How to transform the result of a Pandas `GROUPBY` function to the original dataframe












0














Suppose I have a Pandas DataFrame with 6 columns and a custom function that takes counts of the elements in 2 or 3 columns and produces a boolean output. When a groupby object is created from the original dataframe and the custom function is applied df.groupby('col1').apply(myfunc), the result is a series whose length is equal to the number of categories of col1. How do I expand this output to match the length of the original dataframe? I tried transform, but was not able to use the custom function myfunc with it.



EDIT:



Here is an example code:



A = pd.DataFrame({'X':['a','b','c','a','c'], 'Y':['at','bt','ct','at','ct'], 'Z':['q','q','r','r','s']})

print (A)



def myfunc(df):

    return ((df['Z'].nunique()>=2) and (df['Y'].nunique()<2))



A.groupby('X').apply(myfunc)


Output



I would like to expand this output as a new column Result such that where there is a in column X, the Result will be True.






python pandas dataframe







share|improve this question
























  • Could you show us some of your code?
    – user7374610
    Nov 16 at 3:00










  • @user7374610, I just added a simple sample code.
    – bluetooth
    Nov 16 at 3:25
















0














Suppose I have a Pandas DataFrame with 6 columns and a custom function that takes counts of the elements in 2 or 3 columns and produces a boolean output. When a groupby object is created from the original dataframe and the custom function is applied df.groupby('col1').apply(myfunc), the result is a series whose length is equal to the number of categories of col1. How do I expand this output to match the length of the original dataframe? I tried transform, but was not able to use the custom function myfunc with it.



EDIT:



Here is an example code:



A = pd.DataFrame({'X':['a','b','c','a','c'], 'Y':['at','bt','ct','at','ct'], 'Z':['q','q','r','r','s']})

print (A)



def myfunc(df):

    return ((df['Z'].nunique()>=2) and (df['Y'].nunique()<2))



A.groupby('X').apply(myfunc)


Output



I would like to expand this output as a new column Result such that where there is a in column X, the Result will be True.






python pandas dataframe







share|improve this question
























  • Could you show us some of your code?
    – user7374610
    Nov 16 at 3:00










  • @user7374610, I just added a simple sample code.
    – bluetooth
    Nov 16 at 3:25














0












0








0







Suppose I have a Pandas DataFrame with 6 columns and a custom function that takes counts of the elements in 2 or 3 columns and produces a boolean output. When a groupby object is created from the original dataframe and the custom function is applied df.groupby('col1').apply(myfunc), the result is a series whose length is equal to the number of categories of col1. How do I expand this output to match the length of the original dataframe? I tried transform, but was not able to use the custom function myfunc with it.



EDIT:



Here is an example code:



A = pd.DataFrame({'X':['a','b','c','a','c'], 'Y':['at','bt','ct','at','ct'], 'Z':['q','q','r','r','s']})

print (A)



def myfunc(df):

    return ((df['Z'].nunique()>=2) and (df['Y'].nunique()<2))



A.groupby('X').apply(myfunc)


Output



I would like to expand this output as a new column Result such that where there is a in column X, the Result will be True.






python pandas dataframe







share|improve this question















Suppose I have a Pandas DataFrame with 6 columns and a custom function that takes counts of the elements in 2 or 3 columns and produces a boolean output. When a groupby object is created from the original dataframe and the custom function is applied df.groupby('col1').apply(myfunc), the result is a series whose length is equal to the number of categories of col1. How do I expand this output to match the length of the original dataframe? I tried transform, but was not able to use the custom function myfunc with it.



EDIT:



Here is an example code:



A = pd.DataFrame({'X':['a','b','c','a','c'], 'Y':['at','bt','ct','at','ct'], 'Z':['q','q','r','r','s']})

print (A)



def myfunc(df):

    return ((df['Z'].nunique()>=2) and (df['Y'].nunique()<2))



A.groupby('X').apply(myfunc)


Output



I would like to expand this output as a new column Result such that where there is a in column X, the Result will be True.






python pandas dataframe




python pandas dataframe






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 at 3:24

























asked Nov 16 at 2:58









bluetooth

768




768












  • Could you show us some of your code?
    – user7374610
    Nov 16 at 3:00










  • @user7374610, I just added a simple sample code.
    – bluetooth
    Nov 16 at 3:25


















  • Could you show us some of your code?
    – user7374610
    Nov 16 at 3:00










  • @user7374610, I just added a simple sample code.
    – bluetooth
    Nov 16 at 3:25
















Could you show us some of your code?
– user7374610
Nov 16 at 3:00




Could you show us some of your code?
– user7374610
Nov 16 at 3:00












@user7374610, I just added a simple sample code.
– bluetooth
Nov 16 at 3:25




@user7374610, I just added a simple sample code.
– bluetooth
Nov 16 at 3:25












2 Answers
2






active

oldest

votes


















1














You can map the groupby back to the original dataframe 



A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))


Result would look like:



    X   Y   Z   Result

0   a   at  q   True

1   b   bt  q   False

2   c   ct  r   True

3   a   at  r   True

4   c   ct  s   True





share|improve this answer





























    0














    My solution may not be the best one, which uses a loop, but it's pretty good I think.



    The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.



