CODEWITHSUNDEEP
pythondatetime
Brad Koch
Brad Koch

Reputation: 20267

Parse dates with any separator using python's strptime

I'm parsing a date using Python's datetime.strptime. The dates are in %Y/%m/%d format, but I want to be agnostic about the separator used for the dates - /, -, . and any others are all fine.

Right now, I'm using a try block, but it's a bit verbose.

Is there a simple way to accept any given separator for strptime?

Upvotes: 5

Views: 7721

Answers (4)

dawg
dawg

Reputation: 103744

You can use named groups and a named back-reference:

import re, datetime

txt='''\
1994-8-22
1970 1 1
2014/01/03
2000.1.1
1999 8 27
1991/02-01   bad date'''

for test in txt.splitlines():
    dt=re.match(r'\s*(?P<Y>\d\d\d\d)(?P<sep>\D)(?P<m>\d\d?)(?P=sep)(?P<d>\d\d?)', test)
    if dt:
        print '"{:10}"=>{:20}=>{}'.format(test, 
                                      dt.group(*'Ymd'),
                                      datetime.date(*map(int, dt.group(*'Ymd'))))

That allows you to make sure that the separators are flexible (any non digit) but consistent. ie, this is a good date: 1999/1/2 but this 1999-1/2 maybe is not.

Prints:

"1994-8-22 "=>('1994', '8', '22') =>1994-08-22
"1970 1 1  "=>('1970', '1', '1')  =>1970-01-01
"2014/01/03"=>('2014', '01', '03')=>2014-01-03
"2000.1.1  "=>('2000', '1', '1')  =>2000-01-01
"1999 8 27 "=>('1999', '8', '27') =>1999-08-27

Upvotes: 1

sj7
sj7

Reputation: 1321

from dateutil import parser

Answer = parser.parse(yourtimestring)

Upvotes: 0

Jon Clements
Jon Clements

Reputation: 142116

I'd be tempted to not use strptime, and do something like:

import re
from datetime import datetime

dt = datetime(*map(int, re.findall('\d+', your_string)))

Although your forgo some validation there - such as if there's more than three fields, they'll become arguments to datetime instead of raising an exception that there's unparsed string.

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121416

You'd have to use a regular expression to normalize the separators:

import re

datetimestring = re.sub('[-.:]', '/', datetimestring)

would replace any -, . or : character with a slash.

Alternatively, use the dateutil.parser.parse() function to handle arbitrary date-time formats; it is extremely flexible towards separators.

Upvotes: 6

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