How to split a column into two columns by first and last found pattern in Pandas (Python 3.x)





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2















i have a problem with splitting a column into two columns. I want to split the column by the first and last found pattern '-'. Maybe this is trivial.



Here is my column:



        col1

0       aa-bb-cc-dd

1       aa-bb-cc

2       aa-bb-cc

3       aa-bb-cc-dd


This is the frame i want as result:



        col1           col2

0       bb-cc          dd

1       bb             cc

2       bb             cc

3       bb-cc          dd


Thanks in advance!






python string pandas dataframe split







share|improve this question































    2















    i have a problem with splitting a column into two columns. I want to split the column by the first and last found pattern '-'. Maybe this is trivial.



    Here is my column:



            col1
    
0       aa-bb-cc-dd
    
1       aa-bb-cc
    
2       aa-bb-cc
    
3       aa-bb-cc-dd


    This is the frame i want as result:



            col1           col2
    
0       bb-cc          dd
    
1       bb             cc
    
2       bb             cc
    
3       bb-cc          dd


    Thanks in advance!






    python string pandas dataframe split







    share|improve this question



























      2












      2








      2








      i have a problem with splitting a column into two columns. I want to split the column by the first and last found pattern '-'. Maybe this is trivial.



      Here is my column:



              col1
      
0       aa-bb-cc-dd
      
1       aa-bb-cc
      
2       aa-bb-cc
      
3       aa-bb-cc-dd


      This is the frame i want as result:



              col1           col2
      
0       bb-cc          dd
      
1       bb             cc
      
2       bb             cc
      
3       bb-cc          dd


      Thanks in advance!






      python string pandas dataframe split







      share|improve this question
















      i have a problem with splitting a column into two columns. I want to split the column by the first and last found pattern '-'. Maybe this is trivial.



      Here is my column:



              col1
      
0       aa-bb-cc-dd
      
1       aa-bb-cc
      
2       aa-bb-cc
      
3       aa-bb-cc-dd


      This is the frame i want as result:



              col1           col2
      
0       bb-cc          dd
      
1       bb             cc
      
2       bb             cc
      
3       bb-cc          dd


      Thanks in advance!






      python string pandas dataframe split




      python string pandas dataframe split






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 16 '18 at 12:05









      jpp

      102k2166116




      102k2166116










      asked Nov 16 '18 at 11:26









      Michael GannMichael Gann

      554




      554
























          5 Answers
          5






          active

          oldest

          votes


















          3














          You can use a list comprehension:



          df = pd.DataFrame([i.split('-', 1)[1].rsplit('-', 1) for i in df['col1']],
          
                  columns=['col1', 'col2'])
          

          
print(df)
          

          
    col1 col2
          
0  bb-cc   dd
          
1     bb   cc
          
2     bb   cc
          
3  bb-cc   dd


          Pandas str methods exist primarily for convenience. For clean data, you may find the list comprehension more efficient for larger dataframes.






          share|improve this answer



















          • 1





            This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

            – John Zwinck
            Nov 16 '18 at 11:47





















          1














          If I understand well your question, you need to get rid of the first block delimited by a '-', then split the last '-' block in col2. If that is what you need, you could consider this:



          df= pd.DataFrame({'col1':['aa-bb-cc-dd', 'aa-bb-cc', 'aa-bb-cc', 'aa-bb-cc-dd']})
          
df['col2'] = df['col1'].apply(lambda x: x[x.rfind('-')+1:])
          
df['col1'] =  df['col1'].apply(lambda x: x[x.find('-')+1:x.rfind('-')])
          
print (df)





          share|improve this answer
























          • apply() is usually quite slow. If we want to iterate we can just write for loops.

            – John Zwinck
            Nov 16 '18 at 11:37











          • This solution takes 1.8 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:42





















          1














          First slice and use str.rsplit and rename:



          df = df.col1.str[3:].str.rsplit('-', n=1, expand=True).rename(columns={0:'col1',1:'col2'})
          

          
print(df)
          
    col1 col2
          
0  bb-cc   dd
          
1     bb   cc
          
2     bb   cc
          
3  bb-cc   dd





          share|improve this answer


























          • dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

            – John Zwinck
            Nov 16 '18 at 11:35













          • @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

            – Sandeep Kadapa
            Nov 16 '18 at 11:36











          • If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

            – John Zwinck
            Nov 16 '18 at 11:38











          • @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

            – Sandeep Kadapa
            Nov 16 '18 at 11:41











          • This solution takes 3.9 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:43



















          1














          Here's an idiomatic but slow way to do it:



          df.col1 = df.col1.str.split('-', 1).str[1] # discard first part
          
parts = df.col1.str.rsplit('-', 1).str
          
df.col1 = parts[0]
          
df['col2'] = parts[1]


          While this works, it is not fast: about 4 seconds for 700k rows. Looking at it you'd think this is a good way to do it, but performance-wise it's worse than all the alternatives.






          share|improve this answer

































            -1














            This might help:



            df['col2'] = df['col1'].split('-')[-1]
            
df['col1'] = '-'.join(i for i in df['col1'].split('-')[1:-1])





            share|improve this answer



















            • 1





              You're missing some strs in there.

