如何将以下字符串转换为datetime对象?

"Jun 1 2005  1:33PM"

当前回答

如果您只需要日期格式,则可以通过传递单个字段(如:

>>> import datetime
>>> date = datetime.date(int('2017'),int('12'),int('21'))
>>> date
datetime.date(2017, 12, 21)
>>> type(date)
<type 'datetime.date'>

您可以传递拆分字符串值,将其转换为日期类型,如:

selected_month_rec = '2017-09-01'
date_formate = datetime.date(int(selected_month_rec.split('-')[0]),int(selected_month_rec.split('-')[1]),int(selected_month_rec.split('-')[2]))

您将获得日期格式的结果值。

其他回答

Django时区感知日期时间对象示例。

import datetime
from django.utils.timezone import get_current_timezone
tz = get_current_timezone()

format = '%b %d %Y %I:%M%p'
date_object = datetime.datetime.strptime('Jun 1 2005  1:33PM', format)
date_obj = tz.localize(date_object)

当USE_TZ=True时,这种转换对于Django和Python非常重要:

RuntimeWarning: DateTimeField MyModel.created received a naive datetime (2016-03-04 00:00:00) while time zone support is active.

arrow为日期和时间提供了许多有用的函数。这段代码为这个问题提供了答案,并表明箭头还能够轻松格式化日期并显示其他地区的信息。

>>> import arrow
>>> dateStrings = [ 'Jun 1  2005 1:33PM', 'Aug 28 1999 12:00AM' ]
>>> for dateString in dateStrings:
...     dateString
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').datetime
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').format('ddd, Do MMM YYYY HH:mm')
...     arrow.get(dateString.replace('  ',' '), 'MMM D YYYY H:mmA').humanize(locale='de')
...
'Jun 1  2005 1:33PM'
datetime.datetime(2005, 6, 1, 13, 33, tzinfo=tzutc())
'Wed, 1st Jun 2005 13:33'
'vor 11 Jahren'
'Aug 28 1999 12:00AM'
datetime.datetime(1999, 8, 28, 0, 0, tzinfo=tzutc())
'Sat, 28th Aug 1999 00:00'
'vor 17 Jahren'

看见http://arrow.readthedocs.io/en/latest/了解更多信息。

将yyyy-mm-dd日期字符串映射到datetime.date对象的简短示例:

from datetime import date
date_from_yyyy_mm_dd = lambda δ : date(*[int(_) for _ in δ.split('-')])
date_object = date_from_yyyy_mm_dd('2021-02-15')

使用熊猫时间戳似乎是最快的:

import pandas as pd

N = 1000

l = ['Jun 1 2005  1:33PM'] * N

list(pd.to_datetime(l, format=format))

%timeit _ = list(pd.to_datetime(l, format=format))
1.58 ms ± 21.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

其他解决方案

from datetime import datetime
%timeit _ = list(map(lambda x: datetime.strptime(x, format), l))
9.41 ms ± 95.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

from dateutil.parser import parse
%timeit _ = list(map(lambda x: parse(x), l))
73.8 ms ± 1.14 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

如果字符串是ISO 8601字符串,请使用csio8601:

import ciso8601

l = ['2014-01-09'] * N

%timeit _ = list(map(lambda x: ciso8601.parse_datetime(x), l))
186 µs ± 4.13 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

这里没有提到但很有用的一点:在当天添加后缀。我解耦了后缀逻辑,这样你就可以将它用于任何你喜欢的数字,而不仅仅是日期。

import time

def num_suffix(n):
    '''
    Returns the suffix for any given int
    '''
    suf = ('th','st', 'nd', 'rd')
    n = abs(n) # wise guy
    tens = int(str(n)[-2:])
    units = n % 10
    if tens > 10 and tens < 20:
        return suf[0] # teens with 'th'
    elif units <= 3:
        return suf[units]
    else:
        return suf[0] # 'th'

def day_suffix(t):
    '''
    Returns the suffix of the given struct_time day
    '''
    return num_suffix(t.tm_mday)

# Examples
print num_suffix(123)
print num_suffix(3431)
print num_suffix(1234)
print ''
print day_suffix(time.strptime("1 Dec 00", "%d %b %y"))
print day_suffix(time.strptime("2 Nov 01", "%d %b %y"))
print day_suffix(time.strptime("3 Oct 02", "%d %b %y"))
print day_suffix(time.strptime("4 Sep 03", "%d %b %y"))
print day_suffix(time.strptime("13 Nov 90", "%d %b %y"))
print day_suffix(time.strptime("14 Oct 10", "%d %b %y"))​​​​​​​