Groupby class and count missing values in features












14














I have a problem and I cannot find any solution in the web or documentation, even if I think that it is very trivial.



What do I want to do?



I have a dataframe like this



CLASS FEATURE1 FEATURE2 FEATURE3
X A NaN NaN
X NaN A NaN
B A A A


I want to group by the label(CLASS) and display the number of NaN-Values that are counted in every feature so that it looks like this. The purpose of this is to get a general idea how missing values are distributed over the different classes.



CLASS FEATURE1 FEATURE2 FEATURE3
X 1 1 2
B 0 0 0


I know how to recieve the amount of nonnull-Values - df.groupby['CLASS'].count()



Is there something similar for the NaN-Values?



I tried to subtract the count() from the size() but it returned an unformatted output filled with the value NaN










share|improve this question









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FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    14














    I have a problem and I cannot find any solution in the web or documentation, even if I think that it is very trivial.



    What do I want to do?



    I have a dataframe like this



    CLASS FEATURE1 FEATURE2 FEATURE3
    X A NaN NaN
    X NaN A NaN
    B A A A


    I want to group by the label(CLASS) and display the number of NaN-Values that are counted in every feature so that it looks like this. The purpose of this is to get a general idea how missing values are distributed over the different classes.



    CLASS FEATURE1 FEATURE2 FEATURE3
    X 1 1 2
    B 0 0 0


    I know how to recieve the amount of nonnull-Values - df.groupby['CLASS'].count()



    Is there something similar for the NaN-Values?



    I tried to subtract the count() from the size() but it returned an unformatted output filled with the value NaN










    share|improve this question









    New contributor




    FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      14












      14








      14







      I have a problem and I cannot find any solution in the web or documentation, even if I think that it is very trivial.



      What do I want to do?



      I have a dataframe like this



      CLASS FEATURE1 FEATURE2 FEATURE3
      X A NaN NaN
      X NaN A NaN
      B A A A


      I want to group by the label(CLASS) and display the number of NaN-Values that are counted in every feature so that it looks like this. The purpose of this is to get a general idea how missing values are distributed over the different classes.



      CLASS FEATURE1 FEATURE2 FEATURE3
      X 1 1 2
      B 0 0 0


      I know how to recieve the amount of nonnull-Values - df.groupby['CLASS'].count()



      Is there something similar for the NaN-Values?



      I tried to subtract the count() from the size() but it returned an unformatted output filled with the value NaN










      share|improve this question









      New contributor




      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I have a problem and I cannot find any solution in the web or documentation, even if I think that it is very trivial.



      What do I want to do?



      I have a dataframe like this



      CLASS FEATURE1 FEATURE2 FEATURE3
      X A NaN NaN
      X NaN A NaN
      B A A A


      I want to group by the label(CLASS) and display the number of NaN-Values that are counted in every feature so that it looks like this. The purpose of this is to get a general idea how missing values are distributed over the different classes.



      CLASS FEATURE1 FEATURE2 FEATURE3
      X 1 1 2
      B 0 0 0


      I know how to recieve the amount of nonnull-Values - df.groupby['CLASS'].count()



      Is there something similar for the NaN-Values?



      I tried to subtract the count() from the size() but it returned an unformatted output filled with the value NaN







      python pandas dataframe group-by nan






      share|improve this question









      New contributor




      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited Dec 27 '18 at 16:11









      coldspeed

      120k19119194




      120k19119194






      New contributor




      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked Dec 27 '18 at 15:15









      FelTry2

      955




      955




      New contributor




      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      FelTry2 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.
























          3 Answers
          3






          active

          oldest

          votes


















          14














          Compute a mask with isna, then group and find the sum:



          df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

          CLASS FEATURE1 FEATURE2 FEATURE3
          0 X 1.0 1.0 2.0
          1 B 0.0 0.0 0.0




          Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction:



          df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)


          Or,



          g = df.groupby('CLASS')
          g.count().rsub(g.size(), axis=0)




                 FEATURE1  FEATURE2  FEATURE3
          CLASS
          B 0 0 0
          X 1 1 2




          There are quite a few good answers, so here are some timeits for your perusal:



          df_ = df
          df = pd.concat([df_] * 10000)

          %timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
          %timeit df.set_index('CLASS').isna().sum(level=0)
          %%timeit
          g = df.groupby('CLASS')
          g.count().rsub(g.size(), axis=0)

