How to replace string using regular expression in Python

Question

I need help with replacing characters in a string using regular expressions.

Input :

s3 = ['March/21/2019' , 'Mar/23/2019']

Desired Output :

s3 = ['03/21/2019' , '03/23/2019']

I tried a few things, but none work:

s3[i] = s3[i].replace(r'Mar[a-z]*', '03')
s3[i] = s3[i].replace(r'(?:Mar[a-z]*)', '03')

If you're only working with dates I think you're better off using a datetime: stackoverflow.com/questions/2265357/… — McGlothlin
– McGlothlin, Commented Apr 8, 2019 at 20:10
replace isn't using regexes. You need re.sub(r'Mar[a-z]*', '03',s3[i]) — Jean-François Fabre
– Jean-François Fabre ♦, Commented Apr 8, 2019 at 20:10
If you want to use regex, you need to import it via import re and use it instead of str.replace(), which doesn't support regex patterns. If your data list is part of a pandas dataframe though, you could use df.str.replace(), which in turn supports them. — SpghttCd
– SpghttCd, Commented Apr 8, 2019 at 20:11
This question is similar to: How to input a regex in string.replace?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Pat Myron
– Pat Myron, Commented Jun 12 at 4:43

schilli · Accepted Answer · 2019-04-08 20:20:02Z

5

This works.

import re
s3 = ['March/21/2019' , 'Mar/23/2019']
s3 = [re.sub(r'Mar[a-z]*', '03', item) for item in s3]

# ['03/21/2019', '03/23/2019']

Of course, you can also use a for loop for better readability.

import re
s3 = ['March/21/2019' , 'Mar/23/2019']
for i in range(len(s3)):
    s3[i] = re.sub(r'Mar[a-z]*', '03', s3[i])

# ['03/21/2019', '03/23/2019']

answered Apr 8, 2019 at 20:20

schilli

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Comments

McGlothlin · Accepted Answer · 2019-04-08 20:43:00Z

1

If you're only working with dates, try something like this:

import datetime

s3 = ['March/21/2019' , 'Mar/23/2019']

for i in range(0, len(s3)):
    try:
        newformat = datetime.datetime.strptime(s3[i], '%B/%d/%Y')
    except ValueError:
        newformat = datetime.datetime.strptime(s3[i], '%b/%d/%Y')

    s3[i] = newformat.strftime('%m/%d/%Y')

s3 now contains ['03/21/2019', '03/23/2019']

answered Apr 8, 2019 at 20:43

McGlothlin

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