A pythonic and uFunc-y way to turn pandas column into “increasing” index?











up vote
6
down vote

favorite
2












Let's say I have a pandas df like so:



Index   A     B
0 foo 3
1 foo 2
2 foo 5
3 bar 3
4 bar 4
5 baz 5


What's a good fast way to add a column like so:



Index   A     B    Aidx
0 foo 3 0
1 foo 2 0
2 foo 5 0
3 bar 3 1
4 bar 4 1
5 baz 5 2


I.e. adding an increasing index for each unique value?



I know I could use df.unique(), then use a dict and enumerate to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby with some special function?










share|improve this question


























    up vote
    6
    down vote

    favorite
    2












    Let's say I have a pandas df like so:



    Index   A     B
    0 foo 3
    1 foo 2
    2 foo 5
    3 bar 3
    4 bar 4
    5 baz 5


    What's a good fast way to add a column like so:



    Index   A     B    Aidx
    0 foo 3 0
    1 foo 2 0
    2 foo 5 0
    3 bar 3 1
    4 bar 4 1
    5 baz 5 2


    I.e. adding an increasing index for each unique value?



    I know I could use df.unique(), then use a dict and enumerate to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby with some special function?










    share|improve this question
























      up vote
      6
      down vote

      favorite
      2









      up vote
      6
      down vote

      favorite
      2






      2





      Let's say I have a pandas df like so:



      Index   A     B
      0 foo 3
      1 foo 2
      2 foo 5
      3 bar 3
      4 bar 4
      5 baz 5


      What's a good fast way to add a column like so:



      Index   A     B    Aidx
      0 foo 3 0
      1 foo 2 0
      2 foo 5 0
      3 bar 3 1
      4 bar 4 1
      5 baz 5 2


      I.e. adding an increasing index for each unique value?



      I know I could use df.unique(), then use a dict and enumerate to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby with some special function?










      share|improve this question













      Let's say I have a pandas df like so:



      Index   A     B
      0 foo 3
      1 foo 2
      2 foo 5
      3 bar 3
      4 bar 4
      5 baz 5


      What's a good fast way to add a column like so:



      Index   A     B    Aidx
      0 foo 3 0
      1 foo 2 0
      2 foo 5 0
      3 bar 3 1
      4 bar 4 1
      5 baz 5 2


      I.e. adding an increasing index for each unique value?



      I know I could use df.unique(), then use a dict and enumerate to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby with some special function?







      python pandas






      share|improve this question













      share|improve this question











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      share|improve this question










      asked 1 hour ago









      Lagerbaer

      2,6181124




      2,6181124
























          3 Answers
          3






          active

          oldest

          votes

















          up vote
          6
          down vote













          One way is to use ngroup. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False:



          df['Aidx'] = df.groupby('A',sort=False).ngroup()
          >>> df
          Index A B Aidx
          0 0 foo 3 0
          1 1 foo 2 0
          2 2 foo 5 0
          3 3 bar 3 1
          4 4 bar 4 1
          5 5 baz 5 2





          share|improve this answer






























            up vote
            2
            down vote













            No need groupby using





            Method 1factorize



            pd.factorize(df.A)[0]
            array([0, 0, 0, 1, 1, 2], dtype=int64)
            #df['Aidx']=pd.factorize(df.A)[0]




            Method 2 sklearn



            from sklearn import preprocessing
            le = preprocessing.LabelEncoder()
            le.fit(df.A)
            LabelEncoder()
            le.transform(df.A)
            array([2, 2, 2, 0, 0, 1])




            Method 3 cat.codes



            df.A.astype('category').cat.codes




            Method 4 map + unique



            l=df.A.unique()
            df.A.map(dict(zip(l,range(len(l)))))
            0 0
            1 0
            2 0
            3 1
            4 1
            5 2
            Name: A, dtype: int64





            share|improve this answer






























              up vote
              2
              down vote













              One more method of doing so could be.



              df['C'] = i.ne(df.A.shift()).cumsum()-1
              df


              When we print df value it will be as follows.



