How to convert datetime/date to milliseconds in Python?

Question

How to convert a type date and a datetime millisecond?

>>> ontem # tipo date
datetime.date(2015, 9, 29)
>>> hoje # tipo datetime
datetime.datetime(2015, 9, 30, 18, 15, 36, 856736)

Answer 1

Build a timedelta from the difference between your date and the reference date, then use total_seconds to get the total of seconds contained in the interval. As this result comes expressed in floating point, you can multiply it by 1000 to obtain the result in milliseconds (truncating if necessary):

>>> UNIX_EPOCH = datetime.datetime(1970, 1, 1)
>>> data = datetime.datetime(2015, 9, 30, 18, 15, 36, 856736)
>>> (data - UNIX_EPOCH).total_seconds()*1000
1443636936856.7358
>>> int((data - UNIX_EPOCH).total_seconds()*1000)
1443636936856

^Source

If your date is Timezone-Aware, remember to use the same time zone on the reference date and the date you want to represent in milliseconds (or rather, convert your date to UTC before making this representation, so it’s all canonical). If everything is in UTC, you can also get the value of "Unix Epoch" by:

UNIX_EPOCH = datetime.datetime.utcfromtimestamp(0)

(the utcfromtimestamp expects a value in seconds, and may reject values outside the 1970-2038 range, depending on implementation; in my opinion, it is safer to state the constant explicitly, without depending on this method)

---

Note: it makes no sense to represent a date per millisecond, since a date corresponds to a full day. I suggest choosing [consistently] an instant on that date (for example midnight) and creating a datetime from their date.

Answer 2

It is possible to convert in several ways:

Solution 1

Using datetime.timestamp() present in Python 3.

>>> import datetime
>>> hoje = datetime.datetime(2015, 9, 30)
>>> int(hoje.timestamp() * 1000)
1443582000000

Note that timestamp() only works with datetime and not with date

Then it is necessary to convert to datetime if you are working with date:

>>> hoje = datetime.date(2015, 9, 30)
>>> int(datetime.datetime(*hoje.timetuple()[:3]).timestamp() * 1000)
1443582000000

Solution 2

Using strftime().

>>> import datetime
>>> hoje = datetime.date(2015, 9, 30)
>>> int(hoje.strftime('%s')) * 1000
1441681200000

See that it is %s with s lower case, not upper case. Use s uppercase will return only the seconds of the date (in this case 00) and not the total seconds since 1970.

Remarks:

I found no reference to s lowercase in documentation, but works in Python 3 and 2 on Mac OS.

The strftime('%s') returns only the rounded total seconds, preventing accuracy in milliseconds.

Solution 3

Using mktime() module time:

>>> import datetime, time
>>> hoje = datetime.date(2015, 9, 30)
>>> int(time.mktime(hoje.timetuple()) * 1000)
1443582000000

Note that mktime returns only the rounded total seconds, not allowing accuracy for milliseconds.

Solution 4

Using total_seconds() of timedelta:

By making the difference between two datetime is returned a timedelta and from it it is possible to return the number of total seconds using total_seconds().

>>> import datetime
>>> hoje = datetime.datetime(2015, 9, 30)
>>> (hoje - datetime.datetime.utcfromtimestamp(0)).total_seconds() * 1000
1443582000000

The utcfromtimestamp(0) returns a datetime.datetime(1970, 1, 1, 0, 0) as mentioned by mgibsonbr.

Answer 3

Since the datetime object already has a method to represent a date in seconds since January 1, 1970, the following suffices:

data = datetime.datetime.now()
segundos = data.timestamp()
milissegundos = segundos * 1000

If you have a date object, convert it to datetime:

d = datetime.date(2015, 5, 5)    
data = datetime.datetime.fromordinal(d.toordinal())