How to find an index of an item in a list, searching the item with a regular expression in Python?

I have a list like that:

lst = ['something', 'foo1', 'bar1', 'blabla', 'foo2']

Is it possible to get the index of the first item starting with "foo" (foo1) using regular expressions and lst.index() like:

ind = lst.index("some_regex_for_the_item_startin开发者_运维知识库g_with_foo") ?

I know I can create a counter and a for loop and use method startswith(). I am curious if I miss some shorter and more elegant way.

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I think that it's ok and you can use startswith method if it do what you really want(i am not sure that you really need regEx here - however code below can be easily modified to use regEx):

data = ['text', 'foo2', 'foo1', 'sample']
indeces = (i for i,val in enumerate(data) if val.startswith('foo'))

Or with regex:

from re import match
data = ['text', 'foo2', 'foo1', 'sample']
indeces = (i for i,val in enumerate(data) if match('foo', val))

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No, unfortunately there is no key parameter for list.index. Having that a solution could have been

# warning: NOT working code
result = L.index(True, key=lambda x: regexp.match(x) is not None)

Moreover given that I just discovered that lambda apparently is considered in the python community an abomination I'm not sure if more key parameters are going to be added in the future.

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There is no way to do it using the lst.index, however here is an alternative method that you may find more elegant than a for loop:

try:
    ind = (i for i, v in enumerate(lst) if v.startswith("foo")).next()
except StopIteration:
    ind = -1   # or however you want to say that the item wasn't found

As senderle pointed out in a comment, this can be shortened by using the next() built-in function (2.6+) with a default value to shorten this to one line:

ind = next((i for i, v in enumerate(lst) if v.startswith("foo")), -1)

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l = ['something', 'foo1', 'bar1', 'blabla', 'foo2']
l.index(filter(lambda x:x.startswith('foo'),l)[0])

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It would be kind of cool to have something like this built in. Python doesn't though. There are a few interesting solutions using itertools. (These also made me wish for a itertools.takewhile_false. If it existed, these would be more readable.)

>>> from itertools import takewhile
>>> import re
>>> m = re.compile('foo.*')
>>> print len(tuple(itertools.takewhile(lambda x: not m.match(x), lst)))
1

That was my first idea, but it requires you to create a temporary tuple and take its length. Then it occurred to me that you could just do a simple sum, and avoid the temporary list:

>>> print sum(1 for _ in takewhile(lambda x: not m.match(x), lst))
1

But that's also somewhat cumbersome. I prefer to avoid throw-away variables when possible. Let's try this again.

>>> sum(takewhile(bool, (not m.match(x) for x in lst)))
1

Much better.