如何在python中将2位数字与字符串匹配？

Question

I have following options, but all of them return full string. I need to remove date at the beginning with regex.

d = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', '01.10.2018Any text..')
d = re.match('^[0-9]{2}.[0-9]{2}.[0-9]{4}(.*?)$', '01.10.2018Any text..')

How to do that? Python 3.6

Answer 1

You could use sub to match the date like pattern (Note that that does not validate a date) from the start of the string ^\\d{2}\\.\\d{2}\\.\\d{4} and replace with an empty string.

And as @UnbearableLightness mentioned, you have to escape the dot \\. if you want to match it literally.

import re
result = re.sub(r'^\d{2}\.\d{2}\.\d{4}', '', '01.10.2018Any text..')
print(result) # Any text..

Demo

Answer 2

Grab the first group of the match

>>> d = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', '01.10.2018Any text..').group(1)
>>> print (d)
'Any text..'

If you are not sure, if there would be a match, you have to check it first

>>> s = '01.10.2018Any text..'
>>> match = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', s)
>>> d = match.group(1) if match else s
>>> print(d)
'Any text..'

Answer 3

Use a group to extract the date part:

d = re.match('^(\d{2}.\d{2}.\d{4})(.*?)$', '01.10.2018Any text..')
if d:
    print(d.group(1))
    print(d.group(2))

Group 0 is the whole string, I added a pair of parentheses in the regex around the date. This is group 1. Group 2 is the text after which is what you're after