How can I produce a human readable difference when

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

def humanize_time(amount, units = 'seconds'):    

    def process_time(amount, units):

        INTERVALS = [   1, 60, 
                        60*60, 
                        60*60*24, 
                        60*60*24*7, 
                        60*60*24*7*4, 
                        60*60*24*7*4*12, 
                        60*60*24*7*4*12*100,
                        60*60*24*7*4*12*100*10]
        NAMES = [('second', 'seconds'),
                 ('minute', 'minutes'),
                 ('hour', 'hours'),
                 ('day', 'days'),
                 ('week', 'weeks'),
                 ('month', 'months'),
                 ('year', 'years'),
                 ('century', 'centuries'),
                 ('millennium', 'millennia')]

        result = []

        unit = map(lambda a: a[1], NAMES).index(units)
        # Convert to seconds
        amount = amount * INTERVALS[unit]

        for i in range(len(NAMES)-1, -1, -1):
            a = amount // INTERVALS[i]
            if a > 0: 
                result.append( (a, NAMES[i][1 % a]) )
                amount -= a * INTERVALS[i]

        return result

    rd = process_time(int(amount), units)
    cont = 0
    for u in rd:
        if u[0] > 0:
            cont += 1

    buf = ''
    i = 0
    for u in rd:
        if u[0] > 0:
            buf += "%d %s" % (u[0], u[1])
            cont -= 1

        if i < (len(rd)-1):
            if cont > 1:
                buf += ", "
            else:
                buf += " and "

        i += 1

    return buf

Example of use:

>>> print humanize_time(234567890 - 123456789)
3 years, 9 months, 3 weeks, 5 days, 11 minutes and 41 seconds
>>> humanize_time(9, 'weeks')
2 months and 1 week

Advantage (You don't need third parties!).

Improved from "Liudmil Mitev" algorithm. (Thanks!)

Answer 2

Old question, but I personally like this approach most:

import datetime
import math

def human_time(*args, **kwargs):
    secs  = float(datetime.timedelta(*args, **kwargs).total_seconds())
    units = [("day", 86400), ("hour", 3600), ("minute", 60), ("second", 1)]
    parts = []
    for unit, mul in units:
        if secs / mul >= 1 or mul == 1:
            if mul > 1:
                n = int(math.floor(secs / mul))
                secs -= n * mul
            else:
                n = secs if secs != int(secs) else int(secs)
            parts.append("%s %s%s" % (n, unit, "" if n == 1 else "s"))
    return ", ".join(parts)

human_time(seconds=3721)
# -> "1 hour, 2 minutes, 1 second"

If you want to separate the seconds part with an "and" do:

"%s and %s" % tuple(human_time(seconds=3721).rsplit(", ", 1))
# -> "1 hour, 2 minutes and 1 second"

Answer 3

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), getattr(delta, attr) > 1 and attr or 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']

Answer 4

I had that exact same problem earlier today and I couldn't find anything in the standard libraries that I could use, so this is what I wrote:

humanize_time.py

    #!/usr/bin/env python

    INTERVALS = [1, 60, 3600, 86400, 604800, 2419200, 29030400]
    NAMES = [('second', 'seconds'),
             ('minute', 'minutes'),
             ('hour', 'hours'),
             ('day', 'days'),
             ('week', 'weeks'),
             ('month', 'months'),
             ('year', 'years')]

    def humanize_time(amount, units):
    """
    Divide `amount` in time periods.
    Useful for making time intervals more human readable.

    >>> humanize_time(173, 'hours')
    [(1, 'week'), (5, 'hours')]
    >>> humanize_time(17313, 'seconds')
    [(4, 'hours'), (48, 'minutes'), (33, 'seconds')]
    >>> humanize_time(90, 'weeks')
    [(1, 'year'), (10, 'months'), (2, 'weeks')]
    >>> humanize_time(42, 'months')
    [(3, 'years'), (6, 'months')]
    >>> humanize_time(500, 'days')
    [(1, 'year'), (5, 'months'), (3, 'weeks'), (3, 'days')]
    """
       result = []

       unit = map(lambda a: a[1], NAMES).index(units)
       # Convert to seconds
       amount = amount * INTERVALS[unit]

       for i in range(len(NAMES)-1, -1, -1):
          a = amount / INTERVALS[i]
          if a > 0:
             result.append( (a, NAMES[i][1 % a]) )
             amount -= a * INTERVALS[i]

       return result

    if __name__ == "__main__":
        import doctest
        doctest.testmod()

You can use dateutil.relativedelta() to calculate the accurate time delta and humanize it with this script.

Answer 5

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 6

Here's a shorter one for interval in seconds and within a day (t<86400). Useful if you work with unix timestamps (seconds since epoch, UTC).

t = 45678
print('%d hours, %d minutes, %d seconds' % (t//3600, t%3600//60, t%60))

May be extended further (t//86400, ...).