If heapq.heapify(list) is O(N) and list.sort() is O(NlogN) then why isn't the default sort algorithm heapify?
Question:
If I have a list and need to sort it, is there a good reason to use list.sort() over heapq.heapify(list), given that heapify is O(N) (link) and .sort() is O(NlogN)?
Why isn’t the default sort algorithm heapify if heapify is faster?
Answers:
as guys have said in the comment section, "heapify" rearranges elements of the list to form a heap-oriented binary tree. complete binary trees can be represented using arrays or in other words, any array which is fully filled with comparable values can be logically viewed as a complete binary tree.
by definition from wikipedia, complete binary tree is:
A complete binary tree is a binary tree in which every level, except
possibly the last, is completely filled, and all nodes in the last
level are as far left as possible.
see an example:
a = [5, 3, 8, 2, 4, 6, 10] —> heapq.heapify(a)
to form a heap you need to satisfy 2 rules:
- form a complete binary tree
- sort the tree so that the children of each node are greater than or equal to the parent node for each node in the tree.
that’s exactly what heapify does, it rearranges elements in the array to satisfy those 2 conditions.
as you can see in the picture above, just by taking the top element (root) each time, you’ll access tree elements in a sorted fashion or in other words you can use heaps to sort your array in n*logn time:
import heapq
a = [5, 3, 8, 2, 4, 6, 10]
def sort_with_heap(a):
"""
sort O(nlogn)
"""
heapq.heapify(a) # heapify's the list O(n)
while len(a) > 0: # O(n) times
yield heapq.heappop(a) # O(logn)
print(list(sort_with_heap(a)))
>> [2, 3, 4, 5, 6, 8, 10]
note: that the example above is for the sake of understanding how heaps can be used for sorting and is not space optimal. to do proper sorting using heaps take a look at heapsort (which works with max-oriented heaps) and doesn’t use any additional space than the sorting array itself. i’ve wrote one myself in here.
If I have a list and need to sort it, is there a good reason to use list.sort() over heapq.heapify(list), given that heapify is O(N) (link) and .sort() is O(NlogN)?
Why isn’t the default sort algorithm heapify if heapify is faster?
as guys have said in the comment section, "heapify" rearranges elements of the list to form a heap-oriented binary tree. complete binary trees can be represented using arrays or in other words, any array which is fully filled with comparable values can be logically viewed as a complete binary tree.
by definition from wikipedia, complete binary tree is:
A complete binary tree is a binary tree in which every level, except
possibly the last, is completely filled, and all nodes in the last
level are as far left as possible.
see an example:
a = [5, 3, 8, 2, 4, 6, 10] —> heapq.heapify(a)
to form a heap you need to satisfy 2 rules:
- form a complete binary tree
- sort the tree so that the children of each node are greater than or equal to the parent node for each node in the tree.
that’s exactly what heapify does, it rearranges elements in the array to satisfy those 2 conditions.
as you can see in the picture above, just by taking the top element (root) each time, you’ll access tree elements in a sorted fashion or in other words you can use heaps to sort your array in n*logn time:
import heapq
a = [5, 3, 8, 2, 4, 6, 10]
def sort_with_heap(a):
"""
sort O(nlogn)
"""
heapq.heapify(a) # heapify's the list O(n)
while len(a) > 0: # O(n) times
yield heapq.heappop(a) # O(logn)
print(list(sort_with_heap(a)))
>> [2, 3, 4, 5, 6, 8, 10]
note: that the example above is for the sake of understanding how heaps can be used for sorting and is not space optimal. to do proper sorting using heaps take a look at heapsort (which works with max-oriented heaps) and doesn’t use any additional space than the sorting array itself. i’ve wrote one myself in here.