I have couple of date string with following pattern MM DD(st, nd, rd, th) YYYY HH:MM am. what is the most pythonic way for me to replace (st, nd, rd, th) as empty string ''?
s = ['st', 'nd', 'rd', 'th']
string = 'Mar 1st 2017 00:00 am'
string = 'Mar 2nd 2017 00:00 am'
string = 'Mar 3rd 2017 00:00 am'
string = 'Mar 4th 2017 00:00 am'
for i in s:
a = string.replace(i, '')
a = [string.replace(i, '') for i in s][0]
print(a)
The most pythonic way is to use dateutil.
from dateutil.parser import parse
import datetime
t = parse("Mar 2nd 2017 00:00 am")
# you can access the month, hour, minute, etc:
t.hour # 0
t.minute # 0
t.month # 3
And then, you can use t.strftime(), where the formatting of the resulting string is any of these: http://strftime.org/
If you want a more appropriate representation of the time(like for example in your proper locale), then you could do t.strftime("%c"), or you could easily format it to the answer you wanted above.
This is much safer than a regex match because dateutil is a part of the standard library, and returns to you a concise datetime object.
You could use a regular expression as follows:
import re
strings = ['Mar 1st 2017 00:00 am', 'Mar 2nd 2017 00:00 am', 'Mar 3rd 2017 00:00 am', 'Mar 4th 2017 00:00 am']
for string in strings:
print(re.sub('(.*? \d+)(.*?)( .*)', r'\1\3', string))
This would give you:
Mar 1 2017 00:00 am
Mar 2 2017 00:00 am
Mar 3 2017 00:00 am
Mar 4 2017 00:00 am
If you want to restrict it do just st nd rd th:
print(re.sub('(.*? \d+)(st|nd|rd|th)( .*)', r'\1\3', string))
