How to match any string from a list of strings in regular expressions in python?

regex module has named lists (sets actually):

#!/usr/bin/env python
import regex as re # $ pip install regex

p = re.compile(r"\L<words>", words=['fun', 'dum', 'sun', 'gum'])
if p.search("I love to have fun."):
    print('matched')

Here words is just a name, you can use anything you like instead.
.search() methods is used instead of .* before/after the named list.

To emulate named lists using stdlib's re module:

#!/usr/bin/env python
import re

words = ['fun', 'dum', 'sun', 'gum']
longest_first = sorted(words, key=len, reverse=True)
p = re.compile(r'(?:{})'.format('|'.join(map(re.escape, longest_first))))
if p.search("I love to have fun."):
    print('matched')

re.escape() is used to escape regex meta-characters such as .*? inside individual words (to match the words literally).
sorted() emulates regex behavior and it puts the longest words first among the alternatives, compare:

>>> import re
>>> re.findall("(funny|fun)", "it is funny")
['funny']
>>> re.findall("(fun|funny)", "it is funny")
['fun']
>>> import regex
>>> regex.findall(r"\L<words>", "it is funny", words=['fun', 'funny'])
['funny']
>>> regex.findall(r"\L<words>", "it is funny", words=['funny', 'fun'])
['funny']

Join the list on the pipe character |, which represents different options in regex.

string_lst = ['fun', 'dum', 'sun', 'gum']
x="I love to have fun."

print re.findall(r"(?=("+'|'.join(string_lst)+r"))", x)

Output: ['fun']

You cannot use match as it will match from start. Using search you will get only the first match. So use findall instead.

Also use lookahead if you have overlapping matches not starting at the same point.