Expanding on minitech’s answer:
(start a capture group\da shorthand character class, which matches all numbers; it is the same as[0-9]+one or more of the expression)end a capture group/a literal forward slash
Here is an example:
>>> import re
>>> exp = re.compile('(\d+)/(\d+)')
>>> foo = re.match(exp,'1234/5678')
>>> foo.groups()
('1234', '5678')
If you remove the brackets (), the expression will still match, but you’ll only capture one set:
>>> foo = re.match('\d+/(\d+)','1234/5678')
>>> foo.groups()
('5678',)