Heapreplace and ordering

Question

I have a list (or heap) of elements and I want to replace the element with the lowest value with a new element of higher value. However if there are multiple elements in the heap with this lowest value, I expect to replace the last element.

Example 1 (works as expected):

input: 
n = 7
new_element = 1
hlist = [0 for x in range(n)]
hp.heapreplace(hlist,new_element)
hlist

output:
[0, 0, 0, 0, 0, 0, 1]

Example 2 (doesn't work as expected):

input: 
n = 10
new_element = 1
hlist = [0 for x in range(n)]
hp.heapreplace(hlist,new_element)
hlist

output:
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0]

The expected output is:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 1]

With n in [1,2,...,9,10] it works when n in [1, 2, 3, 6, 7].

Answer 1

The order you see inside a heap, is not necessarily ascending or descending inside the list that stores the values. So you need to sort it out before printing. According to the documentation, you can define a function named heapsort to sort the heap.

from heapq import *

def heapsort(iterable):
    h = []
    for value in iterable:
        heappush(h, value)
    return [heappop(h) for i in range(len(h))]

n = 7
new_element = 1
hlist = [0 for x in range(n)]
heapify(hlist)
heapreplace(hlist,new_element)
print(heapsort(hlist))

and the output will be :

[0, 0, 0, 0, 0, 0, 1]

---

Also you can solve this problem using lists.

n = 7
new_element = 1
hlist = [0 for x in range(n)]
min_value = min(hlist)
for i in range(n-1,-1,-1):
    if hlist[i]==min_value:
        hlist[i] = new_element
        break

and the output will be :

[0, 0, 0, 0, 0, 0, 1]