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




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