                Index  A    B  C
              0 0 foo 3 0
              1 1 foo 2 0
              2 2 foo 5 0
              3 3 bar 3 1
              4 4 bar 4 1
              5 5 baz 5 2


              Explanation of solution: Let's break above solution into parts for understanding purposes.



              1st step: Compare df's A column by shifting its value down to itself as follows.



              i.ne(df.A.shift())


              Output we will get is:



              0     True
              1 False
              2 False
              3 True
              4 False
              5 True


              2nd step: Use of cumsum() function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum() function and its value will be increased.



              i.ne(df.A.shift()).cumsum()-1
              0 0
              1 0
              2 0
              3 1
              4 1
              5 2
              Name: A, dtype: int32


              3rd step: Save command's value into df['C'] which will create a new column named C in df.






              share|improve this answer



















              • 1




                Nice method ve++ for you
                – W-B
                36 mins ago










              • @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                – RavinderSingh13
                34 mins ago













              Your Answer






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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              6
              down vote













              One way is to use ngroup. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False:



              df['Aidx'] = df.groupby('A',sort=False).ngroup()
              >>> df
              Index A B Aidx
              0 0 foo 3 0
              1 1 foo 2 0
              2 2 foo 5 0
              3 3 bar 3 1
              4 4 bar 4 1
              5 5 baz 5 2





              share|improve this answer



























                up vote
                6
                down vote













                One way is to use ngroup. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False:



                df['Aidx'] = df.groupby('A',sort=False).ngroup()
                >>> df
                Index A B Aidx
                0 0 foo 3 0
                1 1 foo 2 0
                2 2 foo 5 0
                3 3 bar 3 1
                4 4 bar 4 1
                5 5 baz 5 2





                share|improve this answer

























                  up vote
                  6
                  down vote










                  up vote
                  6
                  down vote









                  One way is to use ngroup. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False:



                  df['Aidx'] = df.groupby('A',sort=False).ngroup()
                  >>> df
                  Index A B Aidx
                  0 0 foo 3 0
                  1 1 foo 2 0
                  2 2 foo 5 0
                  3 3 bar 3 1
                  4 4 bar 4 1
                  5 5 baz 5 2





                  share|improve this answer














                  One way is to use ngroup. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False:



                  df['Aidx'] = df.groupby('A',sort=False).ngroup()
                  >>> df
                  Index A B Aidx
                  0 0 foo 3 0
                  1 1 foo 2 0
                  2 2 foo 5 0
                  3 3 bar 3 1
                  4 4 bar 4 1
                  5 5 baz 5 2






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 19 mins ago

























                  answered 1 hour ago









                  sacul

                  29.7k41640




                  29.7k41640
























                      up vote
                      2
                      down vote













                      No need groupby using





                      Method 1factorize



                      pd.factorize(df.A)[0]
                      array([0, 0, 0, 1, 1, 2], dtype=int64)
                      #df['Aidx']=pd.factorize(df.A)[0]




                      Method 2 sklearn



                      from sklearn import preprocessing
                      le = preprocessing.LabelEncoder()
                      le.fit(df.A)
                      LabelEncoder()
                      le.transform(df.A)
                      array([2, 2, 2, 0, 0, 1])




                      Method 3 cat.codes



                      df.A.astype('category').cat.codes




                      Method 4 map + unique



                      l=df.A.unique()
                      df.A.map(dict(zip(l,range(len(l)))))
                      0 0
                      1 0
                      2 0
                      3 1
                      4 1
                      5 2
                      Name: A, dtype: int64





                      share|improve this answer



























                        up vote
                        2
                        down vote













                        No need groupby using





                        Method 1factorize



                        pd.factorize(df.A)[0]
                        array([0, 0, 0, 1, 1, 2], dtype=int64)
                        #df['Aidx']=pd.factorize(df.A)[0]




