如何将以下字符串转换为datetime对象?

"Jun 1 2005  1:33PM"

当前回答

Python>=3.7

要将YYYY-MM-DD字符串转换为datetime对象,可以使用datetime.fromisoformat。

from datetime import datetime

date_string = "2012-12-12 10:10:10"
print (datetime.fromisoformat(date_string))
2012-12-12 10:10:10

文档中的注意事项:

这不支持解析任意的ISO 8601字符串-它只是作为datetime.isoformat()的反操作。第三方包dateutil中提供了一个功能更全面的ISO 8602解析器dateutil.parser.isorse。

其他回答

如果您不想明确指定字符串相对于日期时间格式的格式,可以使用此黑客绕过该步骤:

from dateutil.parser import parse

# Function that'll guess the format and convert it into the python datetime format
def update_event(start_datetime=None, end_datetime=None, description=None):
    if start_datetime is not None:
        new_start_time = parse(start_datetime)

        return new_start_time

# Sample input dates in different formats
d = ['06/07/2021 06:40:23.277000', '06/07/2021 06:40', '06/07/2021']

new = [update_event(i) for i in d]

for date in new:
    print(date)
    # Sample output dates in Python datetime object
    #   2014-04-23 00:00:00
    #   2013-04-24 00:00:00
    #   2014-04-25 00:00:00

如果要将其转换为其他日期时间格式,只需使用您喜欢的格式修改最后一行,例如date.strftime(“%Y/%m/%d%H:%m:%S.%f”):

from dateutil.parser import parse

def update_event(start_datetime=None, end_datetime=None, description=None):
    if start_datetime is not None:
        new_start_time = parse(start_datetime)

        return new_start_time

# Sample input dates in different formats
d = ['06/07/2021 06:40:23.277000', '06/07/2021 06:40', '06/07/2021']

# Passing the dates one by one through the function
new = [update_event(i) for i in d]

for date in new:
    print(date.strftime('%Y/%m/%d %H:%M:%S.%f'))
    # Sample output dates in required Python datetime object
    #   2021/06/07 06:40:23.277000
    #   2021/06/07 06:40:00.000000
    #   2021/06/07 00:00:00.000000

尝试运行上面的代码段以获得更好的清晰度。

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)

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。

创建一个小的实用程序函数,如:

def date(datestr="", format="%Y-%m-%d"):
    from datetime import datetime
    if not datestr:
        return datetime.today().date()
    return datetime.strptime(datestr, format).date()

这是足够多功能的:

如果不传递任何参数,它将返回今天的日期。有一个日期格式作为默认值,您可以覆盖它。您可以轻松地修改它以返回日期时间。

这将有助于将字符串转换为datetime和时区:

def convert_string_to_time(date_string, timezone):

    from datetime import datetime
    import pytz

    date_time_obj = datetime.strptime(date_string[:26], '%Y-%m-%d %H:%M:%S.%f')
    date_time_obj_timezone = pytz.timezone(timezone).localize(date_time_obj)

    return date_time_obj_timezone

date = '2018-08-14 13:09:24.543953+00:00'
TIME_ZONE = 'UTC'
date_time_obj_timezone = convert_string_to_time(date, TIME_ZONE)