How to reduce columns in dataframe pandas












7















Here how the datalooks like in df dataframe:



        A   B   C   D

0.js    2   1   1  -1

1.js    3  -5   1  -4

total   5  -4   2  -5


And I would get new dataframe df1:



        A     C

0.js    2     1

1.js    3     1

total   5     2


So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..






python pandas dataframe







share|improve this question




















  • 2





    Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

    – Edeki Okoh
    4 hours ago






  • 2





    Can't you transpose and do the filter?

    – Andrew Naguib
    4 hours ago











  • Your DataFrame is just a single row?

    – ALollz
    4 hours ago






  • 2





    Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

    – ALollz
    4 hours ago








  • 2





    I've updated the question, thanks for the guidance.

    – J Oderberg
    4 hours ago
















7















Here how the datalooks like in df dataframe:



        A   B   C   D

0.js    2   1   1  -1

1.js    3  -5   1  -4

total   5  -4   2  -5


And I would get new dataframe df1:



        A     C

0.js    2     1

1.js    3     1

total   5     2


So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..






python pandas dataframe







share|improve this question




















  • 2





    Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

    – Edeki Okoh
    4 hours ago






  • 2





    Can't you transpose and do the filter?

    – Andrew Naguib
    4 hours ago











  • Your DataFrame is just a single row?

    – ALollz
    4 hours ago






  • 2





    Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

    – ALollz
    4 hours ago








  • 2





    I've updated the question, thanks for the guidance.

    – J Oderberg
    4 hours ago














7












7








7








Here how the datalooks like in df dataframe:



        A   B   C   D

0.js    2   1   1  -1

1.js    3  -5   1  -4

total   5  -4   2  -5


And I would get new dataframe df1:



        A     C

0.js    2     1

1.js    3     1

total   5     2


So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..






python pandas dataframe







share|improve this question
















Here how the datalooks like in df dataframe:



        A   B   C   D

0.js    2   1   1  -1

1.js    3  -5   1  -4

total   5  -4   2  -5


And I would get new dataframe df1:



        A     C

0.js    2     1

1.js    3     1

total   5     2


So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..






python pandas dataframe




python pandas dataframe






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Scott Boston

54.1k73056




54.1k73056










asked 4 hours ago









J OderbergJ Oderberg

404




404








  • 2





    Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

    – Edeki Okoh
    4 hours ago






  • 2





    Can't you transpose and do the filter?

    – Andrew Naguib
    4 hours ago











  • Your DataFrame is just a single row?

    – ALollz
    4 hours ago






  • 2





    Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

    – ALollz
    4 hours ago








  • 2





    I've updated the question, thanks for the guidance.

    – J Oderberg
    4 hours ago














  • 2





    Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

    – Edeki Okoh
    4 hours ago






  • 2





    Can't you transpose and do the filter?

    – Andrew Naguib
    4 hours ago











  • Your DataFrame is just a single row?

    – ALollz
    4 hours ago






  • 2





    Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

    – ALollz
    4 hours ago








  • 2





    I've updated the question, thanks for the guidance.

    – J Oderberg
    4 hours ago








2




2





Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

– Edeki Okoh
4 hours ago





Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.

– Edeki Okoh
4 hours ago




2




2





Can't you transpose and do the filter?

– Andrew Naguib
4 hours ago





Can't you transpose and do the filter?

– Andrew Naguib
4 hours ago













Your DataFrame is just a single row?

– ALollz
4 hours ago





Your DataFrame is just a single row?

– ALollz
4 hours ago




2




2





Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

– ALollz
4 hours ago







Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.

– ALollz
4 hours ago






2




2





I've updated the question, thanks for the guidance.

– J Oderberg
4 hours ago





I've updated the question, thanks for the guidance.

– J Oderberg
4 hours ago












3 Answers
3






active

oldest

votes


















2














You can use, loc with boolean indexing or reindex:



df.loc[:, df.columns[(df.loc['total'] > 0)]]


OR



df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)


Output:



       A  C

0.js   2  1

1.js   3  1

total  5  2





share|improve this answer































    4














    You want to use .loc[:, column_mask] i.e.



