Converting date/time in YYYYMMDD/HHMMSS format to Python datetime

Question

I have a date in YYYYMMDD format and a time in HHMMSS format as strings in the 4th and 5th elements in a list. I.E.:

data[4] = '20100304'
data[5] = '082835'

I am creating an instance of datetime (in a field named generates) like this:

generatedtime = datetime.datetime(int(data[4][:4]),int(data[4][4:6]),int(data[4][6:]),int(data[5][:2]),int(data[5][2:4]),int(data[5][4:6]))

Given that the input format cannot change, is there a cleaner way I should be creating my instance of the datetime object?

Answer 1

You might take a look at time.strptime.

import time
time.strptime('20100304 082835', '%Y%m%d %H%M%S')

The above assumes a 24-hour clock (%H). Use %I instead if using a 12-hour clock.

For a complete list of available format directives, check out the docs for time.strftime

Answer 2

import time, datetime

x = time.strptime("30 Nov 00", "%d %b %y")
datetime.datetime(x)

You can also provide your own format of date string, here is doc.

Answer 3

No need to import time; datetime.datetime.strptime can do it by itself.

import datetime
dt=datetime.datetime.strptime(data[4]+data[5],'%Y%m%d%H%M%S')
print(dt)
# 2010-03-04 08:28:35

For information on the format codes (e.g. %Y%m%d%H%M%S) available, see the docs for strftime.

Answer 4

You could clean up your existing code a bit with a generator (equivalent to a map):

generatedtime = datetime.datetime( *(int(x) for x in
   (data[4][:4], data[4][4:6], data[4][6:], data[5][:2], data[5][2:4],data[5][4:6])
   ) )

or even, if you're crazy like a fox, you could make the datetime statement even cleaner (at the expense of adding a ridiculous line):

slices = ( (4, slice(4)), (4, slice(4,6) ), (4, slice(6,None)), (5, slice(2) ), (5, slice(2,4) ), (5, slice(4,6)) )
generatedtime = datetime.datetime( *(int(data[i][s]) for (i,s) in slices) )