how to efficiently get the k bigger elements of a list in python

Use nlargest from heapq module

from heapq import nlargest
lst = [9,1,6,4,2,8,3,7,5]
nlargest(3, lst) # Gives [9,8,7]

You can also give a key to nlargest in case you wanna change your criteria:

from heapq import nlargest
tags = [ ("python", 30), ("ruby", 25), ("c++", 50), ("lisp", 20) ]
nlargest(2, tags, key=lambda e:e[1]) # Gives [ ("c++", 50), ("python", 30) ]

The simple, O(n log n) way is to sort the list then get the last k elements.

The proper way is to use a selection algorithm, which runs in O(n + k log k) time.

Also, heapq.nlargest takes O(k log n) time on average, which may or may not be good enough.

(If k = O(n), then all 3 algorithms have the same complexity (i.e. don't bother). If k = O(log n), then the selection algorithm as described in Wikipedia is O(n) and heapq.nlargest is O(n log log n), but double logarithm is "constant enough" for most practical n that it doesn't matter.)

l = [9,1,6,4,2,8,3,7,5]

sorted(l)[-k:]