问题:在Python中将字符串日期转换为时间戳
如何将格式转换"%d/%m/%Y"
为时间戳的字符串?
"01/12/2011" -> 1322697600
How to convert a string in the format "%d/%m/%Y"
to timestamp?
"01/12/2011" -> 1322697600
回答 0
>>> import time
>>> import datetime
>>> s = "01/12/2011"
>>> time.mktime(datetime.datetime.strptime(s, "%d/%m/%Y").timetuple())
1322697600.0
>>> import time
>>> import datetime
>>> s = "01/12/2011"
>>> time.mktime(datetime.datetime.strptime(s, "%d/%m/%Y").timetuple())
1322697600.0
回答 1
我使用ciso8601
,它比datetime的strptime快62倍。
t = "01/12/2011"
ts = ciso8601.parse_datetime(t)
# to get time in seconds:
time.mktime(ts.timetuple())
您可以在此处了解更多信息。
I use ciso8601
, which is 62x faster than datetime’s strptime.
t = "01/12/2011"
ts = ciso8601.parse_datetime(t)
# to get time in seconds:
time.mktime(ts.timetuple())
You can learn more here.
回答 2
要将字符串转换为日期对象:
from datetime import date, datetime
date_string = "01/12/2011"
date_object = date(*map(int, reversed(date_string.split("/"))))
assert date_object == datetime.strptime(date_string, "%d/%m/%Y").date()
将日期对象转换为POSIX时间戳的方式取决于时区。从Python中的转换datetime.date
为UTC时间戳:
日期对象代表UTC的午夜
import calendar
timestamp1 = calendar.timegm(utc_date.timetuple())
timestamp2 = (utc_date.toordinal() - date(1970, 1, 1).toordinal()) * 24*60*60
assert timestamp1 == timestamp2
日期对象代表当地时间的午夜
import time
timestamp3 = time.mktime(local_date.timetuple())
assert timestamp3 != timestamp1 or (time.gmtime() == time.localtime())
时间戳是不同的,除非UTC和当地时间的午夜是同一时间实例。
To convert the string into a date object:
from datetime import date, datetime
date_string = "01/12/2011"
date_object = date(*map(int, reversed(date_string.split("/"))))
assert date_object == datetime.strptime(date_string, "%d/%m/%Y").date()
The way to convert the date object into POSIX timestamp depends on timezone. From Converting datetime.date
to UTC timestamp in Python:
date object represents midnight in UTC
import calendar
timestamp1 = calendar.timegm(utc_date.timetuple())
timestamp2 = (utc_date.toordinal() - date(1970, 1, 1).toordinal()) * 24*60*60
assert timestamp1 == timestamp2
date object represents midnight in local time
import time
timestamp3 = time.mktime(local_date.timetuple())
assert timestamp3 != timestamp1 or (time.gmtime() == time.localtime())
The timestamps are different unless midnight in UTC and in local time is the same time instance.
回答 3
>>> int(datetime.datetime.strptime('01/12/2011', '%d/%m/%Y').strftime("%s"))
1322683200
>>> int(datetime.datetime.strptime('01/12/2011', '%d/%m/%Y').strftime("%s"))
1322683200
回答 4
答案还取决于您输入的日期时区。如果您的日期是本地日期,则可以像katrielalex所说的那样使用mktime()-只有我不明白他为什么使用datetime而不是这个较短的版本:
>>> time.mktime(time.strptime('01/12/2011', "%d/%m/%Y"))
1322694000.0
但是请注意,我的结果与他的结果不同,因为我可能位于不同的TZ中(结果是无时区的UNIX时间戳)
现在,如果输入日期已经是UTC,那么我认为正确的解决方案是:
>>> calendar.timegm(time.strptime('01/12/2011', '%d/%m/%Y'))
1322697600
The answer depends also on your input date timezone. If your date is a local date, then you can use mktime() like katrielalex said – only I don’t see why he used datetime instead of this shorter version:
>>> time.mktime(time.strptime('01/12/2011', "%d/%m/%Y"))
1322694000.0
But observe that my result is different than his, as I am probably in a different TZ (and the result is timezone-free UNIX timestamp)
Now if the input date is already in UTC, than I believe the right solution is:
>>> calendar.timegm(time.strptime('01/12/2011', '%d/%m/%Y'))
1322697600
回答 5
只需使用datetime.datetime.strptime
:
import datetime
stime = "01/12/2011"
print(datetime.datetime.strptime(stime, "%d/%m/%Y").timestamp())
结果:
1322697600
要使用UTC代替本地时区,请使用.replace
:
datetime.datetime.strptime(stime, "%d/%m/%Y").replace(tzinfo=datetime.timezone.utc).timestamp()
Simply use datetime.datetime.strptime
:
import datetime
stime = "01/12/2011"
print(datetime.datetime.strptime(stime, "%d/%m/%Y").timestamp())
Result:
1322697600
To use UTC instead of the local timezone use .replace
:
datetime.datetime.strptime(stime, "%d/%m/%Y").replace(tzinfo=datetime.timezone.utc).timestamp()
回答 6
First you must the strptime class to convert the string to a struct_time format.
