In Python, what is the efficiency of the in keyword, such as in:
a = [1, 2, 3]
if 4 in a:
...
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How can a keyword be efficient?Wolf– Wolf2016-11-14 09:53:21 +00:00Commented Nov 14, 2016 at 9:53
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2 Answers
3 Comments
johannestaas
In general, but Felix Kling's answer is best. Someone could implement a
MembershipList class which has a base list and also a membership set instance variable, and override __contains__ to check the set for O(1) time.
PirateApp
what about dictionaries @eduardo :)
It depends on the right hand operand:
The operators
inandnot intest for collection membership. [...] The collection membership test has traditionally been bound to sequences; an object is a member of a collection if the collection is a sequence and contains an element equal to that object. However, it make sense for many other object types to support membership tests without being a sequence. In particular, dictionaries (for keys) and sets support membership testing.
Classes can implement the special method __contains__ to override the default behavior (iterating over the sequence) and thus can provide a more (or less) efficient way to test membership than comparing every element of the container.
The membership test operators (
inandnot in) are normally implemented as an iteration through a sequence. However, container objects can supply the following special method with a more efficient implementation, which also does not require the object be a sequence.
Since you have a list in your example, it is iterated over and each element is compared until a match is found or the list is exhausted. The time complexity is usually O(n).
4 Comments
Karol
I suppose it will be the same for strings?
KocT9H
@Felix Kling If list is converted to a set before checking if element is part of it, will it be faster compared to checking directly in list?
Felix Kling
@KocT9H: Depends on how often you do the check. If it's only once, then there will be no difference since, I assume, converting the the list to a set is also
O(n).