Following on from my previous question, Python time to age, I have now come across a problem regarding the timezone, and it turns out that it’s not always going to be “+0200”. So when strptime tries to parse it as such, it throws up an exception.
I thought about just chopping off the +0200 with [:-6] or whatever, but is there a real way to do this with strptime?
I am using Python 2.5.2 if it matters.
>>> from datetime import datetime
>>> fmt = "%a, %d %b %Y %H:%M:%S +0200"
>>> datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0200", fmt)
datetime.datetime(2008, 7, 22, 8, 17, 41)
>>> datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0300", fmt)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.5/_strptime.py", line 330, in strptime
(data_string, format))
ValueError: time data did not match format: data=Tue, 22 Jul 2008 08:17:41 +0300 fmt=%a, %d %b %Y %H:%M:%S +0200
Answers:
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Method 1
is there a real way to do this with strptime?
No, but since your format appears to be an RFC822-family date, you can read it much more easily using the email library instead:
>>> import email.utils
>>> email.utils.parsedate_tz('Tue, 22 Jul 2008 08:17:41 +0200')
(2008, 7, 22, 8, 17, 41, 0, 1, 0, 7200)
(7200 = timezone offset from UTC in seconds)
Method 2
New in version 2.6.
For a naive object, the %z and %Z
format codes are replaced by empty
strings.
It looks like this is implemented only in >= 2.6, and I think you have to manually parse it.
I can’t see another solution than to remove the time zone data:
from datetime import timedelta,datetime
try:
offset = int("Tue, 22 Jul 2008 08:17:41 +0300"[-5:])
except:
print "Error"
delta = timedelta(hours = offset / 100)
fmt = "%a, %d %b %Y %H:%M:%S"
time = datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0200"[:-6], fmt)
time -= delta
Method 3
You can use the dateutil library which is very useful:
from datetime import datetime
from dateutil.parser import parse
dt = parse("Tue, 22 Jul 2008 08:17:41 +0200")
## datetime.datetime(2008, 7, 22, 8, 17, 41, tzinfo=tzoffset(None, 7200)) <- dt
print dt
2008-07-22 08:17:41+02:00
Method 4
As far as I know, strptime() doesn’t recognize numeric time zone codes. If you know that the string is always going to end with a time zone specification of that form (+ or – followed by 4 digits), just chopping it off and parsing it manually seems like a perfectly reasonable thing to do.
Method 5
It seems that %Z corresponds to time zone names, not offsets.
For example, given:
>>> format = '%a, %d %b %Y %H:%M:%S %Z'
I can parse:
>>> datetime.datetime.strptime('Tue, 22 Jul 2008 08:17:41 GMT', format)
datetime.datetime(2008, 7, 22, 8, 17, 41)
Although it seems that it doesn’t do anything with the time zone, merely observing that it exists and is valid:
>>> datetime.datetime.strptime('Tue, 22 Jul 2008 08:17:41 NZDT', format)
datetime.datetime(2008, 7, 22, 8, 17, 41)
I suppose if you wished, you could locate a mapping of offsets to names, convert your input, and then parse it. It might be simpler to just truncate your input, though.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0