Get the first item from an iterable that matches a condition


Question

I would like to get the first item from a list matching a condition. It's important that the resulting method not process the entire list, which could be quite large. For example, the following function is adequate:

def first(the_iterable, condition = lambda x: True):
    for i in the_iterable:
        if condition(i):
            return i

This function could be used something like this:

>>> first(range(10))
0
>>> first(range(10), lambda i: i > 3)
4

However, I can't think of a good built-in / one-liner to let me do this. I don't particularly want to copy this function around if I don't have to. Is there a built-in way to get the first item matching a condition?

1
250
11/8/2017 5:50:16 PM

Accepted Answer

In Python 2.6 or better:

If you want StopIteration to be raised if no matching element is found:

next(x for x in the_iterable if x > 3)

If you want default_value (e.g. None) to be returned instead:

next( (x for x in the_iterable if x>3), default_value)

Note that you need an extra pair of parentheses around the generator expression in this case - they are needed always when the generator expression isn't the only argument.

I see most answers resolutely ignore the next built-in and so I assume that for some mysterious reason they're 100% focused on versions 2.5 and older -- without mentioning the Python-version issue (but then I don't see that mention in the answers that do mention the next built-in, which is why I thought it necessary to provide an answer myself -- at least the "correct version" issue gets on record this way;-).

In 2.5, the .next() method of iterators immediately raises StopIteration if the iterator immediately finishes -- i.e., for your use case, if no item in the iterable satisfies the condition. If you don't care (i.e., you know there must be at least one satisfactory item) then just use .next() (best on a genexp, line for the next built-in in Python 2.6 and better).

If you do care, wrapping things in a function as you had first indicated in your Q seems best, and while the function implementation you proposed is just fine, you could alternatively use itertools, a for...: break loop, or a genexp, or a try/except StopIteration as the function's body, as various answers suggested. There's not much added value in any of these alternatives so I'd go for the starkly-simple version you first proposed.

386
1/18/2017 10:54:26 AM

As a reusable, documented and tested function

def first(iterable, condition = lambda x: True):
    """
    Returns the first item in the `iterable` that
    satisfies the `condition`.

    If the condition is not given, returns the first item of
    the iterable.

    Raises `StopIteration` if no item satysfing the condition is found.

    >>> first( (1,2,3), condition=lambda x: x % 2 == 0)
    2
    >>> first(range(3, 100))
    3
    >>> first( () )
    Traceback (most recent call last):
    ...
    StopIteration
    """

    return next(x for x in iterable if condition(x))

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