                        Method 2 sklearn



                        from sklearn import preprocessing
                        le = preprocessing.LabelEncoder()
                        le.fit(df.A)
                        LabelEncoder()
                        le.transform(df.A)
                        array([2, 2, 2, 0, 0, 1])




                        Method 3 cat.codes



                        df.A.astype('category').cat.codes




                        Method 4 map + unique



                        l=df.A.unique()
                        df.A.map(dict(zip(l,range(len(l)))))
                        0 0
                        1 0
                        2 0
                        3 1
                        4 1
                        5 2
                        Name: A, dtype: int64





                        share|improve this answer

























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          No need groupby using





                          Method 1factorize



                          pd.factorize(df.A)[0]
                          array([0, 0, 0, 1, 1, 2], dtype=int64)
                          #df['Aidx']=pd.factorize(df.A)[0]




                          Method 2 sklearn



                          from sklearn import preprocessing
                          le = preprocessing.LabelEncoder()
                          le.fit(df.A)
                          LabelEncoder()
                          le.transform(df.A)
                          array([2, 2, 2, 0, 0, 1])




                          Method 3 cat.codes



                          df.A.astype('category').cat.codes




                          Method 4 map + unique



                          l=df.A.unique()
                          df.A.map(dict(zip(l,range(len(l)))))
                          0 0
                          1 0
                          2 0
                          3 1
                          4 1
                          5 2
                          Name: A, dtype: int64





                          share|improve this answer














                          No need groupby using





                          Method 1factorize



                          pd.factorize(df.A)[0]
                          array([0, 0, 0, 1, 1, 2], dtype=int64)
                          #df['Aidx']=pd.factorize(df.A)[0]




                          Method 2 sklearn



                          from sklearn import preprocessing
                          le = preprocessing.LabelEncoder()
                          le.fit(df.A)
                          LabelEncoder()
                          le.transform(df.A)
                          array([2, 2, 2, 0, 0, 1])




                          Method 3 cat.codes



                          df.A.astype('category').cat.codes




                          Method 4 map + unique



                          l=df.A.unique()
                          df.A.map(dict(zip(l,range(len(l)))))
                          0 0
                          1 0
                          2 0
                          3 1
                          4 1
                          5 2
                          Name: A, dtype: int64






                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 28 mins ago

























                          answered 38 mins ago









                          W-B

                          97.8k73162




                          97.8k73162






















                              up vote
                              2
                              down vote













                              One more method of doing so could be.



                              df['C'] = i.ne(df.A.shift()).cumsum()-1
                              df


                              When we print df value it will be as follows.



                                Index  A    B  C
                              0 0 foo 3 0
                              1 1 foo 2 0
                              2 2 foo 5 0
                              3 3 bar 3 1
                              4 4 bar 4 1
                              5 5 baz 5 2


                              Explanation of solution: Let's break above solution into parts for understanding purposes.



                              1st step: Compare df's A column by shifting its value down to itself as follows.



                              i.ne(df.A.shift())


                              Output we will get is:



                              0     True
                              1 False
                              2 False
                              3 True
                              4 False
                              5 True


                              2nd step: Use of cumsum() function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum() function and its value will be increased.



                              i.ne(df.A.shift()).cumsum()-1
                              0 0
                              1 0
                              2 0
                              3 1
                              4 1
                              5 2
                              Name: A, dtype: int32


                              3rd step: Save command's value into df['C'] which will create a new column named C in df.






                              share|improve this answer



















                              • 1




                                Nice method ve++ for you
                                – W-B
                                36 mins ago










                              • @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                                – RavinderSingh13
                                34 mins ago

















                              up vote
                              2
                              down vote













                              One more method of doing so could be.



                              df['C'] = i.ne(df.A.shift()).cumsum()-1
                              df


                              When we print df value it will be as follows.



