python format datetime with "st", "nd", "rd", "th" (english ordinal suffix) like PHP's "S"

Question:

I would like a python datetime object to output (and use the result in django) like this:

Thu the 2nd at 4:30

But I find no way in python to output st, nd, rd, or th like I can with PHP datetime format with the S string (What they call “English Ordinal Suffix”) (http://uk.php.net/manual/en/function.date.php).

Is there a built-in way to do this in django/python? strftime isn’t good enough (http://docs.python.org/library/datetime.html#strftime-strptime-behavior).

Django has a filter which does what I want, but I want a function, not a filter, to do what I want. Either a django or python function will be fine.

Asked By: Alexander Bird

||

Answers:

The django.utils.dateformat has a function format that takes two arguments, the first one being the date (a datetime.date [[or datetime.datetime]] instance, where datetime is the module in Python’s standard library), the second one being the format string, and returns the resulting formatted string. The uppercase-S format item (if part of the format string, of course) is the one that expands to the proper one of ‘st’, ‘nd’, ‘rd’ or ‘th’, depending on the day-of-month of the date in question.

Answered By: Alex Martelli

dont know about built in but I used this…

def ord(n):
    return str(n)+("th" if 4<=n%100<=20 else {1:"st",2:"nd",3:"rd"}.get(n%10, "th"))

and:

def dtStylish(dt,f):
    return dt.strftime(f).replace("{th}", ord(dt.day))

dtStylish can be called as follows to get Thu the 2nd at 4:30. Use {th} where you want to put the day of the month ("%d" python format code)

dtStylish(datetime(2019, 5, 2, 16, 30), '%a the {th} at %I:%M')
Answered By: Frosty Snowman

You can do this simply by using the humanize library

from django.contrib.humanize.templatetags.humanize import ordinal

You can then just give ordinal any integer, ie

ordinal(2) will return 2nd

Answered By: Adrian Krige

I’ve just written a small function to solve the same problem within my own code:

def foo(myDate):
    date_suffix = ["th", "st", "nd", "rd"]

    if myDate % 10 in [1, 2, 3] and myDate not in [11, 12, 13]:
        return date_suffix[myDate % 10]
    else:
        return date_suffix[0]
Answered By: Edster

Following solution I got for above problem:

datetime.strptime(mydate, '%dnd %B %Y')
datetime.strptime(mydate, '%dst %B %Y')
datetime.strptime(mydate, '%dth %B %Y')
datetime.strptime(mydate, '%drd %B %Y')
Answered By: eegloo

My own version. Use python’s strftime’s format codes, substituting {th} where you want to see the ordinal suffix.

def format_date_with_ordinal(d, format_string):
    ordinal = {'1':'st', '2':'nd', '3':'rd'}.get(str(d.day)[-1:], 'th')
    return d.strftime(format_string).replace('{th}', ordinal)

or

format_date_with_ordinal = lambda d,f: d.strftime(f).replace('{th}', {'1':'st', '2':'nd', '3':'rd'}.get(str(d.day)[-1:], 'th'))
Answered By: jobu1342

Here’s something that I use:

def get_ordinal_suffix(day: int) -> str:
    return {1: 'st', 2: 'nd', 3: 'rd'}.get(day % 10, 'th') if day not in (11, 12, 13) else 'th'
Answered By: tschmitz
Categories: questions Tags: , , , ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.