np.logical_and with three tests










1















I am trying to update a column in my dataframe based on the testing and results of the values within three columns (3 tests).



Some sample code:



df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
(None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
df_test.head()


When I try the following df_test['g'] = np.where(np.logical_and(df_test['a'] != 'None', df_test['c'] == 0, df_test['d'] == 0), True, False).astype(int)
I receive the error TypeError: return arrays must be of ArrayType So I try the following:



df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
(None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
df_test['g'] = np.where(np.logical_and(df_test['a'] != None,
np.logical_and(df_test['c'] == 0,
df_test['d'] == 0)),
True, False).astype(int)
df_test.head()


And on row 2, where I would expect to see a 0 I see a 1 while row 4 appears correct. Test 1 (a), should be False while the second (c) and third (d) tests should be True, True. False == True == True is False.



 a b c d e f g
0 ? 2.0 1 0 0 0 0
1 None 2.0 1 0 0 0 0
2 None 2.0 0 0 0 0 1
3 None 2.0 0 1 0 0 0
4 ? 2.0 0 0 0 0 1


I need a way to evaluate 3 tests and return true or false as an int.










share|improve this question


























    1















    I am trying to update a column in my dataframe based on the testing and results of the values within three columns (3 tests).



    Some sample code:



    df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
    (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
    ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
    df_test.head()


    When I try the following df_test['g'] = np.where(np.logical_and(df_test['a'] != 'None', df_test['c'] == 0, df_test['d'] == 0), True, False).astype(int)
    I receive the error TypeError: return arrays must be of ArrayType So I try the following:



    df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
    (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
    ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
    df_test['g'] = np.where(np.logical_and(df_test['a'] != None,
    np.logical_and(df_test['c'] == 0,
    df_test['d'] == 0)),
    True, False).astype(int)
    df_test.head()


    And on row 2, where I would expect to see a 0 I see a 1 while row 4 appears correct. Test 1 (a), should be False while the second (c) and third (d) tests should be True, True. False == True == True is False.



     a b c d e f g
    0 ? 2.0 1 0 0 0 0
    1 None 2.0 1 0 0 0 0
    2 None 2.0 0 0 0 0 1
    3 None 2.0 0 1 0 0 0
    4 ? 2.0 0 0 0 0 1


    I need a way to evaluate 3 tests and return true or false as an int.










    share|improve this question
























      1












      1








      1








      I am trying to update a column in my dataframe based on the testing and results of the values within three columns (3 tests).



      Some sample code:



      df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
      (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
      ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
      df_test.head()


      When I try the following df_test['g'] = np.where(np.logical_and(df_test['a'] != 'None', df_test['c'] == 0, df_test['d'] == 0), True, False).astype(int)
      I receive the error TypeError: return arrays must be of ArrayType So I try the following:



      df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
      (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
      ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
      df_test['g'] = np.where(np.logical_and(df_test['a'] != None,
      np.logical_and(df_test['c'] == 0,
      df_test['d'] == 0)),
      True, False).astype(int)
      df_test.head()


      And on row 2, where I would expect to see a 0 I see a 1 while row 4 appears correct. Test 1 (a), should be False while the second (c) and third (d) tests should be True, True. False == True == True is False.



       a b c d e f g
      0 ? 2.0 1 0 0 0 0
      1 None 2.0 1 0 0 0 0
      2 None 2.0 0 0 0 0 1
      3 None 2.0 0 1 0 0 0
      4 ? 2.0 0 0 0 0 1


      I need a way to evaluate 3 tests and return true or false as an int.










      share|improve this question














      I am trying to update a column in my dataframe based on the testing and results of the values within three columns (3 tests).



      Some sample code:



      df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
      (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
      ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
      df_test.head()


      When I try the following df_test['g'] = np.where(np.logical_and(df_test['a'] != 'None', df_test['c'] == 0, df_test['d'] == 0), True, False).astype(int)
      I receive the error TypeError: return arrays must be of ArrayType So I try the following:



      df_test = pd.DataFrame([('?',2.0,1,0,0,0), (None,2.0,1,0,0,0), 
      (None,2.0,0,0,0,0),(None,2.0,0,1,0,0),
      ('?',2.0,0,0,0,0)], columns=['a','b','c','d','e','f'])
      df_test['g'] = np.where(np.logical_and(df_test['a'] != None,
      np.logical_and(df_test['c'] == 0,
      df_test['d'] == 0)),
      True, False).astype(int)
      df_test.head()


