Python: take max N elements from some list

38
votes

Is there some function which would return me the N highest elements from some list?

I.e. if max(l) returns the single highest element, sth. like max(l, count=10) would return me a list of the 10 highest numbers (or less if l is smaller).

Or what would be an efficient easy way to get these? (Except the obvious canonical implementation; also, no such things which involve sorting the whole list first because that would be inefficient compared to the canonical solution.)

4
pythonmax
Possible duplicate of stackoverflow.com/q/1034846/64633 - Rod
heapq.nlargest is the way to go for really big lists, but on my system, sorted(l)[:count] is faster up until the list reaches ~25000 elements. - Russell Borogove
sorted(l,reverse=True)[0:N] - M.M

4 Answers

57
votes

heapq.nlargest:

>>> import heapq, random
>>> heapq.nlargest(3, (random.gauss(0, 1) for _ in xrange(100)))
[1.9730767232998481, 1.9326532289091407, 1.7762926716966254]
6
votes

The function in the standard library that does this is heapq.nlargest

3
votes

Start with the first 10 from L, call that X. Note the minimum value of X.

Loop over L[i] for i over the rest of L.

If L[i] is greater than min(X), drop min(X) from X and insert L[i]. You may need to keep X as a sorted linked list and do an insertion. Update min(X).

At the end, you have the 10 largest values in X.

I suspect that will be O(kN) (where k is 10 here) since insertion sort is linear. Might be what gsl uses, so if you can read some C code:

http://www.gnu.org/software/gsl/manual/html_node/Selecting-the-k-smallest-or-largest-elements.html

Probably something in numpy that does this.

1
votes

A fairly efficient solution is a variation of quicksort where recursion is limited to the right part of the pivot until the pivot point position is higher than the number of elements required (with a few extra conditions to deal with border cases of course).

The standard library has heapq.nlargest, as pointed out by others here.