datetime to Unix timestamp with millisecond precision

Question

I'm trying to do something really simple, convert a datetime object three days into the future into a Unix UTC timestamp:

import datetime, time
then = datetime.datetime.now() + datetime.timedelta(days=3)
# Method 1
print then.strftime("%s")
# Method 2
print time.mktime(then.timetuple())
# Method 3 
print time.mktime(then.timetuple()) * 1000

Method 1 and 2 give me Unix time in seconds, not milliseconds, and method 3 gives me milliseconds with no actual millisecond precision.

When I simply print then, I get datetime.datetime(2011, 11, 19, 15, 16, 8, 278271), so I know that the precision is available for milliseconds. How can I get a Unix timestamp with actual millisecond precision? If it's returned as a float and I have to flatten it to an an int, that's fine. Is there a solution I'm looking for that does this?

related: Converting datetime.date to UTC timestamp in Python — jfs
– jfs, Commented Mar 22, 2014 at 7:00

Adam Zalcman · Accepted Answer · 2011-11-16 23:44:49Z

49

Datetime objects have a field named microsecond. So one way to achieve what you need is:

time.mktime(then.timetuple())*1e3 + then.microsecond/1e3

This returns milliseconds since UNIX epoch with the required precision.

edited Nov 16, 2011 at 23:44

answered Nov 16, 2011 at 23:39

Adam Zalcman

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6 Comments

Mark Ransom Over a year ago

You're missing the punchline: print time.mktime(then.timetuple()) * 1000 + then.microsecond/1000

Naftuli Kay Over a year ago

Nice solution, especially the ultra-hacker saving of one character by using 1e3 instead of 1000.

Nicholas Hamilton Over a year ago

Naftuli, not really since 1e3 involves 3 distinct characters (1,e,3), and 1000 uses only 2 (1 and 0)

theherk Over a year ago

Maybe I'm missing something, but microseconds are millions of a second not thousandths. Also, I don't know why you multiply then divide. time.mktime(then.timetuple()) + then.microsecond/1e6 works for me.

tobltobs Over a year ago

@TheHerk I am not sure, but I think he wants the result in milliseconds. Your solution gives seconds, with milliseconds as decimal places.

|

Rob Smallshire · Accepted Answer · 2014-03-21 11:38:00Z

28

In Python 3.3 and above, which support the datetime.timestamp() method, you can do this:

from datetime import datetime, timezone, timedelta

(datetime.now(timezone.utc) + timedelta(days=3)).timestamp() * 1e3

answered Mar 21, 2014 at 11:38

Rob Smallshire

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2 Comments

Philippe Remy Over a year ago

This should be the accepted answer.

Benedikt S. Vogler Over a year ago

My measurement shows that using timestamp() is also a lot faster than the solution by Adam Zalcman.

Kyr · Accepted Answer · 2020-02-24 16:45:02Z

10

Assuming someone is interested in UTC, the following in valid after Python 3.3:

from datetime import datetime

now = int(datetime.utcnow().timestamp()*1e3)

Python 3.3 release notes:

New datetime.datetime.timestamp() method: Return POSIX timestamp corresponding to the datetime instance.

Intermediate results:

In [1]: datetime.utcnow().timestamp()
Out[1]: 1582562542.407362

In [2]: datetime.utcnow().timestamp()*1e3
Out[2]: 1582562566701.329

In [3]: int(datetime.utcnow().timestamp()*1e3)
Out[3]: 1582562577296

answered Feb 24, 2020 at 16:45

Kyr

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3 Comments

Håkon T. Over a year ago

Be aware: This is only correct when the local timezone is UTC. datetime.utcnow() returns a naive timestamp (!). When calling method .timestamp() on it, it is assumed to be local timezone - which gives the wrong POSIX timestamp.

Kyr Over a year ago

Thanks @HåkonT.! I will link the explanation coming from this SO answer: stackoverflow.com/questions/28573873/…

Martin H Over a year ago

Using Python 3.12.3, I get a deprecation warning using this answer. The suggested answer is importing UTC from datetime and using datetime.now(UTC) instead of datetime.utcnow(). I.e. from datetime import datetime, UTC and then now = int(datetime.now(UTC).timestamp()*1e3)

wattostudios · Accepted Answer · 2012-12-14 01:22:17Z

4

long((time.time() + 0.5) * 1000)

this has millisecond precision

edited Dec 14, 2012 at 1:22

wattostudios

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answered Dec 14, 2012 at 1:03

Francisco

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1 Comment

arshpreet Over a year ago

what it I have to get timestamp upto 7 decimal places something like this:1469172167.3014405

jfs · Accepted Answer · 2016-12-11 09:35:23Z

4

Day is always 86400 seconds in POSIX time. To get POSIX timestamp 3 days into the future as a float (with fraction of a second):

import time

DAY = 86400 # seconds
future = time.time() + 3 * DAY

It assumes that time.gmtime(0) is 1970 (POSIX Epoch).

If you already have a naive datetime object that represents time in the local timezone then the timestamp may be ambiguous during DST transitions. To avoid ambiguity, you could use a timezone aware datetime object or a naive datetime object that represents time in UTC.

To convert a local datetime dt to seconds since the Epoch:

from datetime import datetime
from time import mktime

timestamp = dt.timestamp() # Python 3.3+
timestamp = mktime(dt.timetuple()) + dt.microsecond / 1e6 # Python 2.7

It may fail if the local timezone had different utc offset in the past and the time implementation has no access to timezone database on the system. Use pytz to handle such cases.

To convert UTC datetime utc_dt to POSIX timestamp:

timestamp = (utc_dt - datetime(1970, 1, 1)).total_seconds()

To get milliseconds, just multiply any of the float number of seconds by 1e3.

edited Dec 11, 2016 at 9:35

answered Mar 22, 2014 at 7:28

jfs

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1 Comment

eraoul Over a year ago

Thanks! Note, should be dt.microsecond, not dt.microseconds

Nick Edgar · Accepted Answer · 2017-01-26 18:44:09Z

2

If sticking with datetimes in UTC, you can keep with that abstraction and leverage timedelta:

from datetime import datetime, timedelta

epoch = datetime.utcfromtimestamp(0)

def dt_from_ms(ms):
    return datetime.utcfromtimestamp(ms / 1000.0)

def dt_to_ms(dt):
    delta = dt - epoch
    return int(delta.total_seconds() * 1000)

then:

now = datetime.utcnow()
now
-> datetime.datetime(2017, 1, 25, 1, 30, 42, 765846)  # note microsecond precision

dt_to_ms(now)
-> 1485307842765

dt_from_ms(1485307842765)
-> datetime.datetime(2017, 1, 25, 1, 30, 42, 765000)  # note millisecond precision

then = now + timedelta(days=3)
dt_to_ms(then)
-> 1485567042765

(1485567042765 - 1485307842765) / (24*60*60*1000.0)
-> 3.0

edited Jan 26, 2017 at 18:44

answered Jan 25, 2017 at 1:38

Nick Edgar

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Comments

Andrew Clark · Accepted Answer · 2011-11-16 23:46:53Z

0

If you are using Python 2.7 or 3.2+, you can use timedelta.total_seconds() to get this fairly easily:

import datetime, time
print time.time() + datetime.timedelta(days=3).total_seconds()

answered Nov 16, 2011 at 23:46

Andrew Clark

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2 Comments

jfs Over a year ago

Day is always 86400 seconds in POSIX time.

Ahmed Over a year ago

Why time delta?

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