Question

Trying to write a RE to recognize date format mm/dd in Python

reg = "(((1[0-2])|(0?[1-9]))/((1[0-9])|(2[0-9])|(3[0-1])|(0?[0-9])))"
match = re.findall(reg, text, re.IGNORECASE)
print match

---

For text = ‘4/13’ it gives me
[('4/13', '4', '', '4', '13', '13', '', '', '')]

Just need the first element. I don’t want inexact matches, how do I remove them.

Thanks,

Cheng

Answer 1

You’re getting all those matches because each set of parentheses in your regular expression generates a match group. You can use a non-grouping match, such as (?:...), if you really don’t want the groups in your result set. You can also simply take the first item from the list and ignore the others.

This would make your expression look like:

reg = "((?:(?:1[0-2])|(?:0?[1-9]))/(?:(?:1[0-9])|(?:2[0-9])|(?:3[0-1])|(?:0?[0-9])))"

See the re documentation for more information.

Here’s a complete example:

>>> text='4/13'
>>> reg = "((?:(?:1[0-2])|(?:0?[1-9]))/(?:(?:1[0-9])|(?:2[0-9])|(?:3[0-1])|(?:0?[0-9])))"
>>> re.findall(reg, text, re.IGNORECASE)
['4/13']