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.)
>>> import heapq, random
>>> heapq.nlargest(3, (random.gauss(0, 1) for _ in xrange(100)))
[1.9730767232998481, 1.9326532289091407, 1.7762926716966254]
The function in the standard library that does this is heapq.nlargest
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.
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.
If you do not mind using pandas then:
import pandas as pd
N = 10
column_name = 0
pd.DataFrame(your_array).nlargest(N, column_name)
The above code will show you the N largest values along with the index position of each value.
