How can I produce a human readable difference when subtracting two UNIX timestamps using Python?

Question

This question is similar to this question about subtracting dates with Python, but not identical. I'm not dealing with strings, I have to figure out the difference between two epoch time stamps and produce the difference in a human readable format.

For instance:

32 Seconds 17 Minutes 22.3 Hours 1.25 Days 3.5 Weeks 2 Months 4.25 Years 

Alternately, I'd like to express the difference like this:

4 years, 6 months, 3 weeks, 4 days, 6 hours 21 minutes and 15 seconds 

I don't think I can use strptime, since I'm working with the difference of two epoch dates. I could write something to do this, but I'm quite sure that there's something already written that I could use.

What module would be appropriate? Am I just missing something in time? My journey into Python is just really beginning, if this is indeed a duplicate it's because I failed to figure out what to search for.

Addendum

For accuracy, I really care most about the current year's calendar.

Answer 1

You can use the wonderful dateutil module and its relativedelta class:

import datetime import dateutil.relativedelta  dt1 = datetime.datetime.fromtimestamp(123456789) # 1973-11-29 22:33:09 dt2 = datetime.datetime.fromtimestamp(234567890) # 1977-06-07 23:44:50 rd = dateutil.relativedelta.relativedelta (dt2, dt1)  print "%d years, %d months, %d days, %d hours, %d minutes and %d seconds" % (rd.years, rd.months, rd.days, rd.hours, rd.minutes, rd.seconds) # 3 years, 6 months, 9 days, 1 hours, 11 minutes and 41 seconds 

It doesn't count weeks, but that shouldn't be too hard to add.

Answer 2

A little improvement over @Schnouki's solution with a single line list comprehension. Also displays the plural in case of plural entities (like hours)

Import relativedelta

>>> from dateutil.relativedelta import relativedelta 

A lambda function

>>> attrs = ['years', 'months', 'days', 'hours', 'minutes', 'seconds'] >>> human_readable = lambda delta: ['%d %s' % (getattr(delta, attr), attr if getattr(delta, attr) > 1 else attr[:-1])  ...     for attr in attrs if getattr(delta, attr)] 

Example usage:

>>> human_readable(relativedelta(minutes=125)) ['2 hours', '5 minutes'] >>> human_readable(relativedelta(hours=(24 * 365) + 1)) ['365 days', '1 hour']