How to parse date days that contain "st", "nd", "rd", or "th"?
date, python, python-2.7
Solution
You can use `regex` to replace `st`, `nd`, `rd`, `th` with an empty string:
import re
def solve(s):
return re.sub(r'(\d)(st|nd|rd|th)', r'\1', s)
Demo:
>>> datetime.strptime(solve('1st January 2014'), "%d %B %Y")
datetime.datetime(2014, 1, 1, 0, 0)
>>> datetime.strptime(solve('3rd March 2014'), "%d %B %Y")
datetime.datetime(2014, 3, 3, 0, 0)
>>> datetime.strptime(solve('2nd June 2014'), "%d %B %Y")
datetime.datetime(2014, 6, 2, 0, 0)
>>> datetime.strptime(solve('1st August 2014'), "%d %B %Y")
datetime.datetime(2014, 8, 1, 0, 0)
Problem
I have a string like this ``` "1st January 2014" ``` I want to parse it into a `datetime.date`. I can do this: If the date is `1 January 2014` I make this: `replace(' ','')` then `datetime.strptime(SecondDateString, "%d%B%Y").date()` But this doesn't work when the day has `st`, `nd`, `rd`, or `th`. Edit: you may say that I myself remove the `st`, `nd`, `rd`, or `th` and then use my own way above, yes this is a solution but I am asking if python has already had something for me.