python – find first sequence item that matches a criterion

The Question :

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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.

The Question Comments :
  • what if you want to get the item and the index?
  • @CharlieParker, to get both the index and the item, use enumerate() – next((idx, obj) for idx, obj in enumerate(objs) if obj.val==5)

The Answer 1

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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).

The Answer 2

5 people think this answer is useful
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))

it’ll return the object if found else it’ll return “not found”

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