Python format timedelta to string

Question

I'm a Python newbie (2 weeks) and I'm having trouble formatting a datetime.timedelta object.

Here's what I'm trying to do: I have a list of objects and one of the members of the class of the object is a timedelta object that shows the duration of an event. I would like to display that duration in the format of hours:minutes.

I have tried a variety of methods for doing this and I'm having difficulty. My current approach is to add methods to the class for my objects that return hours and minutes. I can get the hours by dividing the timedelta.seconds by 3600 and rounding it. I'm having trouble with getting the remainder seconds and converting that to minutes.

By the way, I'm using Google AppEngine with Django Templates for presentation.

If anyone can help or knows of a better way to resolve this, I would be very happy.

Thanks,

Answer 1

I used the humanfriendly python library to do this, it works very well.

import humanfriendly
from datetime import timedelta
delta = timedelta(seconds = 321)
humanfriendly.format_timespan(delta)

'5 minutes and 21 seconds'

Available at https://pypi.org/project/humanfriendly/

Answer 2

You can just convert the timedelta to a string with str(). Here's an example:

import datetime
start = datetime.datetime(2009,2,10,14,00)
end   = datetime.datetime(2009,2,10,16,00)
delta = end-start
print(str(delta))
# prints 2:00:00

Answer 3

def seconds_to_time_left_string(total_seconds):
    s = int(total_seconds)
    years = s // 31104000
    if years > 1:
        return '%d years' % years
    s = s - (years * 31104000)
    months = s // 2592000
    if years == 1:
        r = 'one year'
        if months > 0:
            r += ' and %d months' % months
        return r
    if months > 1:
        return '%d months' % months
    s = s - (months * 2592000)
    days = s // 86400
    if months == 1:
        r = 'one month'
        if days > 0:
            r += ' and %d days' % days
        return r
    if days > 1:
        return '%d days' % days
    s = s - (days * 86400)
    hours = s // 3600
    if days == 1:
        r = 'one day'
        if hours > 0:
            r += ' and %d hours' % hours
        return r 
    s = s - (hours * 3600)
    minutes = s // 60
    seconds = s - (minutes * 60)
    if hours >= 6:
        return '%d hours' % hours
    if hours >= 1:
        r = '%d hours' % hours
        if hours == 1:
            r = 'one hour'
        if minutes > 0:
            r += ' and %d minutes' % minutes
        return r
    if minutes == 1:
        r = 'one minute'
        if seconds > 0:
            r += ' and %d seconds' % seconds
        return r
    if minutes == 0:
        return '%d seconds' % seconds
    if seconds == 0:
        return '%d minutes' % minutes
    return '%d minutes and %d seconds' % (minutes, seconds)

for i in range(10):
    print pow(8, i), seconds_to_time_left_string(pow(8, i))


Output:
1 1 seconds
8 8 seconds
64 one minute and 4 seconds
512 8 minutes and 32 seconds
4096 one hour and 8 minutes
32768 9 hours
262144 3 days
2097152 24 days
16777216 6 months
134217728 4 years

Answer 4

He already has a timedelta object so why not use its built-in method total_seconds() to convert it to seconds, then use divmod() to get hours and minutes?

hours, remainder = divmod(myTimeDelta.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)

# Formatted only for hours and minutes as requested
print '%s:%s' % (hours, minutes)

This works regardless if the time delta has even days or years.

Answer 5

I had a similar problem with the output of overtime calculation at work. The value should always show up in HH:MM, even when it is greater than one day and the value can get negative. I combined some of the shown solutions and maybe someone else find this solution useful. I realized that if the timedelta value is negative most of the shown solutions with the divmod method doesn't work out of the box:

def td2HHMMstr(td):
  '''Convert timedelta objects to a HH:MM string with (+/-) sign'''
  if td < datetime.timedelta(seconds=0):
    sign='-'
    td = -td
  else:
    sign = ''
  tdhours, rem = divmod(td.total_seconds(), 3600)
  tdminutes, rem = divmod(rem, 60)
  tdstr = '{}{:}:{:02d}'.format(sign, int(tdhours), int(tdminutes))
  return tdstr

timedelta to HH:MM string:

td2HHMMstr(datetime.timedelta(hours=1, minutes=45))
'1:54'

td2HHMMstr(datetime.timedelta(days=2, hours=3, minutes=2))
'51:02'

td2HHMMstr(datetime.timedelta(hours=-3, minutes=-2))
'-3:02'

td2HHMMstr(datetime.timedelta(days=-35, hours=-3, minutes=-2))
'-843:02'

Answer 6

I personally use the humanize library for this:

>>> import datetime
>>> humanize.naturalday(datetime.datetime.now())
'today'
>>> humanize.naturalday(datetime.datetime.now() - datetime.timedelta(days=1))
'yesterday'
>>> humanize.naturalday(datetime.date(2007, 6, 5))
'Jun 05'
>>> humanize.naturaldate(datetime.date(2007, 6, 5))
'Jun 05 2007'
>>> humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=1))
'a second ago'
>>> humanize.naturaltime(datetime.datetime.now() - datetime.timedelta(seconds=3600))
'an hour ago'

Of course, it doesn't give you exactly the answer you were looking for (which is, indeed, str(timeA - timeB), but I have found that once you go beyond a few hours, the display becomes quickly unreadable. humanize has support for much larger values that are human-readable, and is also well localized.

It's inspired by Django's contrib.humanize module, apparently, so since you are using Django, you should probably use that.