          11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
          9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
          6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


          Actual performance depends on your data and setup, so your mileage may vary.






          share|improve this answer



















          • 2




            Amazing - thank you very, very much!
            – FelTry2
            Dec 27 '18 at 15:37



















          12














          You can use set_index and sum:



          df.set_index('CLASS').isna().sum(level=0)


          Output:



                 FEATURE1  FEATURE2  FEATURE3
          CLASS
          X 1.0 1.0 2.0
          B 0.0 0.0 0.0





          share|improve this answer





























            7














            Using the diff between count and size



            g=df.groupby('CLASS')

            -g.count().sub(g.size(),0)

            FEATURE1 FEATURE2 FEATURE3
            CLASS
            B 0 0 0
            X 1 1 2


            And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop



            pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
            Out[468]:
            FEATURE1 FEATURE2 FEATURE3
            B 0 0 0
            X 1 1 2





            share|improve this answer



















            • 2




              That's a good one. +1
              – Scott Boston
              Dec 27 '18 at 15:26











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            14














            Compute a mask with isna, then group and find the sum:



            df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

            CLASS FEATURE1 FEATURE2 FEATURE3
            0 X 1.0 1.0 2.0
            1 B 0.0 0.0 0.0




            Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction:



            df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)


            Or,



            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)




                   FEATURE1  FEATURE2  FEATURE3
            CLASS
            B 0 0 0
            X 1 1 2




            There are quite a few good answers, so here are some timeits for your perusal:



            df_ = df
            df = pd.concat([df_] * 10000)

            %timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
            %timeit df.set_index('CLASS').isna().sum(level=0)
            %%timeit
            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)

            11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


            Actual performance depends on your data and setup, so your mileage may vary.






            share|improve this answer



















            • 2




              Amazing - thank you very, very much!
              – FelTry2
              Dec 27 '18 at 15:37
















            14














            Compute a mask with isna, then group and find the sum:



            df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

            CLASS FEATURE1 FEATURE2 FEATURE3
            0 X 1.0 1.0 2.0
            1 B 0.0 0.0 0.0




            Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction:



            df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)


            Or,



            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)




                   FEATURE1  FEATURE2  FEATURE3
            CLASS
            B 0 0 0
            X 1 1 2




            There are quite a few good answers, so here are some timeits for your perusal:



            df_ = df
            df = pd.concat([df_] * 10000)

            %timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
            %timeit df.set_index('CLASS').isna().sum(level=0)
            %%timeit
            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)

            11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


            Actual performance depends on your data and setup, so your mileage may vary.






            share|improve this answer



















            • 2




              Amazing - thank you very, very much!
              – FelTry2
              Dec 27 '18 at 15:37














            14












            14








            14






            Compute a mask with isna, then group and find the sum:



            df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

            CLASS FEATURE1 FEATURE2 FEATURE3
            0 X 1.0 1.0 2.0
            1 B 0.0 0.0 0.0




            Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction:



            df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)


            Or,



            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)




                   FEATURE1  FEATURE2  FEATURE3
            CLASS
            B 0 0 0
            X 1 1 2




            There are quite a few good answers, so here are some timeits for your perusal:



            df_ = df
            df = pd.concat([df_] * 10000)

            %timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
            %timeit df.set_index('CLASS').isna().sum(level=0)
            %%timeit
            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)

            11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


            Actual performance depends on your data and setup, so your mileage may vary.






            share|improve this answer














            Compute a mask with isna, then group and find the sum:



            df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

            CLASS FEATURE1 FEATURE2 FEATURE3
            0 X 1.0 1.0 2.0
            1 B 0.0 0.0 0.0




            Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction:



            df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)


            Or,



            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)




                   FEATURE1  FEATURE2  FEATURE3
            CLASS
            B 0 0 0
            X 1 1 2




            There are quite a few good answers, so here are some timeits for your perusal:



            df_ = df
            df = pd.concat([df_] * 10000)

            %timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
            %timeit df.set_index('CLASS').isna().sum(level=0)
            %%timeit
            g = df.groupby('CLASS')
            g.count().rsub(g.size(), axis=0)

            11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
            6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


