Pandas: assign an index to each group identified b

Question

When using groupby(), how can I create a DataFrame with a new column containing an index of the group number, similar to dplyr::group_indices in R. For example, if I have

>>> df=pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})
>>> df
   a  b
0  1  1
1  1  1
2  1  2
3  2  1
4  2  1
5  2  2

How can I get a DataFrame like

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

(the order of the idx indexes doesn't matter)

Answer 1

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

Answer 2

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.

Answer 3

A way that I believe is faster than the current accepted answer by about an order of magnitude (timing results below):

def create_index_usingduplicated(df, grouping_cols=['a', 'b']):
    df.sort_values(grouping_cols, inplace=True)
    # You could do the following three lines in one, I just thought 
    # this would be clearer as an explanation of what's going on:
    duplicated = df.duplicated(subset=grouping_cols, keep='first')
    new_group = ~duplicated
    return new_group.cumsum()

Timing results:

a = np.random.randint(0, 1000, size=int(1e5))
b = np.random.randint(0, 1000, size=int(1e5))
df = pd.DataFrame({'a': a, 'b': b})

In [6]: %timeit df['idx'] = pd.Categorical(df['a'].astype(str) + df['b'].astype(str)).codes
1 loop, best of 3: 375 ms per loop

In [7]: %timeit df['idx'] = create_index_usingduplicated(df, grouping_cols=['a', 'b'])
100 loops, best of 3: 17.7 ms per loop

Answer 4

Definetely not the most straightforward solution, but here is what I would do (comments in the code):

df=pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})

#create a dummy grouper id by just joining desired rows
df["idx"] = df[["a","b"]].astype(str).apply(lambda x: "".join(x),axis=1)

print df

That would generate an unique idx for each combination of a and b.

   a  b idx
0  1  1  11
1  1  1  11
2  1  2  12
3  2  1  21
4  2  1  21
5  2  2  22

But this is still a rather silly index (think about some more complex values in columns a and b. So let's clear the index:

# create a dictionary of dummy group_ids and their index-wise representation
dict_idx = dict(enumerate(set(df["idx"])))

# switch keys and values, so you can use dict in .replace method
dict_idx = {y:x for x,y in dict_idx.iteritems()}

#replace values with the generated dict
df["idx"].replace(dict_idx,inplace=True)

print df

That would produce the desired output:

   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

Answer 5

I'm not sure this is such a trivial problem. Here is a somewhat convoluted solution that first sorts the grouping columns and then checks whether each row is different than the previous row and if so accumulates by 1. Check further below for an answer with string data.

df.sort_values(['a', 'b']).diff().fillna(0).ne(0).any(1).cumsum().add(1)

Output

0    1
1    1
2    2
3    3
4    3
5    4
dtype: int64

So breaking this up into steps, lets see the output of df.sort_values(['a', 'b']).diff().fillna(0) which checks if each row is different than the previous row. Any non-zero entry indicates a new group.

     a    b
0  0.0  0.0
1  0.0  0.0
2  0.0  1.0
3  1.0 -1.0
4  0.0  0.0
5  0.0  1.0

A new group only need to have a single column different so this is what .ne(0).any(1) checks - not equal to 0 for any of the columns. And then just a cumulative sum to keep track of the groups.

Answer for columns as strings

#create fake data and sort it
df=pd.DataFrame({'a':list('aabbaccdc'),'b':list('aabaacddd')})
df1 = df.sort_values(['a', 'b'])

output of df1

   a  b
0  a  a
1  a  a
4  a  a
3  b  a
2  b  b
5  c  c
6  c  d
8  c  d
7  d  d

Take similar approach by checking if group has changed

df1.ne(df1.shift().bfill()).any(1).cumsum().add(1)

0    1
1    1
4    1
3    2
2    3
5    4
6    5
8    5
7    6

Answer 6

Here is the solution using ngroup from a comment above by Constantino, for those still looking for this function (the equivalent of dplyr::group_indices in R, if you were trying to google with those keywords like me). This is also about 25% faster than the solution given by maxliving according to my own timing.

>>> 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

>>> %timeit df['idx'] = create_index_usingduplicated(df, grouping_cols=['a', 'b'])
1.83 ms ± 67.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit df['idx'] = df.groupby(['a', 'b']).ngroup()
1.38 ms ± 30 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)