Parse date with specific format in python

Question

how would you go about parsing a date like that in python:

     Monday, April 1st

I've tried

    datetime_object = datetime.strptime(date.replace("st","").replace("rd","").replace("th","").replace("nd","").strip(), '%A, %B %d')

But obviously it would remove the "nd" from "Monday" and cause an exception

thanks

Moses Koledoye · Accepted Answer · 2017-09-21 15:05:32Z

2

Don't replace. Strip, from the right using str.rstrip. If the unwanted characters don't exist, the string is returned as is:

>>> from datetime import datetime
>>> s = "Monday, April 1st"
>>> datetime.strptime(s.rstrip('strndh'), '%A, %B %d')
datetime.datetime(1900, 4, 1, 0, 0)

_{Note that the day information here (i.e. Monday) is redundant.}

edited Sep 21, 2017 at 15:05

answered Sep 21, 2017 at 14:57

Moses Koledoye

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1 Comment

fricadelle Over a year ago

yes this is perfect except we need to add 'n' to rstrip :)

Eugene Yarmash · Accepted Answer · 2017-09-21 15:10:54Z

1

You can use the dateutil module (pip install py-dateutil):

>>> from dateutil import parser
>>> parser.parse("Monday, April 1st")
datetime.datetime(2017, 4, 1, 0, 0)

answered Sep 21, 2017 at 15:10

Eugene Yarmash

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Comments

Lupanoide · Accepted Answer · 2017-09-21 15:02:46Z

0

Also if all your string doesn't have the same length:

a = "Monday, April 1st"
if not a[-1].isdigit():
    a = a[:-2]
datetime_object = datetime.strptime(a, '%A, %B %d')

edited Sep 21, 2017 at 15:02

answered Sep 21, 2017 at 14:59

Lupanoide

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