Pandas: assign an index to each group identified by groupby

Here's a concise way using drop_duplicates and merge to get a unique identifier.

group_vars = ['a','b']
df.merge( df.drop_duplicates( group_vars ).reset_index(), on=group_vars )

   a  b  index
0  1  1      0
1  1  1      0
2  1  2      2
3  2  1      3
4  2  1      3
5  2  2      5

The identifier in this case goes 0,2,3,5 (just a residual of original index) but this could be easily changed to 0,1,2,3 with an additional reset_index(drop=True).

Update: Newer versions of pandas (0.20.2) offer a simpler way to do this with the ngroup method as noted in a comment to the question above by @Constantino and a subsequent answer by @CalumYou. I'll leave this here as an alternate approach but ngroup seems like the better way to do this in most cases.

Here is the solution using ngroup (available as of pandas 0.20.2) from a comment above by Constantino.

import pandas as pd
df = pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})
df['idx'] = df.groupby(['a', 'b']).ngroup()
df

   a  b  idx
0  1  1    0
1  1  1    0
2  1  2    1
3  2  1    2
4  2  1    2
5  2  2    3

A simple way to do that would be to concatenate your grouping columns (so that each combination of their values represents a uniquely distinct element), then convert it to a pandas Categorical and keep only its labels:

df['idx'] = pd.Categorical(df['a'].astype(str) + '_' + df['b'].astype(str)).codes
df

    a   b   idx
0   1   1   0
1   1   1   0
2   1   2   1
3   2   1   2
4   2   1   2
5   2   2   3

Edit: changed labels properties to codes as the former seem to be deprecated

Edit2: Added a separator as suggested by Authman Apatira