How to parse datetime that ends with `Z`?

Question

I have the following datetime string s:

2017-10-18T04:46:53.553472514Z

I parese it like that:

t = datetime.strptime(s, '%Y-%m-%dT%H:%M:%SZ')

how to fix ValueError: time data '2017-10-18T04:46:53.553472514Z' does not match format '%Y-%m-%dT%H:%M:%SZ'

Answer 1

In theory,

t = datetime.strptime(s, '%Y-%m-%dT%H:%M:%S.%fZ')

would be the correct format string as you have fractions of second as well. BUT they would then need to be microseconds. Yours are probably nanoseconds, as %f only takes maximum of 6 digits.

So you need to do something like this:

t = datetime.datetime.strptime(s.split(".")[0], '%Y-%m-%dT%H:%M:%S')
t = t + datetime.timedelta(microseconds=int(s.split(".")[1][:-1])/1000)

print (t)

This works but it converts nanoseconds to microseconds. If this is not ok, then you need to do something else.

Answer 2

I think you should use dateutil.parser module

In [23]: s = "2017-10-18T04:46:53.553472514Z"

In [24]: import dateutil.parser as p

In [25]: p.parse(s)
Out[25]: datetime.datetime(2017, 10, 18, 4, 46, 53, 553472, tzinfo=tzutc())

Answer 3

To answer the original question: there's no way to parse "Z" as the timezone using only the standard library.

Here's some discussion on that: https://bugs.python.org/issue35829

Answer 4

I like the mx.DateTime module.

import mx.DateTime as dt
dt.DateTimeFrom('2017-10-18T04:46:53.553472514Z')

Answer 5

datetime is another Python built-in module that is completely superseded by a better extra module: arrow.

You can just do:

import arrow
dt = arrow.get('2017-10-18T04:46:53.553472514Z').datetime

Your date string being in standardized ISO format, it will be parsed without even giving a format string. The parsed df will be:

datetime.datetime(2017, 10, 18, 4, 46, 53, 553472, tzinfo=tzutc())

Or keep the Arrow object in case you want to keep the extra precision