日期序数输出？

Question

我想知道在python中给出一个数字是否有一种快速简便的输出序数的方法.

例如,给定数字1,我想输出"1st",数字2,"2nd"等等等.

这是用于处理面包屑路径中的日期

Home >  Venues >  Bar Academy >  2009 >  April >  01 

是目前显示的

我想要有一些东西

Home >  Venues >  Bar Academy >  2009 >  April >  1st

Answer 1

或者缩短大卫的回答:

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

Answer 2

这是一个更通用的解决方案:

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

Answer 3

5年前当你问这个问题时不确定它是否存在,但是inflect包有一个功能来完成你正在寻找的东西:

>>> import inflect
>>> p = inflect.engine()
>>> for i in range(1,32):
...     print p.ordinal(i)
...
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
11th
12th
13th
14th
15th
16th
17th
18th
19th
20th
21st
22nd
23rd
24th
25th
26th
27th
28th
29th
30th
31st

Answer 4

这些天我会使用 Arrow http://arrow.readthedocs.io/en/latest/（这在 09 年肯定不存在）

>>> import arrow
>>> from datetime import datetime
>>> arrow.get(datetime.utcnow()).format('Do')
'27th'