User-friendly time format in Python?

Question

Python: I need to show file modification times in the \"1 day ago\", \"two hours ago\", format.

Is there something ready to do that? It should be in English.

Answer 1

This is the gist of @sunil 's post

>>> from datetime import datetime
>>> from dateutil.relativedelta import relativedelta
>>> then = datetime(2003, 9, 17, 20, 54, 47, 282310)
>>> relativedelta(then, datetime.now())
relativedelta(years=-11, months=-3, days=-9, hours=-18, minutes=-17, seconds=-8, microseconds=+912664)

Answer 2

DAY_INCREMENTS = [
    [365, "year"],
    [30, "month"],
    [7, "week"],
    [1, "day"],
]

SECOND_INCREMENTS = [
    [3600, "hour"],
    [60, "minute"],
    [1, "second"],
]


def time_ago(dt):
    diff = datetime.now() - dt  # use timezone.now() or equivalent if `dt` is timezone aware
    if diff.days < 0:
        return "in the future?!?"
    for increment, label in DAY_INCREMENTS:
        if diff.days >= increment:
            increment_diff = int(diff.days / increment)
            return str(increment_diff) + " " + label + plural(increment_diff) + " ago"
    for increment, label in SECOND_INCREMENTS:
        if diff.seconds >= increment:
            increment_diff = int(diff.seconds / increment)
            return str(increment_diff) + " " + label + plural(increment_diff) + " ago"
    return "just now"


def plural(num):
    if num != 1:
        return "s"
    return ""

Answer 3

I have written a detailed blog post for the solution on http://sunilarora.org/17329071 I am posting a quick snippet here as well.

from datetime import datetime
from dateutil.relativedelta import relativedelta

def get_fancy_time(d, display_full_version = False):
    """Returns a user friendly date format
    d: some datetime instace in the past
    display_second_unit: True/False
    """
    #some helpers lambda's
    plural = lambda x: 's' if x > 1 else ''
    singular = lambda x: x[:-1]
    #convert pluran (years) --> to singular (year)
    display_unit = lambda unit, name: '%s %s%s'%(unit, name, plural(unit)) if unit > 0 else ''

    #time units we are interested in descending order of significance
    tm_units = ['years', 'months', 'days', 'hours', 'minutes', 'seconds']

    rdelta = relativedelta(datetime.utcnow(), d) #capture the date difference
    for idx, tm_unit in enumerate(tm_units):
        first_unit_val = getattr(rdelta, tm_unit)
        if first_unit_val > 0:
            primary_unit = display_unit(first_unit_val, singular(tm_unit))
            if display_full_version and idx < len(tm_units)-1:
                next_unit = tm_units[idx + 1]
                second_unit_val = getattr(rdelta, next_unit)
                if second_unit_val > 0:
                    secondary_unit = display_unit(second_unit_val, singular(next_unit))
                    return primary_unit + ', '  + secondary_unit
            return primary_unit
    return None

Answer 4

The answer Jed Smith linked to is good, and I used it for a year or so, but I think it could be improved in a few ways:

• It's nice to be able to define each time unit in terms of the preceding unit, instead of having "magic" constants like 3600, 86400, etc. sprinkled throughout the code.
• After much use, I find I don't want to go to the next unit quite so eagerly. Example: both 7 days and 13 days will show as "1 week"; I'd rather see "7 days" or "13 days" instead.

Here's what I came up with:

def PrettyRelativeTime(time_diff_secs):
    # Each tuple in the sequence gives the name of a unit, and the number of
    # previous units which go into it.
    weeks_per_month = 365.242 / 12 / 7
    intervals = [('minute', 60), ('hour', 60), ('day', 24), ('week', 7),
                 ('month', weeks_per_month), ('year', 12)]

    unit, number = 'second', abs(time_diff_secs)
    for new_unit, ratio in intervals:
        new_number = float(number) / ratio
        # If the new number is too small, don't go to the next unit.
        if new_number < 2:
            break
        unit, number = new_unit, new_number
    shown_num = int(number)
    return '{} {}'.format(shown_num, unit + ('' if shown_num == 1 else 's'))

Notice how every tuple in intervals is easy to interpret and check: a 'minute' is 60 seconds; an 'hour' is 60 minutes; etc. The only fudge is setting weeks_per_month to its average value; given the application, that should be fine. (And note that it's clear at a glance that the last three constants multiply out to 365.242, the number of days per year.)

One downside to my function is that it doesn't do anything outside the "## units" pattern: "Yesterday", "just now", etc. are right out. Then again, the original poster didn't ask for these fancy terms, so I prefer my function for its succinctness and the readability of its numerical constants. :)

Answer 5

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.

Answer 6

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'