Find first sequence item that matches a criterion [duplicate]

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).

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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