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.
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
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).
Method 2
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”
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0