这对你的情况应该有效。你需要检测一行相对于下一行的变化，然后在那一行你应该分配日期。

data = {'DATE': {0: '1/1/2023',
  1: '2/1/2023',
  2: '3/1/2023',
  3: '4/1/2023',
  4: '5/1/2023',
  5: '6/1/2023',
  6: '7/1/2023',
  7: '8/1/2023',
  8: '9/1/2023',
  9: '10/1/2023'},
 'column_a': {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 1, 8: 1, 9: 0},
 'column_b': {0: 0, 1: 0, 2: 0, 3: 1, 4: 1, 5: 1, 6: 1, 7: 0, 8: 0, 9: 0},
 'column_c': {0: 1, 1: 1, 2: 1, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0},
 'column_d': {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 1}}

df = pd.DataFrame.from_dict(data)
row_change = df.filter(like='column').diff(-1).eq(-1).any(axis=1)
df.loc[row_change, 'return_date'] = df.loc[row_change, 'DATE']

您可以使用groupby，结合使用.cumcount()来获得所需的索引，然后使用该索引从DATE列拖动日期。

代码：

import pandas as pd

df = pd.DataFrame({
    'DATE': ['1/1/2023', '2/1/2023', '3/1/2023', '4/1/2023', '5/1/2023', '6/1/2023', '7/1/2023', '8/1/2023', '9/1/2023', '10/1/2023'],
    'column_a': [0, 0, 0, 0, 0, 0, 0, 1, 1, 0],
    'column_b': [0, 0, 0, 1, 1, 1, 1, 0, 0, 0],
    'column_c': [1, 1, 1, 0, 0, 0, 0, 0, 0, 0],
    'column_d': [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
})

idx = df[df.groupby(['column_a', 'column_b', 'column_c', 'column_d']).cumcount() == 0].index - 1
df.loc[idx[1:], 'return_date'] = df.loc[idx[1:], 'DATE']

print(df)

输出

DATE  column_a  column_b  column_c  column_d return_date
0   1/1/2023         0         0         1         0         NaN
1   2/1/2023         0         0         1         0         NaN
2   3/1/2023         0         0         1         0    3/1/2023
3   4/1/2023         0         1         0         0         NaN
4   5/1/2023         0         1         0         0         NaN
5   6/1/2023         0         1         0         0         NaN
6   7/1/2023         0         1         0         0    7/1/2023
7   8/1/2023         1         0         0         0         NaN
8   9/1/2023         1         0         0         0    9/1/2023
9  10/1/2023         0         0         0         1         NaN