item = heappop(heap) # pops the smallest item from the heap
item = heap[0] # smallest item on the heap without popping it
heapify(x) # transforms list into a heap, in-place, in linear time
+item = heapreplace(heap, item) # pops and returns smallest item, and adds
+ # new item; the heap size is unchanged
Our API differs from textbook heap algorithms as follows:
returnitem = lastelt
return returnitem
+def heapreplace(heap, item):
+ """Pop and return the current smallest value, and add the new item.
+
+ This is more efficient than heappop() followed by heappush(), and can be
+ more appropriate when using a fixed-size heap. Note that the value
+ returned may be larger than item! That constrains reasonable uses of
+ this routine.
+ """
+
+ if heap:
+ returnitem = heap[0]
+ heap[0] = item
+ _siftup(heap, 0)
+ return returnitem
+ heap.pop() # raise IndexError
+
def heapify(x):
"""Transform list into a heap, in-place, in O(len(heap)) time."""
n = len(x)
from test.test_support import verify, vereq, verbose, TestFailed
-from heapq import heappush, heappop, heapify
+from heapq import heappush, heappop, heapify, heapreplace
import random
def check_invariant(heap):
heapify(heap)
for item in data[10:]:
if item > heap[0]: # this gets rarer the longer we run
- heappop(heap) # we know heap[0] isn't in best 10 anymore
- heappush(heap, item)
+ heapreplace(heap, item)
vereq(list(heapiter(heap)), data_sorted[-10:])
# Make user happy
if verbose: