CODEWITHSUNDEEP
pythondatetime
Daniel Benamy
Daniel Benamy

Reputation: 1153

Date/time conversion using time.mktime seems wrong

>>> import time
>>> time.strptime("01-31-2009", "%m-%d-%Y")
(2009, 1, 31, 0, 0, 0, 5, 31, -1)
>>> time.mktime((2009, 1, 31, 0, 0, 0, 5, 31, -1))
1233378000.0
>>> 60*60*24 # seconds in a day
86400
>>> 1233378000.0 / 86400
14275.208333333334

time.mktime should return the number of seconds since the epoch. Since I'm giving it a time at midnight and the epoch is at midnight, shouldn't the result be evenly divisible by the number of seconds in a day?

Upvotes: 4

Views: 5861

Answers (4)

Philip Reynolds
Philip Reynolds

Reputation: 9392

Short answer: Because of timezones.

The Epoch is in UTC.

For example, I'm on IST (Irish Standard Time) or UTC+1. time.mktime() is relative to my timezone, so on my system this refers to

>>> time.mktime((2009, 1, 31, 0, 0, 0, 5, 31, -1))
1233360000.0

Because you got the result 1233378000, that would suggest that you're 5 hours behind me

>>> (1233378000 - 1233360000) / (60*60)    
5

Have a look at the time.gmtime() function which works off UTC.

Upvotes: 8

Daniel Benamy
Daniel Benamy

Reputation: 1153

Phil's answer really solved it, but I'll elaborate a little more. Since the epoch is in UTC, if I want to compare other times to the epoch, I need to interpret them as UTC as well.

>>> calendar.timegm((2009, 1, 31, 0, 0, 0, 5, 31, -1))
1233360000
>>> 1233360000 / (60*60*24)
14275

By converting the time tuple to a timestamp treating is as UTC time, I get a number which is evenly divisible by the number of seconds in a day.

I can use this to convert a date to a days-from-the-epoch representation which is what I'm ultimately after.

Upvotes: 2

Daren Thomas
Daren Thomas

Reputation: 70314

Interesting. I don't know, but I did try this:

>>> now = time.mktime((2008, 8, 22, 11 ,17, -1, -1, -1, -1))
>>> tomorrow = time.mktime((2008, 8, 23, 11 ,17, -1, -1, -1, -1))
>>> tomorrow - now
86400.0

which is what you expected. My guess? Maybe some time correction was done since the epoch. This could be only a few seconds, something like a leap year. I think I heard something like this before, but can't remember exactly how and when it is done...

Upvotes: 0

Anders Eurenius
Anders Eurenius

Reputation: 4226

mktime(...)
    mktime(tuple) -> floating point number

    Convert a time tuple in local time to seconds since the Epoch.

local time... fancy that.

The time tuple:

The other representation is a tuple of 9 integers giving local time.
The tuple items are:
  year (four digits, e.g. 1998)
  month (1-12)
  day (1-31)
  hours (0-23)
  minutes (0-59)
  seconds (0-59)
  weekday (0-6, Monday is 0)
  Julian day (day in the year, 1-366)
  DST (Daylight Savings Time) flag (-1, 0 or 1)
If the DST flag is 0, the time is given in the regular time zone;
if it is 1, the time is given in the DST time zone;
if it is -1, mktime() should guess based on the date and time.

Incidentally, we seem to be 6 hours apart:

>>> time.mktime((2009, 1, 31, 0, 0, 0, 5, 31, -1))
1233356400.0
>>> (1233378000.0 - 1233356400)/(60*60)
6.0

Upvotes: 3

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