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

Answer 1

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, ...).

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), 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']

Answer 4

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 5

Check out the humanize package

https://github.com/jmoiron/humanize

import datetime

humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=1))

'a second ago'

humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=3600))

'an hour ago'

Answer 6

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.