You need to compile all your regex functions. Check this example:

import re
re1 = r'\d+\.\d*[L][-]\d*\s[A-Z]*[/]\d*'
re2 = '\d*[/]\d*[A-Z]*\d*\s[A-Z]*\d*[A-Z]*'
re3 = '[A-Z]*\d+[/]\d+[A-Z]\d+'
re4 = '\d+[/]\d+[A-Z]*\d+\s\d+[A-Z]\s[A-Z]*'

sentences = [string1, string2, string3, string4]
for sentence in sentences:
    generic_re = re.compile("(%s|%s|%s|%s)" % (re1, re2, re3, re4)).findall(sentence)

To findall with an arbitrary series of REs all you have to do is concatenate the list of matches which each returns:

re_list = [
    '\d+\.\d*[L][-]\d*\s[A-Z]*[/]\d*', # re1 in question,
    ...
    '\d+[/]\d+[A-Z]*\d+\s\d+[A-z]\s[A-Z]*', # re4 in question
]

matches = []
for r in re_list:
   matches += re.findall( r, string)

For efficiency it would be better to use a list of compiled REs.

Alternatively you could join the element RE strings using

generic_re = re.compile( '|'.join( re_list) )

I see lots of people are using pipes, but that seems to only match the first instance. If you want to match all, then try using lookaheads.

Example:

>>> fruit_string = "10a11p" 
>>> fruit_regex = r'(?=.*?(?P<pears>\d+)p)(?=.*?(?P<apples>\d+)a)'
>>> re.match(fruit_regex, fruit_string).groupdict()
{'apples': '10', 'pears': '11'}
>>> re.match(fruit_regex, fruit_string).group(0)
'10a,11p'
>>> re.match(fruit_regex, fruit_string).group(1)
'11'

(?= ...) is a look ahead:

Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.

.*?(?P<pears>\d+)p find a number followed a p anywhere in the string and name the number "pears"