How to add sequential counter column on groups using Pandas groupby

I feel like there is a better way than this:

import pandas as pd
df = pd.DataFrame(
    columns="   index    c1    c2    v1 ".split(),
    data= [
            [       0,  "A",  "X",    3, ],
            [       1,  "A",  "X",    5, ],
            [       2,  "A",  "Y",    7, ],
            [       3,  "A",  "Y",    1, ],
            [       4,  "B",  "X",    3, ],
            [       5,  "B",  "X",    1, ],
            [       6,  "B",  "X",    3, ],
            [       7,  "B",  "Y",    1, ],
            [       8,  "C",  "X",    7, ],
            [       9,  "C",  "Y",    4, ],
            [      10,  "C",  "Y",    1, ],
            [      11,  "C",  "Y",    6, ],]).set_index("index", drop=True)
def callback(x):
    x['seq'] = range(1, x.shape[0] + 1)
    return x
df = df.groupby(['c1', 'c2']).apply(callback)
print df

To achieve this:

   c1 c2  v1  seq
0   A  X   3    1
1   A  X   5    2
2   A  Y   7    1
3   A  Y   1    2
4   B  X   3    1
5   B  X   1    2
6   B  X   3    3
7   B  Y   1    1
8   C  X   7    1
9   C  Y   4    1
10  C  Y   1    2
11  C  Y   6    3

Is there a way to do it that avoids the callback?

Answers:

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Method 1

use cumcount(), see docs here

In [4]: df.groupby(['c1', 'c2']).cumcount()
Out[4]: 
0     0
1     1
2     0
3     1
4     0
5     1
6     2
7     0
8     0
9     0
10    1
11    2
dtype: int64

If you want orderings starting at 1

In [5]: df.groupby(['c1', 'c2']).cumcount()+1
Out[5]: 
0     1
1     2
2     1
3     2
4     1
5     2
6     3
7     1
8     1
9     1
10    2
11    3
dtype: int64

Method 2

This might be useful

df = df.sort_values(['userID', 'date'])
grp = df.groupby('userID')['ItemID'].aggregate(lambda x: '->'.join(tuple(x))).reset_index()
print(grp)

it will create a sequence like this

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0