Relative datetimes in python

reltime.py

def reltime(date, compare_to=None, at='@'):
    r'''Takes a datetime and returns a relative representation of the
    time.
    :param date: The date to render relatively
    :param compare_to: what to compare the date to. Defaults to datetime.now()
    :param at: date/time separator. defaults to "@". "at" is also reasonable.

    >>> from datetime import datetime, timedelta
    >>> today = datetime(2050, 9, 2, 15, 00)
    >>> earlier = datetime(2050, 9, 2, 12)
    >>> reltime(earlier, today)
    'today @ 12pm'
    >>> yesterday = today - timedelta(1)
    >>> reltime(yesterday, compare_to=today)
    'yesterday @ 3pm'
    >>> reltime(datetime(2050, 9, 1, 15, 32), today)
    'yesterday @ 3:32pm'
    >>> reltime(datetime(2050, 8, 31, 16), today)
    'Wednesday @ 4pm (2 days ago)'
    >>> reltime(datetime(2050, 8, 26, 14), today)
    'last Friday @ 2pm (7 days ago)'
    >>> reltime(datetime(2049, 9, 2, 12, 00), today)
    'September 2nd, 2049 @ 12pm (last year)'
    >>> today = datetime(2012, 8, 29, 13, 52)
    >>> last_mon = datetime(2012, 8, 20, 15, 40, 55)
    >>> reltime(last_mon, today)
    'last Monday @ 3:40pm (9 days ago)'
    '''
    def ordinal(n):
        r'''Returns a string ordinal representation of a number
        Taken from: http://stackoverflow.com/a/739301/180718
        '''
        if 10 <= n % 100 < 20:
            return str(n) + 'th'
        else:
            return str(n) + {1 : 'st', 2 : 'nd', 3 : 'rd'}.get(n % 10, "th")

    compare_to = compare_to or datetime.now()
    if date > compare_to:
        return NotImplementedError('reltime only handles dates in the past')
    #get timediff values
    diff = compare_to - date
    if diff.seconds < 60 * 60 * 8: #less than a business day?
        days_ago = diff.days
    else:
        days_ago = diff.days + 1
    months_ago = compare_to.month - date.month
    years_ago = compare_to.year - date.year
    weeks_ago = int(math.ceil(days_ago / 7.0))
    #get a non-zero padded 12-hour hour
    hr = date.strftime('%I')
    if hr.startswith('0'):
        hr = hr[1:]
    wd = compare_to.weekday()
    #calculate the time string
    if date.minute == 0:
        time = '{0}{1}'.format(hr, date.strftime('%p').lower())
    else:
        time = '{0}:{1}'.format(hr, date.strftime('%M%p').lower())
    #calculate the date string
    if days_ago == 0:
        datestr = 'today {at} {time}'
    elif days_ago == 1:
        datestr = 'yesterday {at} {time}'
    elif (wd in (5, 6) and days_ago in (wd+1, wd+2)) or \
            wd + 3 <= days_ago <= wd + 8:
        #this was determined by making a table of wd versus days_ago and
        #divining a relationship based on everyday speech. This is somewhat
        #subjective I guess!
        datestr = 'last {weekday} {at} {time} ({days_ago} days ago)'
    elif days_ago <= wd + 2:
        datestr = '{weekday} {at} {time} ({days_ago} days ago)'
    elif years_ago == 1:
        datestr = '{month} {day}, {year} {at} {time} (last year)'
    elif years_ago > 1:
        datestr = '{month} {day}, {year} {at} {time} ({years_ago} years ago)'
    elif months_ago == 1:
        datestr = '{month} {day} {at} {time} (last month)'
    elif months_ago > 1:
        datestr = '{month} {day} {at} {time} ({months_ago} months ago)'
    else: 
        #not last week, but not last month either
        datestr = '{month} {day} {at} {time} ({days_ago} days ago)'
    return datestr.format(time=time,
                          weekday=date.strftime('%A'),
                          day=ordinal(date.day),
                          days=diff.days,
                          days_ago=days_ago,
                          month=date.strftime('%B'),
                          years_ago=years_ago,
                          months_ago=months_ago,
                          weeks_ago=weeks_ago,
                          year=date.year,
                          at=at)