Get the first item from an iterable that matches a condition
Python 2.6+ and Python 3:
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 whenever 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;-).
Python <= 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.
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
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