python: find first string in string

Question

Given a string and a list of substrings I want to the first position any substring occurs in the string. If no substring occurs, return 0. I want to ignore case.

Is there something more pythonic than:

given = 'Iamfoothegreat'
targets = ['foo', 'bar', 'grea', 'other']
res = len(given)
for t in targets:
    i = given.lower().find(t)
    if i > -1 and i < res:
        res = i

if res == len(given):
    result = 0
else:
    result = res

That code works, but seems inefficient.

Answer 1

Use regex

Another example just use regex, cause think the python regex implementation is super fast. Not my regex function is

import re

given = 'IamFoothegreat'
targets = ['foo', 'bar', 'grea', 'other']

targets = [re.escape(x) for x in targets]    
pattern = r"%(pattern)s" % {'pattern' : "|".join(targets)}
firstMatch = next(re.finditer(pattern, given, re.IGNORECASE),None)
if firstMatch:
    print firstMatch.start()
    print firstMatch.group()

Output is

3
foo

If nothing is found output is nothing. Should be self explained to check if nothing is found.

Much more normal not really pythonic

Give you the matched string, too

given = 'Iamfoothegreat'.lower()
targets = ['foo', 'bar', 'grea', 'other']

dct = {'pos' : - 1, 'string' : None};
given = given.lower()

for t in targets:
    i = given.find(t)
    if i > -1 and (i < list['pos'] or list['pos'] == -1):
        dct['pos'] = i;
        dct['string'] = t;

print dct

Output is:

{'pos': 3, 'string': 'foo'}

If element is not found:

{'pos': -1, 'string': None}

Performance Comparision of both

with this string and pattern

given = "hello world" * 5000
given += "grea" + given
targets = ['foo', 'bar', 'grea', 'other']

1000 loops with timeit:

regex approach: 4.08629107475 sec for 1000
normal approach: 1.80048894882 sec for 1000

10 loops. Now with much bigger targets (targets * 1000):

normal approach: 4.06895017624 for 10
regex approach: 34.8153910637 for 10