New column in pandas - adding series to dataframe by applying a list groupby

Question

Give the following df

  Id other  concat
0  A     z       1
1  A     y       2
2  B     x       3
3  B     w       4
4  B     v       5
5  B     u       6

I want the result with new column with grouped values as list

  Id other  concat           new
0  A     z       1        [1, 2]
1  A     y       2        [1, 2]
2  B     x       3  [3, 4, 5, 6]
3  B     w       4  [3, 4, 5, 6]
4  B     v       5  [3, 4, 5, 6]
5  B     u       6  [3, 4, 5, 6]

This is similar to these questions:

grouping rows in list in pandas groupby

Replicating GROUP_CONCAT for pandas.DataFrame

However, it is apply the grouping you get from df.groupby('Id')['concat'].apply(list), which is a Series of smaller size than the dataframe, to the original dataframe.

I have tried the code below, but it does not apply this to the dataframe:

import pandas as pd
df = pd.DataFrame( {'Id':['A','A','B','B','B','C'], 'other':['z','y','x','w','v','u'], 'concat':[1,2,5,5,4,6]})
df.groupby('Id')['concat'].apply(list)

I know that transform can be used to apply groupings to dataframes, but it does not work in this case.

>>> df['new_col'] = df.groupby('Id')['concat'].transform(list)
>>> df
  Id  concat other  new_col
0  A       1     z        1
1  A       2     y        2
2  B       5     x        5
3  B       5     w        5
4  B       4     v        4
5  C       6     u        6
>>> df['new_col'] = df.groupby('Id')['concat'].apply(list)
>>> df
  Id  concat other new_col
0  A       1     z     NaN
1  A       2     y     NaN
2  B       5     x     NaN
3  B       5     w     NaN
4  B       4     v     NaN
5  C       6     u     NaN

Answer 1

groupby with join

df.join(df.groupby('Id').concat.apply(list).to_frame('new'), on='Id')

Answer 2

Less elegant (and slower..) solution, but let it be here just as an alternative.

def func(gr):
    gr['new'] = [list(gr.concat)] * len(gr.index)
    return gr
df.groupby('Id').apply(func)

%timeit df.groupby('Id').apply(func)
100 loops, best of 3: 4.18 ms per loop

%timeit df.join(df.groupby('Id').concat.apply(list).to_frame('new'), on='Id')
1000 loops, best of 3: 1.69 ms per loop