Question

_{这个问题与this类似，我最初用这个解决方案回答了问题，但事实证明我误解了这个问题。但是，我觉得我的answer对于略有不同的用例很有用，所以我在这里发布。}

给定一个文本文件：

04/20/2009; 04/20/09; 4/20/09; 4/3/09
Mar-20-2009; Mar 20, 2009; March 20, 2009; Mar. 20, 2009; Mar 20 2009;
20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009
Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009
Feb 2009; Sep 2009; Oct 2010
6/2008; 12/2009
2009; 2010

包含不同格式的已提取日期...任务是将它们读入数据框然后对它们进行排序，然后以MM / DD / YYYY格式显示输出。

预期产出：

0     06/01/2008
1     01/01/2009
2     02/01/2009
3     03/20/2009
4     03/20/2009
5     03/20/2009
6     03/20/2009
7     03/20/2009
8     03/20/2009
9     03/20/2009
10    03/20/2009
11    03/20/2009
12    03/20/2009
13    03/21/2009
14    03/22/2009
15    04/03/2009
16    04/20/2009
17    04/20/2009
18    04/20/2009
19    09/01/2009
20    12/01/2009
21    01/01/2010
22    10/01/2010

如何在熊猫中完成？

注意：如果缺少这一天，请考虑第1天，如果缺少月份，请考虑1月。

Answer 1

可重复设置（轻松实现MCVE）：

import pandas as pd
import io

text = '''04/20/2009; 04/20/09; 4/20/09; 4/3/09
Mar-20-2009; Mar 20, 2009; March 20, 2009; Mar. 20, 2009; Mar 20 2009;
20 Mar 2009; 20 March 2009; 20 Mar. 2009; 20 March, 2009
Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009
Feb 2009; Sep 2009; Oct 2010
6/2008; 12/2009
2009; 2010'''

buf = io.stringIO(text)

df = pd.read_csv(buf, engine='python', delimiter=';\s+', header=None).reset_index()

df

            index               0               1               2  \
0      04/20/2009        04/20/09         4/20/09          4/3/09   
1     Mar-20-2009    Mar 20, 2009  March 20, 2009   Mar. 20, 2009   
2     20 Mar 2009   20 March 2009    20 Mar. 2009  20 March, 2009   
3  Mar 20th, 2009  Mar 21st, 2009  Mar 22nd, 2009            None   
4        Feb 2009        Sep 2009        Oct 2010            None   
5          6/2008         12/2009            None            None   
6            2009            2010            None            None   

              3  
0          None  
1  Mar 20 2009;  
2          None  
3          None  
4          None  
5          None  
6          None

将buf替换为文本文件的名称。

您可以使用df.apply和df.stack，然后使用pd.Series.sort_values。

out = df.stack().apply(pd.to_datetime)\
        .reset_index(drop=1)\
        .sort_values().dt.strftime('%m/%d/%Y')\
        .reset_index(drop=1)
print(out)

0     06/01/2008
1     01/01/2009
2     02/01/2009
3     03/20/2009
4     03/20/2009
5     03/20/2009
6     03/20/2009
7     03/20/2009
8     03/20/2009
9     03/20/2009
10    03/20/2009
11    03/20/2009
12    03/20/2009
13    03/21/2009
14    03/22/2009
15    04/03/2009
16    04/20/2009
17    04/20/2009
18    04/20/2009
19    09/01/2009
20    12/01/2009
21    01/01/2010
22    10/01/2010

Answer 2

Simplier应该只省略apply和reset_index一次：

在我看来，drop=1的可读性更差，如drop=True。

out = pd.to_datetime(df.stack()).sort_values().dt.strftime('%m/%d/%Y').reset_index(drop=True)
print(out)
0     06/01/2008
1     01/01/2009
2     02/01/2009
3     03/20/2009
4     03/20/2009
5     03/20/2009
6     03/20/2009
7     03/20/2009
8     03/20/2009
9     03/20/2009
10    03/20/2009
11    03/20/2009
12    03/20/2009
13    03/21/2009
14    03/22/2009
15    04/03/2009
16    04/20/2009
17    04/20/2009
18    04/20/2009
19    09/01/2009
20    12/01/2009
21    01/01/2010
22    10/01/2010
dtype: object

从文件中读取不同格式的日期并对其进行排序

2 个答案: