python - 当 'ID' 为 1 时，如何创建一个新列插入分组列 'interaction'(及时)的单元格值

Question

我有三个相关列:时间、ID 和交互。我如何创建一个新列，其 id 值在给定时间窗口中的“交互”列中为“1”？

应该看起来像这样:

time   id   vec_len  quadrant   interaction    Paired with

1    3271    0.9    7   0 
1    3229    0.1    0   0
1    4228    0.5    0   0
1    2778   -0.3    5   0
2    4228    0.2    0   0
2    3271    0.1    6   0
2    3229    -0.7   5   1    [2778, 4228]
2    3229    -0.3   2   0
2    4228    -0.8   5   1    [2778, 3229]
2    2778   -0.6    5   1    [4228, 3229]
3    4228    0.2    0   0
3    3271    0.1    6   0
3    4228    -0.7   5   1    [3271]
3    3229    -0.3   2   0
3    3271    -0.8   5   1    [4228]

谢谢你的帮助!!

Answer 1

import numpy as np

# initialize dict for all time blocks
dict_time_ids = dict.fromkeys(df.time.unique(), set())

# populate dictionary with ids for each time block where interaction == 1
dict_time_ids.update(df.query('interaction == 1').groupby('time').id.apply(set).to_dict())

# make new column with set of corresponding ids where interaction == 1
df['paired'] = np.where(df.interaction == 1, df.time.apply(lambda x: dict_time_ids[x]), set())

# remove the id from the set and convert to list
df.paired = df.apply(lambda x: list(x.paired - {x.id}), axis=1)

# Out:
    time    id  interaction        paired
0      1  3271            0            []
1      1  3229            0            []
2      1  4228            0            []
3      1  2778            0            []
4      2  4228            0            []
5      2  3271            0            []
6      2  3229            1  [2778, 4228]
7      2  3229            0            []
8      2  4228            1  [2778, 3229]
9      2  2778            1  [4228, 3229]
10     3  4228            0            []
11     3  3271            0            []
12     3  4228            1        [3271]
13     3  3229            0            []
14     3  3271            1        [4228]