Then just use mktime from there to get your float.
回答 7
这些答案中有很多都不必考虑日期是幼稚的
为正确起见,您需要先将天真日期设为可识别时区的日期时间
import datetime
import pytz
# naive datetime
d = datetime.datetime.strptime('01/12/2011', '%d/%m/%Y')
>>> datetime.datetime(2011, 12, 1, 0, 0)
# add proper timezone
pst = pytz.timezone('America/Los_Angeles')
d = pst.localize(d)
>>> datetime.datetime(2011, 12, 1, 0, 0,
tzinfo=<DstTzInfo 'America/Los_Angeles' PST-1 day, 16:00:00 STD>)
# convert to UTC timezone
utc = pytz.UTC
d = d.astimezone(utc)
>>> datetime.datetime(2011, 12, 1, 8, 0, tzinfo=<UTC>)
# epoch is the beginning of time in the UTC timestamp world
epoch = datetime.datetime(1970,1,1,0,0,0,tzinfo=pytz.UTC)
>>> datetime.datetime(1970, 1, 1, 0, 0, tzinfo=<UTC>)
# get the total second difference
ts = (d - epoch).total_seconds()
>>> 1322726400.0
也:
请注意,在许多时区中都使用pytz
“ 不工作”。请参见带有pytz时区的datetime。不同的偏移量取决于tzinfo的设置方式tzinfo
datetime.datetime
# Don't do this:
d = datetime.datetime(2011, 12, 1,0,0,0, tzinfo=pytz.timezone('America/Los_Angeles'))
>>> datetime.datetime(2011, 1, 12, 0, 0,
tzinfo=<DstTzInfo 'America/Los_Angeles' LMT-1 day, 16:07:00 STD>)
# tzinfo in not PST but LMT here, with a 7min offset !!!
# when converting to UTC:
d = d.astimezone(pytz.UTC)
>>> datetime.datetime(2011, 1, 12, 7, 53, tzinfo=<UTC>)
# you end up with an offset
https://zh.wikipedia.org/wiki/Local_mean_time
A lot of these answers don’t bother to consider that the date is naive to begin with
To be correct, you need to make the naive date a timezone aware datetime first
import datetime
import pytz
# naive datetime
d = datetime.datetime.strptime('01/12/2011', '%d/%m/%Y')
>>> datetime.datetime(2011, 12, 1, 0, 0)
# add proper timezone
pst = pytz.timezone('America/Los_Angeles')
d = pst.localize(d)
>>> datetime.datetime(2011, 12, 1, 0, 0,
tzinfo=<DstTzInfo 'America/Los_Angeles' PST-1 day, 16:00:00 STD>)
# convert to UTC timezone
utc = pytz.UTC
d = d.astimezone(utc)
>>> datetime.datetime(2011, 12, 1, 8, 0, tzinfo=<UTC>)
# epoch is the beginning of time in the UTC timestamp world
epoch = datetime.datetime(1970,1,1,0,0,0,tzinfo=pytz.UTC)
>>> datetime.datetime(1970, 1, 1, 0, 0, tzinfo=<UTC>)
# get the total second difference
ts = (d - epoch).total_seconds()
>>> 1322726400.0
Also:
Be careful, using pytz
for tzinfo
in a datetime.datetime
DOESN’T WORK for many timezones. See datetime with pytz timezone. Different offset depending on how tzinfo is set
# Don't do this:
d = datetime.datetime(2011, 12, 1,0,0,0, tzinfo=pytz.timezone('America/Los_Angeles'))
>>> datetime.datetime(2011, 1, 12, 0, 0,
tzinfo=<DstTzInfo 'America/Los_Angeles' LMT-1 day, 16:07:00 STD>)