    Here is an example:



    import pandas as pd
    
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
    
gp = df.groupby('a')  # group
    
s = gp.apply(sum)['a'] # apply a func
    
adf = 
    

    
# then create a new dataframe
    
for i, gdf in gp:
    
    tdf = gdf.copy()
    
    tdf.loc[:,'c'] = s.loc[i]
    
    adf.append(tdf)
    
pd.concat(adf)


    from:



        a   b
    
0   1   a
    
1   2   b
    
2   1   c
    
3   2   d


    to:



        a   b   c
    
0   1   a   2
    
2   1   c   2
    
1   2   b   4
    
3   2   d   4





    share|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      You can map the groupby back to the original dataframe 



      A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))


      Result would look like:



          X   Y   Z   Result
      
0   a   at  q   True
      
1   b   bt  q   False
      
2   c   ct  r   True
      
3   a   at  r   True
      
4   c   ct  s   True





      share|improve this answer


























        1














        You can map the groupby back to the original dataframe 



        A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))


        Result would look like:



            X   Y   Z   Result
        
0   a   at  q   True
        
1   b   bt  q   False
        
2   c   ct  r   True
        
3   a   at  r   True
        
4   c   ct  s   True





        share|improve this answer
























          1












          1








          1






          You can map the groupby back to the original dataframe 



          A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))


          Result would look like:



              X   Y   Z   Result
          
0   a   at  q   True
          
1   b   bt  q   False
          
2   c   ct  r   True
          
3   a   at  r   True
          
4   c   ct  s   True





          share|improve this answer












          You can map the groupby back to the original dataframe 



          A['Result'] = A['X'].map(A.groupby('X').apply(myfunc))


          Result would look like:



              X   Y   Z   Result
          
0   a   at  q   True
          
1   b   bt  q   False
          
2   c   ct  r   True
          
3   a   at  r   True
          
4   c   ct  s   True






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 16 at 3:32









          user7374610

          6981422




          6981422

























              0














              My solution may not be the best one, which uses a loop, but it's pretty good I think.



              The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.



              Here is an example:



              import pandas as pd
              
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
              
gp = df.groupby('a')  # group
              
s = gp.apply(sum)['a'] # apply a func
              
adf = 
              

              
# then create a new dataframe
              
for i, gdf in gp:
              
    tdf = gdf.copy()
              
    tdf.loc[:,'c'] = s.loc[i]
              
    adf.append(tdf)
              
pd.concat(adf)


              from:



                  a   b
              
0   1   a
              
1   2   b
              
2   1   c
              
3   2   d


              to:



                  a   b   c
              
0   1   a   2
              
2   1   c   2
              
1   2   b   4
              
3   2   d   4





              share|improve this answer




























                0














                My solution may not be the best one, which uses a loop, but it's pretty good I think.



                The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.



                Here is an example:



                import pandas as pd
                
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
                
gp = df.groupby('a')  # group
                
s = gp.apply(sum)['a'] # apply a func
                
adf = 
                

                
# then create a new dataframe
                
for i, gdf in gp:
                
    tdf = gdf.copy()
                
    tdf.loc[:,'c'] = s.loc[i]
                
    adf.append(tdf)
                
pd.concat(adf)


                from:



                    a   b
                
0   1   a
                
1   2   b
                
2   1   c
                
3   2   d


                to:



                    a   b   c
                
0   1   a   2
                
2   1   c   2
                
1   2   b   4
                
3   2   d   4





                share|improve this answer


























                  0












                  0








                  0






                  My solution may not be the best one, which uses a loop, but it's pretty good I think.



                  The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.



                  Here is an example:



                  import pandas as pd
                  
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
                  
gp = df.groupby('a')  # group
                  
s = gp.apply(sum)['a'] # apply a func
                  
adf = 
                  

                  
# then create a new dataframe
                  
for i, gdf in gp:
                  
    tdf = gdf.copy()
                  
    tdf.loc[:,'c'] = s.loc[i]
                  
    adf.append(tdf)
                  
pd.concat(adf)


                  from:



                      a   b
                  
0   1   a
                  
1   2   b
                  
2   1   c
                  
3   2   d


                  to:



                      a   b   c
                  
0   1   a   2
                  
2   1   c   2
                  
1   2   b   4
                  
3   2   d   4





                  share|improve this answer














                  My solution may not be the best one, which uses a loop, but it's pretty good I think.



                  The core idea is you can traverse all the sub-dataframe (gdf) by for i, gdf in gp. Then add the column result (in my example it is c) for each sub-dataframe. Finally concat all the sub-dataframe into one.



                  Here is an example:



                  import pandas as pd
                  
df = pd.DataFrame({'a':[1,2,1,2],'b':['a','b','c','d']})
                  
gp = df.groupby('a')  # group
                  
s = gp.apply(sum)['a'] # apply a func
                  
adf = 
                  

                  
# then create a new dataframe
                  
for i, gdf in gp:
                  
    tdf = gdf.copy()
                  
    tdf.loc[:,'c'] = s.loc[i]
                  
    adf.append(tdf)
                  
pd.concat(adf)


                  from:



                      a   b
                  
0   1   a
                  
1   2   b
                  
2   1   c
                  
3   2   d


                  to:



                      a   b   c
                  
0   1   a   2
                  
2   1   c   2
                  
1   2   b   4
                  
3   2   d   4






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 16 at 3:39

























                  answered Nov 16 at 3:32









                  Zealseeker

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