问题

我想编写一个函数，如果日期是星期二、星期三或星期四，则将其转换为本周的星期一；如果日期是星期五、星期六或星期日，则将其转换为下周的星期一。如果日期是星期一，则保持不变。

示例输入数据框如下：

理想的输出如下：

根据我指定的规则，所有日期标头都转换为星期一。

非常感谢您的帮助！

答案1

得分: 1

import pandas as pd
data1 = {'Model#': {0: 'A', 1: 'B', 2: 'C'},
 'Order Category': {0: 'Open', 1: 'Close', 2: 'Open'},
 '2022/4/18': {0: 0, 1: 1, 2: 0},
 '2022/5/10': {0: 0, 1: 1, 2: 0},
 '2022/5/18': {0: 0, 1: 0, 2: 1},
 '2022/5/26': {0: 0, 1: 0, 2: 1},
 '2022/6/24': {0: 1, 1: 1, 2: 0},
 '2022/7/16': {0: 1, 1: 2, 2: 1},
 '2022/7/24': {0: 2, 1: 2, 2: 2}}
df1 = pd.DataFrame(data1)

idx = pd.to_datetime(df1.columns[2:], errors='coerce').to_period('W').to_timestamp()

idx

DatetimeIndex(['2022-04-18', '2022-05-09', '2022-05-16', '2022-05-23',
               '2022-06-20', '2022-07-11', '2022-07-18'],
              dtype='datetime64[ns]', freq=None)

(df1
 .set_index(['Model#', 'Order Category'])
 .set_axis(idx.astype('str'), axis=1)
 .reset_index())

Output:

   Model# Order Category  2022-04-18  2022-05-09  2022-05-16  2022-05-23  2022-06-20  2022-07-11  2022-07-18
0      A           Open           0           0           0           0           1           1           2
1      B          Close           1           1           0           0           1           2           2
2      C           Open           0           0           1           1           0           1           2

import pandas as pd
data1 = {'Model#': {0: 'A', 1: 'B', 2: 'C'},
 'Order Category': {0: 'Open', 1: 'Close', 2: 'Open'},
 '2022/4/18': {0: 0, 1: 1, 2: 0},
 '2022/5/10': {0: 0, 1: 1, 2: 0},
 '2022/5/18': {0: 0, 1: 0, 2: 1},
 '2022/5/26': {0: 0, 1: 0, 2: 1},
 '2022/6/24': {0: 1, 1: 1, 2: 0},
 '2022/7/16': {0: 1, 1: 2, 2: 1},
 '2022/7/24': {0: 2, 1: 2, 2: 2}}
df1 = pd.DataFrame(data1)

idx = pd.to_datetime(df1.columns[2:], errors='coerce').to_period('W').to_timestamp()

DatetimeIndex(['2022-04-18', '2022-05-09', '2022-05-16', '2022-05-23',
               '2022-06-20', '2022-07-11', '2022-07-18'],
              dtype='datetime64[ns]', freq=None)

(df1
 .set_index(['Model#', 'Order Category'])
 .set_axis(idx.astype('str'), axis=1)
 .reset_index())

Model#	Order Category	2022-04-18	2022-05-09	2022-05-16	2022-05-23	2022-06-20	2022-07-11	2022-07-18
0	A	Open	        0	        0	        0	        0	        1	        1	        2
1	B	Close	        1	        1	        0	        0	        1	        2	        2
2	C	Open	        0	        0	        1	        1	        0	        1	        2

答案2

得分: 1

你可以使用以下代码：

import numpy as np

# 将列转换为日期并提取星期几
dates = pd.to_datetime(df.columns[2:])
dow = dates.day_of_week

# 创建时间差，星期一至星期四为负数，星期五至星期日为正数
offsets = pd.to_timedelta(np.where(dow < 4, -dow, 7-dow), unit='D')

# 调整日期，然后重新索引列
dates = (dates + offsets).strftime('%Y/%-m/%-d')
df.columns = df.columns[:2].append(dates)

输出：

中间结果：

<details>
<summary>英文:</summary>

You can use:

import numpy as np

Convert as datetime and extract day of week

dates = pd.to_datetime(df.columns[2:])
dow = dates.day_of_week

Create timedelta Mon, Thu, Wed, Thu are negative, Fri, Sat, Sun are positive

offsets = pd.to_timedelta(np.where(dow < 4, -dow, 7-dow), unit='D')

Adjust the dates then reindex your columns

dates = (dates + offsets).strftime('%Y/%-m/%-d')
df.columns = df.columns[:2].append(dates)

Output:

| Model#   | Order Category   |   2022/4/18 |   2022/5/9 |   2022/5/16 |   2022/5/23 |   2022/6/27 |   2022/7/18 |   2022/7/25 |
|:---------|:-----------------|------------:|-----------:|------------:|------------:|------------:|------------:|------------:|
| A        | Open             |           0 |          0 |           0 |           0 |           1 |           1 |           2 |
| B        | Close            |           1 |          1 |           0 |           0 |           1 |           2 |           2 |
| C        | Open             |           0 |          0 |           1 |           1 |           0 |           1 |           2 |

Intermediates:

| Before     | Offset   | After      |
|:-----------|:---------|:-----------|
| 2022-04-18 | 0 days   | 2022-04-18 |
| 2022-05-10 | -1 days  | 2022-05-09 |
| 2022-05-18 | -2 days  | 2022-05-16 |
| 2022-05-26 | -3 days  | 2022-05-23 |
| 2022-06-24 | 3 days   | 2022-06-27 |
| 2022-07-16 | 2 days   | 2022-07-18 |
| 2022-07-24 | 1 days   | 2022-07-25 |


</details>