User-friendly time format in Python?

If you happen to be using Django, then new in version 1.4 is the naturaltime template filter.

To use it, first add 'django.contrib.humanize' to your INSTALLED_APPS setting in settings.py, and {% load humanize %} into the template you're using the filter in.

Then, in your template, if you have a datetime variable my_date, you can print its distance from the present by using {{ my_date|naturaltime }}, which will be rendered as something like 4 minutes ago.

Other new things in Django 1.4.

Documentation for naturaltime and other filters in the django.contrib.humanize set.

You can also do that with arrow package

From github page:

>>> import arrow
>>> utc = arrow.utcnow()
>>> utc = utc.shift(hours=-1)
>>> utc.humanize()
'an hour ago'

The code was originally published on a blog post "Python Pretty Date function" (http://evaisse.com/post/93417709/python-pretty-date-function)

It is reproduced here as the blog account has been suspended and the page is no longer available.

def pretty_date(time=False):
    """
    Get a datetime object or a int() Epoch timestamp and return a
    pretty string like 'an hour ago', 'Yesterday', '3 months ago',
    'just now', etc
    """
    from datetime import datetime
    now = datetime.now()
    if type(time) is int:
        diff = now - datetime.fromtimestamp(time)
    elif isinstance(time, datetime):
        diff = now - time
    elif not time:
        diff = 0
    second_diff = diff.seconds
    day_diff = diff.days

    if day_diff < 0:
        return ''

    if day_diff == 0:
        if second_diff < 10:
            return "just now"
        if second_diff < 60:
            return str(second_diff) + " seconds ago"
        if second_diff < 120:
            return "a minute ago"
        if second_diff < 3600:
            return str(second_diff // 60) + " minutes ago"
        if second_diff < 7200:
            return "an hour ago"
        if second_diff < 86400:
            return str(second_diff // 3600) + " hours ago"
    if day_diff == 1:
        return "Yesterday"
    if day_diff < 7:
        return str(day_diff) + " days ago"
    if day_diff < 31:
        return str(day_diff // 7) + " weeks ago"
    if day_diff < 365:
        return str(day_diff // 30) + " months ago"
    return str(day_diff // 365) + " years ago"

In looking for the same thing with the additional requirement that it handle future dates, I found this: http://pypi.python.org/pypi/py-pretty/1

Example code (from site):

from datetime import datetime, timedelta
now = datetime.now()
hrago = now - timedelta(hours=1)
yesterday = now - timedelta(days=1)
tomorrow = now + timedelta(days=1)
dayafter = now + timedelta(days=2)

import pretty
print pretty.date(now)                      # 'now'
print pretty.date(hrago)                    # 'an hour ago'
print pretty.date(hrago, short=True)        # '1h ago'
print pretty.date(hrago, asdays=True)       # 'today'
print pretty.date(yesterday, short=True)    # 'yest'
print pretty.date(tomorrow)                 # 'tomorrow'