    In [11]: df.loc[:, df.sum() > 0]
    
Out[11]:
    
       A  C
    
total  5  2
    

    
# or
    

    
In [12]: df.loc[:, df.iloc[0] > 0]
    
Out[12]:
    
       A  C
    
total  5  2





    share|improve this answer































      4














      Use .where to set negative values to NaN and then dropna setting axis = 1: 



      df.where(df.gt(0)).dropna(axis=1)
      

      
       A  C
      
total  5  2





      share|improve this answer





















      • 1





        Okay, that was clever +1

        – harvpan
        4 hours ago











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      You can use, loc with boolean indexing or reindex:



      df.loc[:, df.columns[(df.loc['total'] > 0)]]


      OR



      df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)


      Output:



             A  C
      
0.js   2  1
      
1.js   3  1
      
total  5  2





      share|improve this answer




























        2














        You can use, loc with boolean indexing or reindex:



        df.loc[:, df.columns[(df.loc['total'] > 0)]]


        OR



        df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)


        Output:



               A  C
        
0.js   2  1
        
1.js   3  1
        
total  5  2





        share|improve this answer


























          2












          2








          2







          You can use, loc with boolean indexing or reindex:



          df.loc[:, df.columns[(df.loc['total'] > 0)]]


          OR



          df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)


          Output:



                 A  C
          
0.js   2  1
          
1.js   3  1
          
total  5  2





          share|improve this answer













          You can use, loc with boolean indexing or reindex:



          df.loc[:, df.columns[(df.loc['total'] > 0)]]


          OR



          df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)


          Output:



                 A  C
          
0.js   2  1
          
1.js   3  1
          
total  5  2






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          Scott BostonScott Boston

          54.1k73056




          54.1k73056

























              4














              You want to use .loc[:, column_mask] i.e.



              In [11]: df.loc[:, df.sum() > 0]
              
Out[11]:
              
       A  C
              
total  5  2
              

              
# or
              

              
In [12]: df.loc[:, df.iloc[0] > 0]
              
Out[12]:
              
       A  C
              
total  5  2





              share|improve this answer




























                4














                You want to use .loc[:, column_mask] i.e.



                In [11]: df.loc[:, df.sum() > 0]
                
Out[11]:
                
       A  C
                
total  5  2
                

                
# or
                

                
In [12]: df.loc[:, df.iloc[0] > 0]
                
Out[12]:
                
       A  C
                
total  5  2





                share|improve this answer


























                  4












                  4








                  4







                  You want to use .loc[:, column_mask] i.e.



                  In [11]: df.loc[:, df.sum() > 0]
                  
Out[11]:
                  
       A  C
                  
total  5  2
                  

                  
# or
                  

                  
In [12]: df.loc[:, df.iloc[0] > 0]
                  
Out[12]:
                  
       A  C
                  
total  5  2





                  share|improve this answer













                  You want to use .loc[:, column_mask] i.e.



                  In [11]: df.loc[:, df.sum() > 0]
                  
Out[11]:
                  
       A  C
                  
total  5  2
                  

                  
# or
                  

                  
In [12]: df.loc[:, df.iloc[0] > 0]
                  
Out[12]:
                  
       A  C
                  
total  5  2






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 4 hours ago









                  Andy HaydenAndy Hayden

                  182k52429417




                  182k52429417























                      4














                      Use .where to set negative values to NaN and then dropna setting axis = 1: 



                      df.where(df.gt(0)).dropna(axis=1)
                      

                      
       A  C
                      
total  5  2





                      share|improve this answer





















                      • 1





                        Okay, that was clever +1

                        – harvpan
                        4 hours ago
















                      4














                      Use .where to set negative values to NaN and then dropna setting axis = 1: 



                      df.where(df.gt(0)).dropna(axis=1)
                      

                      
       A  C
                      
total  5  2





                      share|improve this answer





















                      • 1





                        Okay, that was clever +1

                        – harvpan
                        4 hours ago














                      4












                      4








                      4







                      Use .where to set negative values to NaN and then dropna setting axis = 1: 



                      df.where(df.gt(0)).dropna(axis=1)
                      

                      
       A  C
                      
total  5  2





                      share|improve this answer















                      Use .where to set negative values to NaN and then dropna setting axis = 1: 



                      df.where(df.gt(0)).dropna(axis=1)
                      

                      
       A  C
                      
total  5  2






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 4 hours ago

























                      answered 4 hours ago









                      yatuyatu

                      8,2011926




                      8,2011926








                      • 1





                        Okay, that was clever +1

                        – harvpan
                        4 hours ago














                      • 1





                        Okay, that was clever +1

                        – harvpan
                        4 hours ago








                      1




                      1





                      Okay, that was clever +1

                      – harvpan
                      4 hours ago





                      Okay, that was clever +1

                      – harvpan
                      4 hours ago


















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