                                Index  A    B  C
                              0 0 foo 3 0
                              1 1 foo 2 0
                              2 2 foo 5 0
                              3 3 bar 3 1
                              4 4 bar 4 1
                              5 5 baz 5 2


                              Explanation of solution: Let's break above solution into parts for understanding purposes.



                              1st step: Compare df's A column by shifting its value down to itself as follows.



                              i.ne(df.A.shift())


                              Output we will get is:



                              0     True
                              1 False
                              2 False
                              3 True
                              4 False
                              5 True


                              2nd step: Use of cumsum() function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum() function and its value will be increased.



                              i.ne(df.A.shift()).cumsum()-1
                              0 0
                              1 0
                              2 0
                              3 1
                              4 1
                              5 2
                              Name: A, dtype: int32


                              3rd step: Save command's value into df['C'] which will create a new column named C in df.






                              share|improve this answer



















                              • 1




                                Nice method ve++ for you
                                – W-B
                                36 mins ago










                              • @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                                – RavinderSingh13
                                34 mins ago















                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              One more method of doing so could be.



                              df['C'] = i.ne(df.A.shift()).cumsum()-1
                              df


                              When we print df value it will be as follows.



                                Index  A    B  C
                              0 0 foo 3 0
                              1 1 foo 2 0
                              2 2 foo 5 0
                              3 3 bar 3 1
                              4 4 bar 4 1
                              5 5 baz 5 2


                              Explanation of solution: Let's break above solution into parts for understanding purposes.



                              1st step: Compare df's A column by shifting its value down to itself as follows.



                              i.ne(df.A.shift())


                              Output we will get is:



                              0     True
                              1 False
                              2 False
                              3 True
                              4 False
                              5 True


                              2nd step: Use of cumsum() function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum() function and its value will be increased.



                              i.ne(df.A.shift()).cumsum()-1
                              0 0
                              1 0
                              2 0
                              3 1
                              4 1
                              5 2
                              Name: A, dtype: int32


                              3rd step: Save command's value into df['C'] which will create a new column named C in df.






                              share|improve this answer














                              One more method of doing so could be.



                              df['C'] = i.ne(df.A.shift()).cumsum()-1
                              df


                              When we print df value it will be as follows.



                                Index  A    B  C
                              0 0 foo 3 0
                              1 1 foo 2 0
                              2 2 foo 5 0
                              3 3 bar 3 1
                              4 4 bar 4 1
                              5 5 baz 5 2


                              Explanation of solution: Let's break above solution into parts for understanding purposes.



                              1st step: Compare df's A column by shifting its value down to itself as follows.



                              i.ne(df.A.shift())


                              Output we will get is:



                              0     True
                              1 False
                              2 False
                              3 True
                              4 False
                              5 True


                              2nd step: Use of cumsum() function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum() function and its value will be increased.



                              i.ne(df.A.shift()).cumsum()-1
                              0 0
                              1 0
                              2 0
                              3 1
                              4 1
                              5 2
                              Name: A, dtype: int32


                              3rd step: Save command's value into df['C'] which will create a new column named C in df.







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 23 mins ago

























                              answered 42 mins ago









                              RavinderSingh13

                              25k41437




                              25k41437








                              • 1




                                Nice method ve++ for you
                                – W-B
                                36 mins ago










                              • @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                                – RavinderSingh13
                                34 mins ago
















                              • 1




                                Nice method ve++ for you
                                – W-B
                                36 mins ago










                              • @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                                – RavinderSingh13
                                34 mins ago










                              1




                              1




                              Nice method ve++ for you
                              – W-B
                              36 mins ago




                              Nice method ve++ for you
                              – W-B
                              36 mins ago












                              @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                              – RavinderSingh13
                              34 mins ago






                              @W-B, thank you for encouragement sir, ++ve for your unique style already :)
                              – RavinderSingh13
                              34 mins ago




















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