              – John Zwinck
              Nov 16 '18 at 11:49












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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            You can use a list comprehension:



            df = pd.DataFrame([i.split('-', 1)[1].rsplit('-', 1) for i in df['col1']],
            
                  columns=['col1', 'col2'])
            

            
print(df)
            

            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd


            Pandas str methods exist primarily for convenience. For clean data, you may find the list comprehension more efficient for larger dataframes.






            share|improve this answer



















            • 1





              This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

              – John Zwinck
              Nov 16 '18 at 11:47


















            3














            You can use a list comprehension:



            df = pd.DataFrame([i.split('-', 1)[1].rsplit('-', 1) for i in df['col1']],
            
                  columns=['col1', 'col2'])
            

            
print(df)
            

            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd


            Pandas str methods exist primarily for convenience. For clean data, you may find the list comprehension more efficient for larger dataframes.






            share|improve this answer



















            • 1





              This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

              – John Zwinck
              Nov 16 '18 at 11:47
















            3












            3








            3







            You can use a list comprehension:



            df = pd.DataFrame([i.split('-', 1)[1].rsplit('-', 1) for i in df['col1']],
            
                  columns=['col1', 'col2'])
            

            
print(df)
            

            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd


            Pandas str methods exist primarily for convenience. For clean data, you may find the list comprehension more efficient for larger dataframes.






            share|improve this answer













            You can use a list comprehension:



            df = pd.DataFrame([i.split('-', 1)[1].rsplit('-', 1) for i in df['col1']],
            
                  columns=['col1', 'col2'])
            

            
print(df)
            

            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd


            Pandas str methods exist primarily for convenience. For clean data, you may find the list comprehension more efficient for larger dataframes.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 16 '18 at 11:37









            jppjpp

            102k2166116




            102k2166116








            • 1





              This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

              – John Zwinck
              Nov 16 '18 at 11:47
















            • 1





              This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

              – John Zwinck
              Nov 16 '18 at 11:47










            1




            1





            This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

            – John Zwinck
            Nov 16 '18 at 11:47







            This solution takes 1.1 seconds for 700k rows--the fastest I've tested so far! 3x faster than using Series.str.split().

            – John Zwinck
            Nov 16 '18 at 11:47















            1














            If I understand well your question, you need to get rid of the first block delimited by a '-', then split the last '-' block in col2. If that is what you need, you could consider this:



            df= pd.DataFrame({'col1':['aa-bb-cc-dd', 'aa-bb-cc', 'aa-bb-cc', 'aa-bb-cc-dd']})
            
df['col2'] = df['col1'].apply(lambda x: x[x.rfind('-')+1:])
            
df['col1'] =  df['col1'].apply(lambda x: x[x.find('-')+1:x.rfind('-')])
            
print (df)





            share|improve this answer
























            • apply() is usually quite slow. If we want to iterate we can just write for loops.

              – John Zwinck
              Nov 16 '18 at 11:37











            • This solution takes 1.8 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:42


















            1














            If I understand well your question, you need to get rid of the first block delimited by a '-', then split the last '-' block in col2. If that is what you need, you could consider this:



            df= pd.DataFrame({'col1':['aa-bb-cc-dd', 'aa-bb-cc', 'aa-bb-cc', 'aa-bb-cc-dd']})
            
df['col2'] = df['col1'].apply(lambda x: x[x.rfind('-')+1:])
            
df['col1'] =  df['col1'].apply(lambda x: x[x.find('-')+1:x.rfind('-')])
            
print (df)





            share|improve this answer
























            • apply() is usually quite slow. If we want to iterate we can just write for loops.

