Python - Datetime not accounting for leap second properly?

Question

I am parsing some data that has the leapsecond timestampe datetime 2012-06-30T23:59:60.209215. I used following code to parse that string and convert to a datetime object:

    nofrag, frag = t.split('.')
    nofrag_dt = datetime.datetime.strptime(nofrag, "%Y-%m-%dT%H:%M:%S")
    dt = nofrag_dt.replace(microsecond=int(frag))

Python documentation claims that this shouldn't be an issue as %S accepts [0, 61]. But, I get this error with the above timestamp

nofrag_dt = datetime.datetime.strptime(nofrag, "%Y-%m-%dT%H:%M:%S")
ValueError: second must be in 0..59

Thanks

Answer 1

Do this:

import time
import datetime 
t = '2012-06-30T23:59:60.209215'
nofrag, frag = t.split('.')
nofrag_dt = time.strptime(nofrag, "%Y-%m-%dT%H:%M:%S")
ts = datetime.datetime.fromtimestamp(time.mktime(nofrag_dt))
dt = ts.replace(microsecond=int(frag))
print(dt)

Output is:

2012-07-01 00:00:00.209215

Answer 2

The documentation for %S says:

Unlike the time module, the datetime module does not support leap seconds.

The time string "2012-06-30T23:59:60.209215" implies that the time is in UTC (it is the last leap second at the moment):

import time
from calendar import timegm
from datetime import datetime, timedelta

time_string = '2012-06-30T23:59:60.209215'
time_string, dot, us = time_string.partition('.')
utc_time_tuple = time.strptime(time_string, "%Y-%m-%dT%H:%M:%S")
dt = datetime(1970, 1, 1) + timedelta(seconds=timegm(utc_time_tuple))
if dot:
    dt = dt.replace(microsecond=datetime.strptime(us, '%f').microsecond)
print(dt)
# -> 2012-07-01 00:00:00.209215