[python] Get the first item from an iterable that matches a condition

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?

The itertools module contains a filter function for iterators. The first element of the filtered iterator can be obtained by calling next() on it:

from itertools import ifilter

print ifilter((lambda i: i > 3), range(10)).next()

You could also use the argwhere function in Numpy. For example:

i) Find the first "l" in "helloworld":

import numpy as np
l = list("helloworld") # Create list
i = np.argwhere(np.array(l)=="l") # i = array([[2],[3],[8]])
index_of_first = i.min()

ii) Find first random number > 0.1

import numpy as np
r = np.random.rand(50) # Create random numbers
i = np.argwhere(r>0.1)
index_of_first = i.min()

iii) Find the last random number > 0.1

import numpy as np
r = np.random.rand(50) # Create random numbers
i = np.argwhere(r>0.1)
index_of_last = i.max()

In Python 3:

a = (None, False, 0, 1)
assert next(filter(None, a)) == 1

In Python 2.6:

a = (None, False, 0, 1)
assert next(iter(filter(None, a))) == 1

EDIT: I thought it was obvious, but apparently not: instead of None you can pass a function (or a lambda) with a check for the condition:

a = [2,3,4,5,6,7,8]
assert next(filter(lambda x: x%2, a)) == 3

The most efficient way in Python 3 are one of the following (using a similar example):

With "comprehension" style:

next(i for i in range(100000000) if i == 1000)

WARNING: The expression works also with Python 2, but in the example is used range that returns an iterable object in Python 3 instead of a list like Python 2 (if you want to construct an iterable in Python 2 use xrange instead).

Note that the expression avoid to construct a list in the comprehension expression next([i for ...]), that would cause to create a list with all the elements before filter the elements, and would cause to process the entire options, instead of stop the iteration once i == 1000.

With "functional" style:

next(filter(lambda i: i == 1000, range(100000000)))

WARNING: This doesn't work in Python 2, even replacing range with xrange due that filter create a list instead of a iterator (inefficient), and the next function only works with iterators.

Default value

As mentioned in other responses, you must add a extra-parameter to the function next if you want to avoid an exception raised when the condition is not fulfilled.

"functional" style:

next(filter(lambda i: i == 1000, range(100000000)), False)

"comprehension" style:

With this style you need to surround the comprehension expression with () to avoid a SyntaxError: Generator expression must be parenthesized if not sole argument:

next((i for i in range(100000000) if i == 1000), False)

For older versions of Python where the next built-in doesn't exist:

(x for x in range(10) if x > 3).next()

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))

Version with default argument

@zorf suggested a version of this function where you can have a predefined return value if the iterable is empty or has no items matching the condition:

def first(iterable, default = None, 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.

    If the `default` argument is given and the iterable is empty,
    or if it has no items matching the condition, the `default` argument
    is returned if it matches the condition.

    The `default` argument being None is the same as it not being given.

    Raises `StopIteration` if no item satisfying the condition is found
    and default is not given or doesn't satisfy the condition.

    >>> first( (1,2,3), condition=lambda x: x % 2 == 0)
    2
    >>> first(range(3, 100))
    3
    >>> first( () )
    Traceback (most recent call last):
    ...
    StopIteration
    >>> first([], default=1)
    1
    >>> first([], default=1, condition=lambda x: x % 2 == 0)
    Traceback (most recent call last):
    ...
    StopIteration
    >>> first([1,3,5], default=1, condition=lambda x: x % 2 == 0)
    Traceback (most recent call last):
    ...
    StopIteration
    """

    try:
        return next(x for x in iterable if condition(x))
    except StopIteration:
        if default is not None and condition(default):
            return default
        else:
            raise

Oneliner:

thefirst = [i for i in range(10) if i > 3][0]

If youre not sure that any element will be valid according to the criteria, you should enclose this with try/except since that [0] can raise an IndexError.

For anyone using Python 3.8 or newer I recommend using "Assignment Expressions" as described in PEP 572 -- Assignment Expressions.

if any((match := i) > 3 for i in range(10)):
    print(match)

Damn Exceptions!

I love this answer. However, since next() raise a StopIteration exception when there are no items, i would use the following snippet to avoid an exception:

a = []
item = next((x for x in a), None)

For example,

a = []
item = next(x for x in a)

Will raise a StopIteration exception;

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

By using

(index for index, value in enumerate(the_iterable) if condition(value))

one can check the condition of the value of the first item in the_iterable, and obtain its index without the need to evaluate all of the items in the_iterable.

The complete expression to use is

first_index = next(index for index, value in enumerate(the_iterable) if condition(value))

Here first_index assumes the value of the first value identified in the expression discussed above.

I would write this

next(x for x in xrange(10) if x > 3)

This question already has great answers. I'm only adding my two cents because I landed here trying to find a solution to my own problem, which is very similar to the OP.

If you want to find the INDEX of the first item matching a criteria using generators, you can simply do:

next(index for index, value in enumerate(iterable) if condition)

Similar to using ifilter, you could use a generator expression:

>>> (x for x in xrange(10) if x > 5).next()
6

In either case, you probably want to catch StopIteration though, in case no elements satisfy your condition.

Technically speaking, I suppose you could do something like this:

>>> foo = None
>>> for foo in (x for x in xrange(10) if x > 5): break
... 
>>> foo
6

It would avoid having to make a try/except block. But that seems kind of obscure and abusive to the syntax.