How to efficiently get the k bigger elements of a list?
Question:
What´s the most efficient, elegant and pythonic way of solving this problem?
Given a list (or set or whatever) of n elements, we want to get the k biggest ones. ( You can assume k<n/2 without loss of generality, I guess)
For example, if the list were:
l = [9,1,6,4,2,8,3,7,5]
n = 9, and let’s say k = 3.
What’s the most efficient algorithm for retrieving the 3 biggest ones?
In this case we should get [9,8,7], in no particular order.
Thanks!
Manuel
Answers:
l = [9,1,6,4,2,8,3,7,5]
sorted(l)[-k:]
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) ]
You can use the heapq module.
>>> from heapq import heapify, nlargest
>>> l = [9,1,6,4,2,8,3,7,5]
>>> heapify(l)
>>> nlargest(3, l)
[9, 8, 7]
>>>
sorted(l, reverse=True)[:k]
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(n log k) 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.)
What´s the most efficient, elegant and pythonic way of solving this problem?
Given a list (or set or whatever) of n elements, we want to get the k biggest ones. ( You can assume k<n/2 without loss of generality, I guess)
For example, if the list were:
l = [9,1,6,4,2,8,3,7,5]
n = 9, and let’s say k = 3.
What’s the most efficient algorithm for retrieving the 3 biggest ones?
In this case we should get [9,8,7], in no particular order.
Thanks!
Manuel
l = [9,1,6,4,2,8,3,7,5]
sorted(l)[-k:]
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) ]
You can use the heapq module.
>>> from heapq import heapify, nlargest
>>> l = [9,1,6,4,2,8,3,7,5]
>>> heapify(l)
>>> nlargest(3, l)
[9, 8, 7]
>>>
sorted(l, reverse=True)[:k]
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(n log k) 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.)