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











up vote
7
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
    7
    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
      7
      down vote

      favorite
      2









      up vote
      7
      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











      share|improve this question




      share|improve this question










      asked 2 hours ago









      Lagerbaer

      2,6231124




      2,6231124
























          3 Answers
          3






          active

          oldest

          votes

















          up vote
          7
          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













            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




            Method 5 np.unique



            x,y=np.unique(df.A.values,return_inverse=True)
            y
            array([2, 2, 2, 0, 0, 1], dtype=int64)





            share|improve this answer























            • Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
              – Vaishali
              38 mins ago












            • @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
              – W-B
              25 mins ago


















            up vote
            4
            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
              1 hour ago










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













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            7
            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
              7
              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
                7
                down vote










                up vote
                7
                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 1 hour ago

























                answered 1 hour ago









                sacul

                29.7k41640




                29.7k41640
























                    up vote
                    6
                    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




                    Method 5 np.unique



                    x,y=np.unique(df.A.values,return_inverse=True)
                    y
                    array([2, 2, 2, 0, 0, 1], dtype=int64)





                    share|improve this answer























                    • Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                      – Vaishali
                      38 mins ago












                    • @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                      – W-B
                      25 mins ago















                    up vote
                    6
                    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




                    Method 5 np.unique



                    x,y=np.unique(df.A.values,return_inverse=True)
                    y
                    array([2, 2, 2, 0, 0, 1], dtype=int64)





                    share|improve this answer























                    • Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                      – Vaishali
                      38 mins ago












                    • @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                      – W-B
                      25 mins ago













                    up vote
                    6
                    down vote










                    up vote
                    6
                    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




                    Method 5 np.unique



                    x,y=np.unique(df.A.values,return_inverse=True)
                    y
                    array([2, 2, 2, 0, 0, 1], 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




                    Method 5 np.unique



                    x,y=np.unique(df.A.values,return_inverse=True)
                    y
                    array([2, 2, 2, 0, 0, 1], dtype=int64)






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 28 mins ago

























                    answered 1 hour ago









                    W-B

                    97.8k73162




                    97.8k73162












                    • Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                      – Vaishali
                      38 mins ago












                    • @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                      – W-B
                      25 mins ago


















                    • Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                      – Vaishali
                      38 mins ago












                    • @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                      – W-B
                      25 mins ago
















                    Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                    – Vaishali
                    38 mins ago






                    Good solutions, they should be really fast as well. May be add time comparison as OP is looking for the most efficient solution
                    – Vaishali
                    38 mins ago














                    @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                    – W-B
                    25 mins ago




                    @Vaishali sorry it is hard for me to get the timing , would you mind add that for me , thanks a lot
                    – W-B
                    25 mins ago










                    up vote
                    4
                    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
                      1 hour ago










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

















                    up vote
                    4
                    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
                      1 hour ago










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















                    up vote
                    4
                    down vote










                    up vote
                    4
                    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 1 hour ago

























                    answered 1 hour ago









                    RavinderSingh13

                    25k41437




                    25k41437








                    • 1




                      Nice method ve++ for you
                      – W-B
                      1 hour ago










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
















                    • 1




                      Nice method ve++ for you
                      – W-B
                      1 hour ago










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










                    1




                    1




                    Nice method ve++ for you
                    – W-B
                    1 hour ago




                    Nice method ve++ for you
                    – W-B
                    1 hour ago












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






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




















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