我正在尝试将带有日期的凌乱笔记本转换为熊猫中的排序日期系列。
0 03/25/93 Total time of visit (in minutes):\n
1 6/18/85 Primary Care Doctor:\n
2 sshe plans to move as of 7/8/71 In-Home Servic...
3 7 on 9/27/75 Audit C Score Current:\n
4 2/6/96 sleep studyPain Treatment Pain Level (N...
5 .Per 7/06/79 Movement D/O note:\n
6 4, 5/18/78 Patient's thoughts about current su...
7 10/24/89 CPT Code: 90801 - Psychiatric Diagnos...
8 3/7/86 SOS-10 Total Score:\n
9 (4/10/71)Score-1Audit C Score Current:\n
10 (5/11/85) Crt-1.96, BUN-26; AST/ALT-16/22; WBC...
11 4/09/75 SOS-10 Total Score:\n
12 8/01/98 Communication with referring physician...
13 1/26/72 Communication with referring physician...
14 5/24/1990 CPT Code: 90792: With medical servic...
15 1/25/2011 CPT Code: 90792: With medical servic...
我有多种日期格式，例如04/20/2009;零九年四月二十零日; 09年4月20日; 09年4月3日。我想把所有这些转换成mm / dd / yyyy到一个新列。
到目前为止，我已经完成了
df2['date']= df2['text'].str.extractall(r'(\d{1,2}[/-]\d{1,2}[/-]\d{2,})')
此外，我不知道如何在不干扰上述代码的情况下提取仅具有mm / yy或yyyy格式日期的所有行。请记住，如果缺少日期或月份，我会将第1和第1个视为默认值。
答案 0 :(得分：1)
您可以将var res = Regex.Replace(s, @"[\x0A\x0D\x09]", " ");
与正则表达式一起使用，然后应用pd.Series.str.extract：
pd.to_datetime df['Date'] = df.Text.str.extract(r'(?P<Date>\d+(?:\/\d+){2})', expand=False)\
.apply(pd.to_datetime)
df
Text Date
0
0 03/25/93 Total time of visit (in minutes):\n 1993-03-25
1 6/18/85 Primary Care Doctor:\n 1985-06-18
2 sshe plans to move as of 7/8/71 In-Home Servic... 1971-07-08
3 7 on 9/27/75 Audit C Score Current:\n 1975-09-27
4 2/6/96 sleep studyPain Treatment Pain Level (N... 1996-02-06
5 .Per 7/06/79 Movement D/O note:\n 1979-07-06
6 4, 5/18/78 Patient's thoughts about current su... 1978-05-18
7 10/24/89 CPT Code: 90801 - Psychiatric Diagnos... 1989-10-24
8 3/7/86 SOS-10 Total Score:\n 1986-03-07
9 (4/10/71)Score-1Audit C Score Current:\n 1971-04-10
10 (5/11/85) Crt-1.96, BUN-26; AST/ALT-16/22; WBC... 1985-05-11
11 4/09/75 SOS-10 Total Score:\n 1975-04-09
12 8/01/98 Communication with referring physician... 1998-08-01
13 1/26/72 Communication with referring physician... 1972-01-26
14 5/24/1990 CPT Code: 90792: With medical servic... 1990-05-24
15 1/25/2011 CPT Code: 90792: With medical servic... 2011-01-25
返回一系列如下所示的字符串：
str.extract正则表达式详细信息
array(['03/25/93', '6/18/85', '7/8/71', '9/27/75', '2/6/96', '7/06/79',
'5/18/78', '10/24/89', '3/7/86', '4/10/71', '5/11/85', '4/09/75',
'8/01/98', '1/26/72', '5/24/1990', '1/25/2011'], dtype=object)
(?P<Date>\d+(?:\/\d+){2})
- 命名捕获组(?P<Date>....)一个或多个数字\d+ - 非捕获组重复两次，其中
(?:\/\d+){2} - 转发正斜杠\/ - 转发器（两次）缺少天数的正则表达式
要处理可选的{2}，需要稍加修改的正则表达式：
days详细
(?P<Date>(?:\d+\/)?\d+/\d+)
- 命名捕获组(?P<Date>....) - 嵌套组（非捕获）(?:\d+\/)?是可选的。\d+\/一个或多个数字\d+逃脱正斜线其余的都是一样的。用这个正则表达式代替当前的正则表达式。 \/将处理遗失的日子。