      And on row 2, where I would expect to see a 0 I see a 1 while row 4 appears correct. Test 1 (a), should be False while the second (c) and third (d) tests should be True, True. False == True == True is False.



       a b c d e f g
      0 ? 2.0 1 0 0 0 0
      1 None 2.0 1 0 0 0 0
      2 None 2.0 0 0 0 0 1
      3 None 2.0 0 1 0 0 0
      4 ? 2.0 0 0 0 0 1


      I need a way to evaluate 3 tests and return true or false as an int.







      python pandas numpy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 12 '18 at 16:59









      EricEric

      1471214




      1471214






















          2 Answers
          2






          active

          oldest

          votes


















          3














          For 3 or more conditions, use np.logical_and.reduce and pass a list of masks;



          mask = np.logical_and.reduce([
          df_test['a'].notna(), df_test['c'].eq(0), df_test['d'].eq(0)])
          df_test['g'] = mask.astype(int)

          print(df_test)
          a b c d e f g
          0 ? 2.0 1 0 0 0 0
          1 None 2.0 1 0 0 0 0
          2 None 2.0 0 0 0 0 0
          3 None 2.0 0 1 0 0 0
          4 ? 2.0 0 0 0 0 1


          The subsequent np.where is superfluous here.






          share|improve this answer






























            1














            Try:



            df_test['htest' ] = (df_test['a'].isna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)



            df_test['htest' ] = (df_test['a'].notna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)


            Output:



             a b c d e f htest
            0 ? 2.0 1 0 0 0 0
            1 None 2.0 1 0 0 0 0
            2 None 2.0 0 0 0 0 0
            3 None 2.0 0 1 0 0 0
            4 ? 2.0 0 0 0 0 1





            share|improve this answer

























            • Think the first condition should be notna.

              – coldspeed
              Nov 12 '18 at 17:14











            • yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

              – Eric
              Nov 12 '18 at 17:19











            • Ah.. Got thanks.

              – Scott Boston
              Nov 12 '18 at 17:22










            Your Answer






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            For 3 or more conditions, use np.logical_and.reduce and pass a list of masks;



            mask = np.logical_and.reduce([
            df_test['a'].notna(), df_test['c'].eq(0), df_test['d'].eq(0)])
            df_test['g'] = mask.astype(int)

            print(df_test)
            a b c d e f g
            0 ? 2.0 1 0 0 0 0
            1 None 2.0 1 0 0 0 0
            2 None 2.0 0 0 0 0 0
            3 None 2.0 0 1 0 0 0
            4 ? 2.0 0 0 0 0 1


            The subsequent np.where is superfluous here.






            share|improve this answer



























              3














              For 3 or more conditions, use np.logical_and.reduce and pass a list of masks;



              mask = np.logical_and.reduce([
              df_test['a'].notna(), df_test['c'].eq(0), df_test['d'].eq(0)])
              df_test['g'] = mask.astype(int)

              print(df_test)
              a b c d e f g
              0 ? 2.0 1 0 0 0 0
              1 None 2.0 1 0 0 0 0
              2 None 2.0 0 0 0 0 0
              3 None 2.0 0 1 0 0 0
              4 ? 2.0 0 0 0 0 1


              The subsequent np.where is superfluous here.






              share|improve this answer

























                3












                3








                3







                For 3 or more conditions, use np.logical_and.reduce and pass a list of masks;



                mask = np.logical_and.reduce([
                df_test['a'].notna(), df_test['c'].eq(0), df_test['d'].eq(0)])
                df_test['g'] = mask.astype(int)

                print(df_test)
                a b c d e f g
                0 ? 2.0 1 0 0 0 0
                1 None 2.0 1 0 0 0 0
                2 None 2.0 0 0 0 0 0
                3 None 2.0 0 1 0 0 0
                4 ? 2.0 0 0 0 0 1


                The subsequent np.where is superfluous here.






                share|improve this answer













                For 3 or more conditions, use np.logical_and.reduce and pass a list of masks;



                mask = np.logical_and.reduce([
                df_test['a'].notna(), df_test['c'].eq(0), df_test['d'].eq(0)])
                df_test['g'] = mask.astype(int)

                print(df_test)
                a b c d e f g
                0 ? 2.0 1 0 0 0 0
                1 None 2.0 1 0 0 0 0
                2 None 2.0 0 0 0 0 0
                3 None 2.0 0 1 0 0 0
                4 ? 2.0 0 0 0 0 1


                The subsequent np.where is superfluous here.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 '18 at 17:10









                coldspeedcoldspeed

                134k23145230




                134k23145230























                    1














                    Try:



                    df_test['htest' ] = (df_test['a'].isna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)



                    df_test['htest' ] = (df_test['a'].notna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)


                    Output:



                     a b c d e f htest
                    0 ? 2.0 1 0 0 0 0
                    1 None 2.0 1 0 0 0 0
                    2 None 2.0 0 0 0 0 0
                    3 None 2.0 0 1 0 0 0
                    4 ? 2.0 0 0 0 0 1





                    share|improve this answer

























                    • Think the first condition should be notna.

                      – coldspeed
                      Nov 12 '18 at 17:14











                    • yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                      – Eric
                      Nov 12 '18 at 17:19











                    • Ah.. Got thanks.

                      – Scott Boston
                      Nov 12 '18 at 17:22















                    1














                    Try:



                    df_test['htest' ] = (df_test['a'].isna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)



                    df_test['htest' ] = (df_test['a'].notna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)


                    Output:



                     a b c d e f htest
                    0 ? 2.0 1 0 0 0 0
                    1 None 2.0 1 0 0 0 0
                    2 None 2.0 0 0 0 0 0
                    3 None 2.0 0 1 0 0 0
                    4 ? 2.0 0 0 0 0 1





                    share|improve this answer

























                    • Think the first condition should be notna.

                      – coldspeed
                      Nov 12 '18 at 17:14











                    • yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                      – Eric
                      Nov 12 '18 at 17:19











                    • Ah.. Got thanks.

                      – Scott Boston
                      Nov 12 '18 at 17:22













                    1












                    1








                    1







                    Try:



                    df_test['htest' ] = (df_test['a'].isna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)



                    df_test['htest' ] = (df_test['a'].notna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)


                    Output:



                     a b c d e f htest
                    0 ? 2.0 1 0 0 0 0
                    1 None 2.0 1 0 0 0 0
                    2 None 2.0 0 0 0 0 0
                    3 None 2.0 0 1 0 0 0
                    4 ? 2.0 0 0 0 0 1





                    share|improve this answer















                    Try:



                    df_test['htest' ] = (df_test['a'].isna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)



                    df_test['htest' ] = (df_test['a'].notna() & (df_test['c'] == 0) & (df_test['d'] == 0)).astype(int)


                    Output:



                     a b c d e f htest
                    0 ? 2.0 1 0 0 0 0
                    1 None 2.0 1 0 0 0 0
                    2 None 2.0 0 0 0 0 0
                    3 None 2.0 0 1 0 0 0
                    4 ? 2.0 0 0 0 0 1






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 12 '18 at 17:21

























                    answered Nov 12 '18 at 17:11









                    Scott BostonScott Boston

                    56.2k73157




                    56.2k73157












                    • Think the first condition should be notna.

                      – coldspeed
                      Nov 12 '18 at 17:14











                    • yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                      – Eric
                      Nov 12 '18 at 17:19











                    • Ah.. Got thanks.

                      – Scott Boston
                      Nov 12 '18 at 17:22

















                    • Think the first condition should be notna.

                      – coldspeed
                      Nov 12 '18 at 17:14











                    • yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                      – Eric
                      Nov 12 '18 at 17:19











                    • Ah.. Got thanks.

                      – Scott Boston
                      Nov 12 '18 at 17:22
















                    Think the first condition should be notna.

                    – coldspeed
                    Nov 12 '18 at 17:14





                    Think the first condition should be notna.

                    – coldspeed
                    Nov 12 '18 at 17:14













                    yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                    – Eric
                    Nov 12 '18 at 17:19





                    yes this works correctly in terms of multiple tests, changing isna() to notna() works as I would expect.

                    – Eric
                    Nov 12 '18 at 17:19













                    Ah.. Got thanks.

                    – Scott Boston
                    Nov 12 '18 at 17:22





                    Ah.. Got thanks.

                    – Scott Boston
                    Nov 12 '18 at 17:22

















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