如何将datetime对象格式化为带有毫秒的字符串?


当前回答

使用[:-3]删除最后3个字符,因为%f代表微秒:

>>> from datetime import datetime
>>> datetime.now().strftime('%Y/%m/%d %H:%M:%S.%f')[:-3]
'2013/12/04 16:50:03.141'

其他回答

@Cabbi提出了一个问题,在一些系统上(带有Python 2.7的Windows),微秒格式%f可能会错误地给出“0”,所以简单地修改最后三个字符是不可移植的。这样的系统不遵循文档中指定的行为:

Directive Meaning Example
%f Microsecond as a decimal number, zero-padded to 6 digits. 000000, 000001, …, 999999

下面的代码小心地以毫秒为单位格式化时间戳:

>>> from datetime import datetime
>>> (dt, micro) = datetime.utcnow().strftime('%Y-%m-%d %H:%M:%S.%f').split('.')
>>> "%s.%03d" % (dt, int(micro) / 1000)
'2016-02-26 04:37:53.133'

为了得到OP想要的准确输出,我们必须去掉标点符号:

>>> from datetime import datetime
>>> (dt, micro) = datetime.utcnow().strftime('%Y%m%d%H%M%S.%f').split('.')
>>> "%s%03d" % (dt, int(micro) / 1000)
'20160226043839901'
python -c "from datetime import datetime; print str(datetime.now())[:-3]"
2017-02-09 10:06:37.006

使用Python 3.6+,你可以设置isoformat的timespec:

>>> from datetime import datetime
>>> datetime.utcnow().isoformat(sep=' ', timespec='milliseconds')
'2019-05-10 09:08:53.155'

使用strftime:

>>> from datetime import datetime
>>> datetime.utcnow().strftime('%Y%m%d%H%M%S%f')
'20220402055654344968'

我假设您的意思是您正在寻找比datetime.datetime.strftime()更快的东西,并且实际上是从utc时间戳中剥离非alpha字符。

你的方法稍微快一点,我认为你可以通过切片字符串来加快速度:

>>> import timeit
>>> t=timeit.Timer('datetime.utcnow().strftime("%Y%m%d%H%M%S%f")','''
... from datetime import datetime''')
>>> t.timeit(number=10000000)
116.15451288223267

>>> def replaceutc(s):
...     return s\
...         .replace('-','') \
...         .replace(':','') \
...         .replace('.','') \
...         .replace(' ','') \
...         .strip()
... 
>>> t=timeit.Timer('replaceutc(str(datetime.datetime.utcnow()))','''
... from __main__ import replaceutc
... import datetime''')
>>> t.timeit(number=10000000)
77.96774983406067

>>> def sliceutc(s):
...     return s[:4] + s[5:7] + s[8:10] + s[11:13] + s[14:16] + s[17:19] + s[20:]
... 
>>> t=timeit.Timer('sliceutc(str(datetime.utcnow()))','''
... from __main__ import sliceutc
... from datetime import datetime''')
>>> t.timeit(number=10000000)
62.378515005111694