如何实现原dataframe每一行的组键？ （'by' 是 pandas 石斑鱼）

Question

I would like to materialize for each row of a dataframe the corresponding group key it would get if I was using a groupby operation with a pandas Grouper .

import pandas as pd

# Test data
ts = [pd.Timestamp('2022/03/01 09:00'),
      pd.Timestamp('2022/03/01 10:00'),
      pd.Timestamp('2022/03/01 10:30'),
      pd.Timestamp('2022/03/01 15:00')]
df = pd.DataFrame({'a':range(len(ts)), 'ts': ts})

grouper = pd.Grouper(key='ts', freq='2H', sort=False, origin='start_day')

Is there any way to get for each row the corresponding groupkey? The result I am looking for could be either a list, or a pandas Series or Index, or numpy array, the same length as the initial dataframe, and would then contain following values.

result = pd.Series([pd.Timestamp('2022-03-01 08:00:00'),
                    pd.Timestamp('2022-03-01 10:00:00'),
                    pd.Timestamp('2022-03-01 10:00:00'),
                    pd.Timestamp('2022-03-01 14:00:00')])

Thanks for your help! Bests

Answer 1

Similar idea to @Andrej, just creates a table with a new column

pd.concat(g.assign(grouper_val = i) for i,g in df.groupby(grouper))

Answer 2

Not directly using the groupby but you can use:

df['ts'].dt.floor('2H')

With the groupby :

df.groupby(grouper)['ts'].transform(lambda g: g.name)

Output:

0   2022-03-01 08:00:00
1   2022-03-01 10:00:00
2   2022-03-01 10:00:00
3   2022-03-01 14:00:00
Name: ts, dtype: datetime64[ns]

Answer 3

Given:

   a                  ts
0  0 2022-03-01 09:00:00
1  1 2022-03-01 10:00:00
2  2 2022-03-01 10:30:00
3  3 2022-03-01 15:00:00

Doing:

pd.Series(df.resample('2H', origin='start_day', on='ts').groups)

Output:

2022-03-01 08:00:00    1
2022-03-01 10:00:00    3
2022-03-01 12:00:00    3
2022-03-01 14:00:00    4
dtype: int64