Get first list index containing sub string in Python

Question

For lists the method list.index(x) returns the index in the list of the first item whose value is x. But if I want to look inside the items, and not just at the whole items, how do I make the slickest Python method for this?

For example, with

l = ['the cat ate the mouse','the tiger ate the chicken','the horse ate the straw']

this function would return 1 provided with the argument tiger.

Answer 1

A non-slicky method:

def index_containing_substring(the_list, substring):
    for i, s in enumerate(the_list):
        if substring in s:
              return i
    return -1

Answer 2

def find(l, s):
    for i in range(len(l)):
        if l[i].find(s)!=-1:
            return i
    return None # Or -1

Answer 3

This is quite slick and fairly efficient.

>>> def find(lst, predicate):
...     return (i for i, j in enumerate(lst) if predicate(j)).next()
... 
>>> l = ['the cat ate the mouse','the tiger ate the chicken','the horse ate the straw']
>>> find(l, lambda x: 'tiger' in x)
1

Only problem is that it will raise StopIteration if the item is not found (though that is easily remedied).

Answer 4

def first_substring(strings, substring):
    return min(i for i, string in enumerate(strings) if substring in string)

Note: This will raise ValueError in case no match is found, which is better in my opinion.

Answer 5

Variation of abyx solution (optimised to stop when the match is found)

def first_substring(strings, substring):
    return next(i for i, string in enumerate(strings) if substring in string)

If you are pre 2.6 you'll need to put the next() at the end

def first_substring(strings, substring):
    return (i for i, string in enumerate(strings) if substring in string).next()