find first sequence item that matches a criterion

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"

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).

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