> I understand that heapq is not that efficient to implement timeouts as
> I thought at first.
> It would have been perfect if there were functions to remove arbitrary
> elements withouth needing to re-heapify() the heap every time.
It is efficient for that - you just need to use it correctly.
To remove the nth element from a heap, replace it with the last element,
and then _siftup that element:
def remove_at_index_n(heap, n):
if n == len(heap)-1:
return heap.pop()
result = heap[n]
heap[n] = heap.pop()
heapq._siftup(heap, n)
return result
The time complexity for that operation is O(log(len(heap))).
HTH,
Martin
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