如何将datetime对象格式化为带有毫秒的字符串?


当前回答

@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'

其他回答

from datetime import datetime
from time import clock

t = datetime.utcnow()
print 't == %s    %s\n\n' % (t,type(t))

n = 100000

te = clock()
for i in xrange(1):
    t_stripped = t.strftime('%Y%m%d%H%M%S%f')
print clock()-te
print t_stripped," t.strftime('%Y%m%d%H%M%S%f')"

print

te = clock()
for i in xrange(1):
    t_stripped = str(t).replace('-','').replace(':','').replace('.','').replace(' ','')
print clock()-te
print t_stripped," str(t).replace('-','').replace(':','').replace('.','').replace(' ','')"

print

te = clock()
for i in xrange(n):
    t_stripped = str(t).translate(None,' -:.')
print clock()-te
print t_stripped," str(t).translate(None,' -:.')"

print

te = clock()
for i in xrange(n):
    s = str(t)
    t_stripped = s[:4] + s[5:7] + s[8:10] + s[11:13] + s[14:16] + s[17:19] + s[20:] 
print clock()-te
print t_stripped," s[:4] + s[5:7] + s[8:10] + s[11:13] + s[14:16] + s[17:19] + s[20:] "

结果

t == 2011-09-28 21:31:45.562000    <type 'datetime.datetime'>


3.33410112179
20110928212155046000  t.strftime('%Y%m%d%H%M%S%f')

1.17067364707
20110928212130453000 str(t).replace('-','').replace(':','').replace('.','').replace(' ','')

0.658806915404
20110928212130453000 str(t).translate(None,' -:.')

0.645189262881
20110928212130453000 s[:4] + s[5:7] + s[8:10] + s[11:13] + s[14:16] + s[17:19] + s[20:]

translate()和slicing方法的使用同时运行 Translate()的优点是可以在一行中使用

在第一个的基础上比较时间:

1.000 * t.strftime('%Y%m%d%H% m% S%f') 0.351 * str (t) .replace(‘-’,”).replace(‘:’,”).replace(‘。’,”).replace (' “,”) 0.198 * str(t)。翻译(,):。) 0.194 * s[4] +[7] +[8:10] +年代[11:13]+[十四16]+年代(十七19)+ 年代(20:)

字段宽度格式规范

UNIX date命令允许指定%3将精度降低到3位:

$ date '+%Y-%m-%d %H:%M:%S.%3N'
2022-01-01 00:01:23.456

下面是一个自定义函数,它可以在Python中做到这一点:

from datetime import datetime

def strftime_(fmt: str, dt: datetime) -> str:
    tokens = fmt.split("%")
    tokens[1:] = [_format_token(dt, x) for x in tokens[1:]]
    return "".join(tokens)

def _format_token(dt: datetime, token: str) -> str:
    if len(token) == 0:
        return ""
    if token[0].isnumeric():
        width = int(token[0])
        s = dt.strftime(f"%{token[1]}")[:width]
        return f"{s}{token[2:]}"
    return dt.strftime(f"%{token}")

使用示例:

>>> strftime_("%Y-%m-%d %H:%M:%S.%3f", datetime.now())
'2022-01-01 00:01:23.456'

注:%%不支持。

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'
datetime
t = datetime.datetime.now()
ms = '%s.%i' % (t.strftime('%H:%M:%S'), t.microsecond/1000)
print(ms)
14:44:37.134

我处理过同样的问题,但在我的情况下,毫秒是四舍五入而不是截断的,这很重要

from datetime import datetime, timedelta

def strftime_ms(datetime_obj):
    y,m,d,H,M,S = datetime_obj.timetuple()[:6]
    ms = timedelta(microseconds = round(datetime_obj.microsecond/1000.0)*1000)
    ms_date = datetime(y,m,d,H,M,S) + ms
    return ms_date.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]