# tzinfo in not PST but LMT here, with a 7min offset !!!
# when converting to UTC:
d = d.astimezone(pytz.UTC)
>>> datetime.datetime(2011, 1, 12, 7, 53, tzinfo=<UTC>)
# you end up with an offset
https://en.wikipedia.org/wiki/Local_mean_time
回答 8
我建议dateutil:
import dateutil.parser
dateutil.parser.parse("01/12/2011", dayfirst=True).timestamp()
I would suggest dateutil:
import dateutil.parser
dateutil.parser.parse("01/12/2011", dayfirst=True).timestamp()
回答 9
似乎相当有效:
import datetime
day, month, year = '01/12/2011'.split('/')
datetime.datetime(int(year), int(month), int(day)).timestamp()
每个循环1.61 µs±120 ns(平均±标准偏差,共运行7次,每个循环100000个)
Seems to be quite efficient:
import datetime
day, month, year = '01/12/2011'.split('/')
datetime.datetime(int(year), int(month), int(day)).timestamp()
1.61 µs ± 120 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
回答 10
只需使用datetime.timestamp(您的datetime实例),datetime实例将包含时区信息,因此该时间戳将是标准utc时间戳。如果将datetime转换为timetuple,它将丢失其时区,因此结果将是错误的。如果要提供接口,则应这样编写:int(datetime.timestamp(time_instance))* 1000
just use datetime.timestamp(your datetime instanse), datetime instance contains the timezone infomation, so the timestamp will be a standard utc timestamp. if you transform the datetime to timetuple, it will lose it’s timezone, so the result will be error.
if you want to provide an interface, you should write like this:
int(datetime.timestamp(time_instance)) * 1000
回答 11
回答 12
您可以转换为isoformat
my_date = '2020/08/08'
my_date = my_date.replace('/','-') # just to adapte to your question
date_timestamp = datetime.datetime.fromisoformat(my_date).timestamp()
you can convert to isoformat
my_date = '2020/08/08'
my_date = my_date.replace('/','-') # just to adapte to your question
date_timestamp = datetime.datetime.fromisoformat(my_date).timestamp()
回答 13
一个获取UNIX Epoch时间的简单函数。
注意:此功能假定输入的日期时间为UTC格式(请参阅此处的注释)。
def utctimestamp(ts: str, DATETIME_FORMAT: str = "%d/%m/%Y"):
import datetime, calendar
ts = datetime.datetime.utcnow() if ts is None else datetime.datetime.strptime(ts, DATETIME_FORMAT)
return calendar.timegm(ts.utctimetuple())
用法:
>>> utctimestamp("01/12/2011")
1322697600
>>> utctimestamp("2011-12-01", "%Y-%m-%d")
1322697600
A simple function to get UNIX Epoch time.
NOTE: This function assumes the input date time is in UTC format (Refer to comments here).
def utctimestamp(ts: str, DATETIME_FORMAT: str = "%d/%m/%Y"):
import datetime, calendar
ts = datetime.datetime.utcnow() if ts is None else datetime.datetime.strptime(ts, DATETIME_FORMAT)
return calendar.timegm(ts.utctimetuple())
Usage:
>>> utctimestamp("01/12/2011")
1322697600
>>> utctimestamp("2011-12-01", "%Y-%m-%d")
1322697600