Reputation: 13699
I was under the impression that the first value was what determined a values position in the heap, however that doesn't seem to be the case.
from __future__ import print_function
import heapq
q = []
heapq.heappush(q, (10, 11))
heapq.heappush(q, (11, 12))
heapq.heappush(q, (9, 10))
print(q)
This gives me an output of
[(9, 10), (11, 12), (10, 11)]
However I was expecting an output like
[(9, 10), (10, 11), (11, 12)]
Upvotes: 3
Views: 1724
Reputation: 3644
The condition on heapq
is not a "sort guarantee" over the provided list. Instead, it guarantees q[k] <= q[2*k+1]
and q[k] <= q[2*k+2]
(using q
as in your example).
This is due that it is managed internally as a binary tree.
If you simply expect to use the sorted list, you can use the heappop
as shown here. In your specific example you could:
sorted_q = [heappop(q) for i in range(len(q))
and the result, as you expected, will be:
>>> print sorted_q
[(9, 10), (10, 11), (11, 12)]
The theory is explained here in the docs. Relevant is the following line:
The interesting property of a heap is that a[0] is always its smallest element.
Which is a direct result of the condition q[k] <= q[2*k+1]
and q[k] <= q[2*k+2]
, which is a condition of the heap.
However, there are no further guarantees about the order on the rest of the array. And, in fact, both following trees are valid heaps:
0
1 2
2 5 3 4
and
0
2 1
5 3 4 2
Which are stored, respectively, as
[0, 1, 2, 2, 5, 3, 4]
and
[0, 2, 1, 5, 3, 4, 2]
Upvotes: 4