Display the date, like “May 5th”, using pythons strftime? [duplicate]

Answer 1

This seems to add the appropriate suffix, and remove the ugly leading zeroes in the day number:

#!/usr/bin/python

import time

day_endings = {
    1: 'st',
    2: 'nd',
    3: 'rd',
    21: 'st',
    22: 'nd',
    23: 'rd',
    31: 'st'
}

def custom_strftime(format, t):
    return time.strftime(format, t).replace('{TH}', str(t[2]) + day_endings.get(t[2], 'th'))

print custom_strftime('%B {TH}, %Y', time.localtime())

Answer 2

You cannot. The time.strftime function and the datetime.datetime.strftime method both (usually) use the platform C library's strftime function, and it (usually) does not offer that format. You would need to use a third-party library, like dateutil.

Answer 3

strftime doesn't allow you to format a date with a suffix.

Here's a way to get the correct suffix:

if 4 <= day <= 20 or 24 <= day <= 30:
    suffix = "th"
else:
    suffix = ["st", "nd", "rd"][day % 10 - 1]

found here

Update:

Combining a more compact solution based on Jochen's comment with gsteff's answer:

from datetime import datetime as dt

def suffix(d):
    return 'th' if 11<=d<=13 else {1:'st',2:'nd',3:'rd'}.get(d%10, 'th')

def custom_strftime(format, t):
    return t.strftime(format).replace('{S}', str(t.day) + suffix(t.day))

print custom_strftime('%B {S}, %Y', dt.now())

Gives:

May 5th, 2011

Answer 4

from time import strftime

print strftime('%A %B %dth')

EDIT:

Correcting after having seen the answers of gurus:

from time import strftime

def special_strftime(dic = {'01':'st','21':'st','31':'st',
                            '02':'nd','22':'nd',
                            '03':'rd','23':'rd'}):
    x = strftime('%A %B %d')
    return x + dic.get(x[-2:],'th')


print special_strftime()

.

EDIT 2

Also:

from time import strftime


def special_strftime(dic = {'1':'st','2':'nd','3':'rd'}):

    x = strftime('%A %B %d')
    return x + ('th' if x[-2:] in ('11','12','13')
                else dic.get(x[-1],'th')

print special_strftime()

.

EDIT 3

Finally, it can be simplified:

from time import strftime

def special_strftime(dic = {'1':'st','2':'nd','3':'rd'}):

    x = strftime('%A %B %d')
    return x + ('th' if x[-2]=='1' else dic.get(x[-1],'th')

print special_strftime()

Answer 5

"%s%s"%(day, 'trnshddt'[0xc0006c000000006c>>2*day&3::4])

But seriously, this is locale specific, so you should be doing it during internationalisation