            Actual performance depends on your data and setup, so your mileage may vary.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 27 '18 at 15:29

























            answered Dec 27 '18 at 15:16









            coldspeed

            120k19119194




            120k19119194








            • 2




              Amazing - thank you very, very much!
              – FelTry2
              Dec 27 '18 at 15:37














            • 2




              Amazing - thank you very, very much!
              – FelTry2
              Dec 27 '18 at 15:37








            2




            2




            Amazing - thank you very, very much!
            – FelTry2
            Dec 27 '18 at 15:37




            Amazing - thank you very, very much!
            – FelTry2
            Dec 27 '18 at 15:37













            12














            You can use set_index and sum:



            df.set_index('CLASS').isna().sum(level=0)


            Output:



                   FEATURE1  FEATURE2  FEATURE3
            CLASS
            X 1.0 1.0 2.0
            B 0.0 0.0 0.0





            share|improve this answer


























              12














              You can use set_index and sum:



              df.set_index('CLASS').isna().sum(level=0)


              Output:



                     FEATURE1  FEATURE2  FEATURE3
              CLASS
              X 1.0 1.0 2.0
              B 0.0 0.0 0.0





              share|improve this answer
























                12












                12








                12






                You can use set_index and sum:



                df.set_index('CLASS').isna().sum(level=0)


                Output:



                       FEATURE1  FEATURE2  FEATURE3
                CLASS
                X 1.0 1.0 2.0
                B 0.0 0.0 0.0





                share|improve this answer












                You can use set_index and sum:



                df.set_index('CLASS').isna().sum(level=0)


                Output:



                       FEATURE1  FEATURE2  FEATURE3
                CLASS
                X 1.0 1.0 2.0
                B 0.0 0.0 0.0






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Dec 27 '18 at 15:18









                Scott Boston

                51.3k72955




                51.3k72955























                    7














                    Using the diff between count and size



                    g=df.groupby('CLASS')

                    -g.count().sub(g.size(),0)

                    FEATURE1 FEATURE2 FEATURE3
                    CLASS
                    B 0 0 0
                    X 1 1 2


                    And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop



                    pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
                    Out[468]:
                    FEATURE1 FEATURE2 FEATURE3
                    B 0 0 0
                    X 1 1 2





                    share|improve this answer



















                    • 2




                      That's a good one. +1
                      – Scott Boston
                      Dec 27 '18 at 15:26
















                    7














                    Using the diff between count and size



                    g=df.groupby('CLASS')

                    -g.count().sub(g.size(),0)

                    FEATURE1 FEATURE2 FEATURE3
                    CLASS
                    B 0 0 0
                    X 1 1 2


                    And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop



                    pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
                    Out[468]:
                    FEATURE1 FEATURE2 FEATURE3
                    B 0 0 0
                    X 1 1 2





                    share|improve this answer



















                    • 2




                      That's a good one. +1
                      – Scott Boston
                      Dec 27 '18 at 15:26














                    7












                    7








                    7






                    Using the diff between count and size



                    g=df.groupby('CLASS')

                    -g.count().sub(g.size(),0)

                    FEATURE1 FEATURE2 FEATURE3
                    CLASS
                    B 0 0 0
                    X 1 1 2


                    And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop



                    pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
                    Out[468]:
                    FEATURE1 FEATURE2 FEATURE3
                    B 0 0 0
                    X 1 1 2





                    share|improve this answer














                    Using the diff between count and size



                    g=df.groupby('CLASS')

                    -g.count().sub(g.size(),0)

                    FEATURE1 FEATURE2 FEATURE3
                    CLASS
                    B 0 0 0
                    X 1 1 2


                    And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop



                    pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
                    Out[468]:
                    FEATURE1 FEATURE2 FEATURE3
                    B 0 0 0
                    X 1 1 2






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 27 '18 at 16:25

























                    answered Dec 27 '18 at 15:19









                    W-B

                    101k73163




                    101k73163








                    • 2




                      That's a good one. +1
                      – Scott Boston
                      Dec 27 '18 at 15:26














                    • 2




                      That's a good one. +1
                      – Scott Boston
                      Dec 27 '18 at 15:26








                    2




                    2




                    That's a good one. +1
                    – Scott Boston
                    Dec 27 '18 at 15:26




                    That's a good one. +1
                    – Scott Boston
                    Dec 27 '18 at 15:26










                    FelTry2 is a new contributor. Be nice, and check out our Code of Conduct.










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