Alternative to list comprehension if there will be only one result

Question

I'm starting to get used to list comprehension in Python but I'm afraid I'm using it somewhat improperly. I've run into a scenario a few times where I'm using list comprehension but immediately taking the first (and only) item from the list that is generated. Here is an example:

actor = [actor for actor in self.actors if actor.name==actorName][0]

(self.actors contains a list of objects and I'm trying to get to the one with a specific (string) name, which is in actorName.)

I'm trying to pull out the object from the list that matches the parameter I'm looking for. Is this method unreasonable? The dangling [0] makes me feel a bit insecure.

Don't be afraid, It's not _that_ terrible :) One of the answers below that stop looping when the match is found are probably a better choice. `next()` is not too bad, but can become ugly if you try to make it work in 80 char lines. If you are looking up a bunch of actors by name it's going to be a better idea to make a `dict` so you can look them up directly. — John La Rooy, Aug 10 '11 at 07:05
Fortunately this is not something I am doing frequently or on large sets. I'm fairly good with dictionaries, and will keep that technique in mind. — timfreilly, Aug 10 '11 at 07:46
Possible duplicate of [What is the best way to get the first item from an iterable matching a condition?](http://stackoverflow.com/questions/2361426/what-is-the-best-way-to-get-the-first-item-from-an-iterable-matching-a-condition) — erickrf, Mar 15 '17 at 15:08

score 141 · Accepted Answer · edited May 23 '17 at 12:10

141

You could use a generator expression and next instead. This would be more efficient as well, since an intermediate list is not created and iteration can stop once a match has been found:

actor = next(actor for actor in self.actors if actor.name==actorName)

And as senderle points out, another advantage to this approach is that you can specify a default if no match is found:

actor = next((actor for actor in self.actors if actor.name==actorName), None)

edited May 23 '17 at 12:10

Community

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answered Aug 10 '11 at 06:47

Zach Kelling

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More efficent -- and allows you to specify a default if no match is found. – senderle Aug 10 '11 at 06:50
Thanks. Could you clarify "does not need to be consumed"? Does that mean it will stop iterating once it finds the match? – timfreilly Aug 10 '11 at 07:17
Clarified wording in that regard. – Zach Kelling Aug 10 '11 at 07:20
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Thank you. This is definitely useful information and clearly superior to my method. – timfreilly Aug 10 '11 at 07:45
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Loving Python more and more with these one-liners ! :] – Kostas Demiris Jun 09 '17 at 09:41
Can you check for multiple conditions with this method? – Neal Dec 05 '20 at 15:52

Beni Cherniavsky-Paskin · Answer 2 · 2020-07-26T09:30:20.040

If you want to take the first match of potentially many, next(...) is great. But if you expect exactly one, consider writing it defensively:

[actor] = [actor for actor in self.actors if actor.name==actorName]

This always scans to the end, but unlike [0], the destructuring assignment into [actor] throws a ValueError if there are 0 or more than one match. Perhaps even more important then catching bugs, this communicates your assumption to the reader.

If you want a default for 0 matches, but still catch >1 matches:

[actor] = [actor for actor in self.actors if actor.name==actorName] or [default]

P.S. it's also possible to use a generator expression on right side:

[actor] = (actor for actor in self.actors if actor.name==actorName)

which may be a tiny bit more efficient (?). You could also use tuple syntax on the left side — looks more symmetric but the comma is ugly and too easy to miss IMHO:

(actor,) = (actor for actor in self.actors if actor.name==actorName)
actor, = (actor for actor in self.actors if actor.name==actorName)

(anyway list vs tuple syntax on left side is purely cosmetic doesn't affect behavior)

You've probably seen this: `[a, b] = [b, a]` for swapping, though it's usually shown without brackets. It's called "destructuring assignment", or "unpacking". The brackets/parents on the left side are optional, but are needed for the single-element case. The right side can be any python expression, in this case a list comprehension. — Beni Cherniavsky-Paskin, Aug 11 '11 at 15:27
I see. Thanks for the terminology. I've used packing/unpacking in other situations but have never used it with a single variable to force a single result. — timfreilly, Aug 11 '11 at 23:58

score 2 · Answer 3 · answered Aug 10 '11 at 06:48

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This post has a custom find() function which works quite well, and a commenter there also linked to this method based on generators. Basically, it sounds like there's no single great way to do this — but these solutions aren't bad.

answered Aug 10 '11 at 06:48

jtbandes

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Interesting that there are blog posts on this subject. Thanks for the overview, I had difficulty finding information on this question. – timfreilly Aug 10 '11 at 07:17

score 1 · Answer 4 · answered Aug 10 '11 at 06:47

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Personally I'd to this in a proper loop.

actor = None
for actor in self.actors:
    if actor.name == actorName:
        break

It's quite a bit longer, but it does have the advantage that it stops looping as soon as a match is found.

answered Aug 10 '11 at 06:47

Daniel Roseman

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as does the `next()` solution. – glglgl Aug 10 '11 at 07:21

Alternative to list comprehension if there will be only one result

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