python - Find first sequence item that matches a criterion

346
votes

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.

2
pythonlist
what if you want to get the item and the index? - Charlie Parker
@CharlieParker, to get both the index and the item, use enumerate() - next((idx, obj) for idx, obj in enumerate(objs) if obj.val==5) - Captain Lepton

2 Answers

632
votes

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 58.3µs (100 times faster because the loop 100 times shorter).

5
votes
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"