Find first sequence item that matches a criterion

Question:

What would be the most elegant and efficient way of finding/returning the first list item that matches a certain criterion?

For example, if I have a list of objects and I would like to get the first object of those with attribute obj.val==5. I could of course use list comprehension, but that would incur O(n) and if n is large, it’s wasteful. I could also use a loop with break once the criterion was met, but I thought there could be a more pythonic/elegant solution.

Asked By: Jonathan Livni

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Answers:

If you don’t have any other indexes or sorted information for your objects, then you will have to iterate until such an object is found:

next(obj for obj in objs if obj.val == 5)

This is however faster than a complete list comprehension. Compare these two:

[i for i in xrange(100000) if i == 1000][0]

next(i for i in xrange(100000) if i == 1000)

The first one needs 5.75ms, the second one 58.3┬Ás (100 times faster because the loop 100 times shorter).

Answered By: eumiro

This will return the object if found, else it will return "not found"

a = [100, 200, 300, 400, 500]

def search(b):
    try:
        k = a.index(b)
        return a[k] 
    except ValueError:
        return 'not found'

print(search(500))
Answered By: Ashwini Chaudhary
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