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regexpython-3.xregex-lookaroundsregex-groupregex-greedy

RegEx for matching the month, day and year

I'm trying to find a regular expression to extract the month, day and year from a datetime stamp in this format: 

01/20/2019 12:34:54

It should return a list:

['01', '20', '2019']

I know this can be solved using: 

dt.split(' ')[0].split('/')

But, I'm trying to find a regex to do it:

[^\/\s]+

But, I need it to exclude everything after the space.

Solution

  • As you are expecting the date month and year to be returned as a list, you can use this Python code,

    import re

s = '01/20/2019 12:34:54'
print(re.findall(r'\d+(?=[ /])', s))

    Prints,

    ['01', '20', '2019']

    Otherwise, you can better write your regex as,

    (\d{2})/(\d{2})/(\d{4})

    And get date, month and year from group1, group2 and group3

    Regex Demo

    Python code in this way should be,

    import re

s = '01/20/2019 12:34:54'
m = re.search(r'(\d{2})/(\d{2})/(\d{4})', s)
if m:
 print([m.group(1), m.group(2), m.group(3)])

    Prints,

    ['01', '20', '2019']