Question

I m试图过滤从C.csv档案中检索的一个日期,但我似乎没有尝试合并。日期为“2011-10-01 19:25:01”或“年月:分钟:无”。

I want just the year, month and date but I get can t seem to get ride of the time from the string:

date = bug[2] # Column in which the date is located  
date = date.replace( " ,  ) #getting rid of the quotations  
mdate = date.replace( : ,  )  
re.split( $[d]+ ,mdate) # trying to get rid of the trailing set of number (from the time)

提前获得咨询意见。

Answer 1

如果你的来源是扼杀,你或许会更好地利用strptime 。

import datetime
string = "2011-10-01 19:25:01"
dt = datetime.datetime.strptime(string, "%Y-%m-%d %H:%M:%S")

之后使用

dt.year
dt.month
dt.day

查阅你想要的数据。

Answer 2

Use datetime to parse your input as a datetime object, then output it in whatever format you like: http://docs.python.org/library/datetime.html

Answer 3

我认为,你重新混淆了一条线和美元线的开端。 Try ^[d-]+。

Answer 4

如果格式总是<代码>,YYYYY-MM-DD HH:mm:ss”,则尝试:

date = date[1:11]

In a prompt:

>>> date =  "2012-01-12 15:13:20" 
>>> date[1:11]
 2012-01-12 
>>>

无需reg

Answer 5

>>> date =  "2011-10-01 19:25:01" 
>>> date.strip( " ).split()[0]
 2011-10-01

Answer 6

贵组织守则的一个问题是,在你最后的定期表述中,$与扼杀的结束相吻合,这样经常表达永远不会与任何内容相匹配。你可以更简单地通过空间分化来做到这一点,只取得初步结果。删除引号后,线

date.split()

<[>2011-10-01”、“19:25:01”>>>,因此该清单的第一个要素是您的需要。

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