Python human readable date difference

Question

I have given date strings like these:

Mon Jun 28 10:51:07 2010
Fri Jun 18 10:18:43 2010
Wed Dec 15 09:18:43 2010

What is a handy python way to calculate the difference in days? Assuming the time zone is the same.

The strings were returned by linux commands.

Edit: Thank you, so many good answers

Answer 1

Use strptime.

Sample usage:

from datetime import datetime

my_date = datetime.strptime('Mon Jun 28 10:51:07 2010', '%a %b %d %H:%M:%S %Y')
print my_date

EDIT:

You could also print the time difference in a human readable form, like so:

from time import strptime
from datetime import datetime

def date_diff(older, newer):
    """
    Returns a humanized string representing time difference

    The output rounds up to days, hours, minutes, or seconds.
    4 days 5 hours returns '4 days'
    0 days 4 hours 3 minutes returns '4 hours', etc...
    """

    timeDiff = newer - older
    days = timeDiff.days
    hours = timeDiff.seconds/3600
    minutes = timeDiff.seconds%3600/60
    seconds = timeDiff.seconds%3600%60

    str = ""
    tStr = ""
    if days > 0:
        if days == 1:   tStr = "day"
        else:           tStr = "days"
        str = str + "%s %s" %(days, tStr)
        return str
    elif hours > 0:
        if hours == 1:  tStr = "hour"
        else:           tStr = "hours"
        str = str + "%s %s" %(hours, tStr)
        return str
    elif minutes > 0:
        if minutes == 1:tStr = "min"
        else:           tStr = "mins"           
        str = str + "%s %s" %(minutes, tStr)
        return str
    elif seconds > 0:
        if seconds == 1:tStr = "sec"
        else:           tStr = "secs"
        str = str + "%s %s" %(seconds, tStr)
        return str
    else:
        return None

older = datetime.strptime('Mon Jun 28 10:51:07 2010', '%a %b %d %H:%M:%S %Y')
newer = datetime.strptime('Tue Jun 28 10:52:07 2010', '%a %b %d %H:%M:%S %Y')
print date_diff(older, newer)

Original source for the time snippet.

Answer 2

>>> import datetime
>>> a = datetime.datetime.strptime("Mon Jun 28 10:51:07 2010", "%a %b %d %H:%M:%S %Y")
>>> b = datetime.datetime.strptime("Fri Jun 18 10:18:43 2010", "%a %b %d %H:%M:%S %Y")
>>> c = a-b
>>> c.days
10

Answer 3

Try this:

>>> (datetime.datetime.strptime("Mon Jun 28 10:51:07 2010", "%a %b %d %H:%M:%S %Y") - datetime.datetime.strptime("Fri Jun 18 10:18:43 2010", "%a %b %d %H:%M:%S %Y")).days
10

Answer 4

#!/usr/bin/env python

import datetime

def hrdd(d1, d2):
    """
    Human-readable date difference.
    """
    _d1 = datetime.datetime.strptime(d1, "%a %b %d %H:%M:%S %Y")
    _d2 = datetime.datetime.strptime(d2, "%a %b %d %H:%M:%S %Y")
    diff = _d2 - _d1
    return diff.days # <-- alternatively: diff.seconds 

if __name__ == '__main__':
    d1 = "Mon Jun 28 10:51:07 2010"
    d2 = "Fri Jun 18 10:18:43 2010"
    d3 = "Wed Dec 15 09:18:43 2010"

    print hrdd(d1, d2)
    # ==> -11
    print hrdd(d2, d1)
    # ==> 10
    print hrdd(d1, d3)
    # ==> 169
    # ...

Answer 5

from datetime import datetime

resp = raw_input("What is the first date ?")
date1 = datetime.strptime(resp,"%a %b %d %H:%M:%S %Y")
resp2 = raw_input("What is the second date ?")
date2 = datetime.strptime(resp2,"%a %b %d %H:%M:%S %Y")
res = date2-date1
print str(res)

For details on how to print a timedelta object better, you can see this previous post.