              – John Zwinck
              Nov 16 '18 at 11:37











            • This solution takes 1.8 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:42
















            1












            1








            1







            If I understand well your question, you need to get rid of the first block delimited by a '-', then split the last '-' block in col2. If that is what you need, you could consider this:



            df= pd.DataFrame({'col1':['aa-bb-cc-dd', 'aa-bb-cc', 'aa-bb-cc', 'aa-bb-cc-dd']})
            
df['col2'] = df['col1'].apply(lambda x: x[x.rfind('-')+1:])
            
df['col1'] =  df['col1'].apply(lambda x: x[x.find('-')+1:x.rfind('-')])
            
print (df)





            share|improve this answer













            If I understand well your question, you need to get rid of the first block delimited by a '-', then split the last '-' block in col2. If that is what you need, you could consider this:



            df= pd.DataFrame({'col1':['aa-bb-cc-dd', 'aa-bb-cc', 'aa-bb-cc', 'aa-bb-cc-dd']})
            
df['col2'] = df['col1'].apply(lambda x: x[x.rfind('-')+1:])
            
df['col1'] =  df['col1'].apply(lambda x: x[x.find('-')+1:x.rfind('-')])
            
print (df)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 16 '18 at 11:32









            Matina GMatina G

            629213




            629213













            • apply() is usually quite slow. If we want to iterate we can just write for loops.

              – John Zwinck
              Nov 16 '18 at 11:37











            • This solution takes 1.8 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:42





















            • apply() is usually quite slow. If we want to iterate we can just write for loops.

              – John Zwinck
              Nov 16 '18 at 11:37











            • This solution takes 1.8 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:42



















            apply() is usually quite slow. If we want to iterate we can just write for loops.

            – John Zwinck
            Nov 16 '18 at 11:37





            apply() is usually quite slow. If we want to iterate we can just write for loops.

            – John Zwinck
            Nov 16 '18 at 11:37













            This solution takes 1.8 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:42







            This solution takes 1.8 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:42













            1














            First slice and use str.rsplit and rename:



            df = df.col1.str[3:].str.rsplit('-', n=1, expand=True).rename(columns={0:'col1',1:'col2'})
            

            
print(df)
            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd





            share|improve this answer


























            • dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

              – John Zwinck
              Nov 16 '18 at 11:35













            • @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

              – Sandeep Kadapa
              Nov 16 '18 at 11:36











            • If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

              – John Zwinck
              Nov 16 '18 at 11:38











            • @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

              – Sandeep Kadapa
              Nov 16 '18 at 11:41











            • This solution takes 3.9 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:43
















            1














            First slice and use str.rsplit and rename:



            df = df.col1.str[3:].str.rsplit('-', n=1, expand=True).rename(columns={0:'col1',1:'col2'})
            

            
print(df)
            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd





            share|improve this answer


























            • dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

              – John Zwinck
              Nov 16 '18 at 11:35













            • @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

              – Sandeep Kadapa
              Nov 16 '18 at 11:36











            • If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

              – John Zwinck
              Nov 16 '18 at 11:38











            • @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

              – Sandeep Kadapa
              Nov 16 '18 at 11:41











            • This solution takes 3.9 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:43














            1












            1








            1







            First slice and use str.rsplit and rename:



            df = df.col1.str[3:].str.rsplit('-', n=1, expand=True).rename(columns={0:'col1',1:'col2'})
            

            
print(df)
            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd





            share|improve this answer















            First slice and use str.rsplit and rename:



            df = df.col1.str[3:].str.rsplit('-', n=1, expand=True).rename(columns={0:'col1',1:'col2'})
            

            
print(df)
            
    col1 col2
            
0  bb-cc   dd
            
1     bb   cc
            
2     bb   cc
            
3  bb-cc   dd






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 16 '18 at 11:37

























            answered Nov 16 '18 at 11:34









            Sandeep KadapaSandeep Kadapa

            7,408831




            7,408831













            • dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

              – John Zwinck
              Nov 16 '18 at 11:35













            • @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

              – Sandeep Kadapa
              Nov 16 '18 at 11:36











            • If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

              – John Zwinck
              Nov 16 '18 at 11:38











            • @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

              – Sandeep Kadapa
              Nov 16 '18 at 11:41











            • This solution takes 3.9 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:43



















            • dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

              – John Zwinck
              Nov 16 '18 at 11:35













            • @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

              – Sandeep Kadapa
              Nov 16 '18 at 11:36











            • If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

              – John Zwinck
              Nov 16 '18 at 11:38











            • @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

              – Sandeep Kadapa
              Nov 16 '18 at 11:41











            • This solution takes 3.9 seconds for 700k rows.

              – John Zwinck
              Nov 16 '18 at 11:43

















            dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

            – John Zwinck
            Nov 16 '18 at 11:35







            dict(zip([0,1],['col1','col2'])) is just {0: 'col1', 1: 'col2'}.

