如何将以下字符串转换为datetime对象?
"Jun 1 2005 1:33PM"
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。
其他回答
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)
我已经完成了一个项目,可以转换一些非常整洁的表达式。查看时间字符串。
以下是一些示例:
pip install timestring
>>> import timestring
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm')
<timestring.Date 2015-08-15 20:40:00 4491909392>
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm').date
datetime.datetime(2015, 8, 15, 20, 40)
>>> timestring.Range('next week')
<timestring.Range From 03/10/14 00:00:00 to 03/03/14 00:00:00 4496004880>
>>> (timestring.Range('next week').start.date, timestring.Range('next week').end.date)
(datetime.datetime(2014, 3, 10, 0, 0), datetime.datetime(2014, 3, 14, 0, 0))
许多时间戳都有一个隐含的时区。为了确保您的代码在每个时区都有效,您应该在内部使用UTC,并在每次外来对象进入系统时附加一个时区。
Python 3.2+:
>>> datetime.datetime.strptime(
... "March 5, 2014, 20:13:50", "%B %d, %Y, %H:%M:%S"
... ).replace(tzinfo=datetime.timezone(datetime.timedelta(hours=-3)))
这假设您知道偏移量。如果您不知道,但您知道例如位置,您可以使用pytz包查询IANA时区数据库中的偏移量。我将在这里以德黑兰为例,因为它有半小时的偏移量:
>>> tehran = pytz.timezone("Asia/Tehran")
>>> local_time = tehran.localize(
... datetime.datetime.strptime("March 5, 2014, 20:13:50",
... "%B %d, %Y, %H:%M:%S")
... )
>>> local_time
datetime.datetime(2014, 3, 5, 20, 13, 50, tzinfo=<DstTzInfo 'Asia/Tehran' +0330+3:30:00 STD>)
如您所见,pytz已确定在特定日期的偏移量为+3:30。您现在可以将其转换为UTC时间,它将应用偏移量:
>>> utc_time = local_time.astimezone(pytz.utc)
>>> utc_time
datetime.datetime(2014, 3, 5, 16, 43, 50, tzinfo=<UTC>)
请注意,采用时区之前的日期会给您带来奇怪的偏移。这是因为IANA决定使用本地平均时间:
>>> chicago = pytz.timezone("America/Chicago")
>>> weird_time = chicago.localize(
... datetime.datetime.strptime("November 18, 1883, 11:00:00",
... "%B %d, %Y, %H:%M:%S")
... )
>>> weird_time.astimezone(pytz.utc)
datetime.datetime(1883, 11, 18, 7, 34, tzinfo=<UTC>)
奇怪的“7小时34分钟”源自芝加哥的经度。我使用这个时间戳是因为它正好在芝加哥采用标准时间之前。
Use:
emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv")
emp.info()
它显示“开始日期时间”列和“上次登录时间”都是数据帧中的“对象=字符串”:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name 933 non-null object
Gender 855 non-null object
Start Date 1000 non-null object
Last Login Time 1000 non-null object
Salary 1000 non-null int64
Bonus % 1000 non-null float64
Senior Management 933 non-null object
Team 957 non-null object
dtypes: float64(1), int64(1), object(6)
memory usage: 62.6+ KB
通过使用read_csv中的parse_dates选项,可以将字符串datetime转换为panda datetime格式。
emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv", parse_dates=["Start Date", "Last Login Time"])
emp.info()
输出:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name 933 non-null object
Gender 855 non-null object
Start Date 1000 non-null datetime64[ns]
Last Login Time 1000 non-null datetime64[ns]
Salary 1000 non-null int64
Bonus % 1000 non-null float64
Senior Management 933 non-null object
Team 957 non-null object
dtypes: datetime64[ns](2), float64(1), int64(1), object(4)
memory usage: 62.6+ KB
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。