            – John Zwinck
            Nov 16 '18 at 11:35















            @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

            – Sandeep Kadapa
            Nov 16 '18 at 11:36





            @JohnZwinck I thought to create dict dynamically but here only 2 columns not needed and thank you.

            – Sandeep Kadapa
            Nov 16 '18 at 11:36













            If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

            – John Zwinck
            Nov 16 '18 at 11:38





            If you needed to do it dynamically, zip([0,1],['col1','col2']) is better written as enumerate(['col1', 'col2']).

            – John Zwinck
            Nov 16 '18 at 11:38













            @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

            – Sandeep Kadapa
            Nov 16 '18 at 11:41





            @JohnZwinck It's good, I thought to use range but I will use enumerate trick from now.

            – Sandeep Kadapa
            Nov 16 '18 at 11:41













            This solution takes 3.9 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:43





            This solution takes 3.9 seconds for 700k rows.

            – John Zwinck
            Nov 16 '18 at 11:43











            1














            Here's an idiomatic but slow way to do it:



            df.col1 = df.col1.str.split('-', 1).str[1] # discard first part
            
parts = df.col1.str.rsplit('-', 1).str
            
df.col1 = parts[0]
            
df['col2'] = parts[1]


            While this works, it is not fast: about 4 seconds for 700k rows. Looking at it you'd think this is a good way to do it, but performance-wise it's worse than all the alternatives.






            share|improve this answer






























              1














              Here's an idiomatic but slow way to do it:



              df.col1 = df.col1.str.split('-', 1).str[1] # discard first part
              
parts = df.col1.str.rsplit('-', 1).str
              
df.col1 = parts[0]
              
df['col2'] = parts[1]


              While this works, it is not fast: about 4 seconds for 700k rows. Looking at it you'd think this is a good way to do it, but performance-wise it's worse than all the alternatives.






              share|improve this answer




























                1












                1








                1







                Here's an idiomatic but slow way to do it:



                df.col1 = df.col1.str.split('-', 1).str[1] # discard first part
                
parts = df.col1.str.rsplit('-', 1).str
                
df.col1 = parts[0]
                
df['col2'] = parts[1]


                While this works, it is not fast: about 4 seconds for 700k rows. Looking at it you'd think this is a good way to do it, but performance-wise it's worse than all the alternatives.






                share|improve this answer















                Here's an idiomatic but slow way to do it:



                df.col1 = df.col1.str.split('-', 1).str[1] # discard first part
                
parts = df.col1.str.rsplit('-', 1).str
                
df.col1 = parts[0]
                
df['col2'] = parts[1]


                While this works, it is not fast: about 4 seconds for 700k rows. Looking at it you'd think this is a good way to do it, but performance-wise it's worse than all the alternatives.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 16 '18 at 11:52

























                answered Nov 16 '18 at 11:35









                John ZwinckJohn Zwinck

                155k17180298




                155k17180298























                    -1














                    This might help:



                    df['col2'] = df['col1'].split('-')[-1]
                    
df['col1'] = '-'.join(i for i in df['col1'].split('-')[1:-1])





                    share|improve this answer



















                    • 1





                      You're missing some strs in there.

                      – John Zwinck
                      Nov 16 '18 at 11:49
















                    -1














                    This might help:



                    df['col2'] = df['col1'].split('-')[-1]
                    
df['col1'] = '-'.join(i for i in df['col1'].split('-')[1:-1])





                    share|improve this answer



















                    • 1





                      You're missing some strs in there.

                      – John Zwinck
                      Nov 16 '18 at 11:49














                    -1












                    -1








                    -1







                    This might help:



                    df['col2'] = df['col1'].split('-')[-1]
                    
df['col1'] = '-'.join(i for i in df['col1'].split('-')[1:-1])





                    share|improve this answer













                    This might help:



                    df['col2'] = df['col1'].split('-')[-1]
                    
df['col1'] = '-'.join(i for i in df['col1'].split('-')[1:-1])






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 16 '18 at 11:41









                    specbugspecbug

                    310310




                    310310








                    • 1





                      You're missing some strs in there.

                      – John Zwinck
                      Nov 16 '18 at 11:49














                    • 1





                      You're missing some strs in there.

                      – John Zwinck
                      Nov 16 '18 at 11:49








                    1




                    1





                    You're missing some strs in there.

                    – John Zwinck
                    Nov 16 '18 at 11:49





                    You're missing some strs in there.

                    – John Zwinck
                    Nov 16 '18 at 11:49


















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