#
heapq
#
heapq.heappush(heap, item)
Push the value item
onto the heap, maintaining the heap invariant.
#
heapq.heappop(heap)
Pop and return the smallest item from the heap, maintaining the heap invariant. If the heap is empty, IndexError is raised. To access the smallest item without popping it, use heap[0]
.
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heapq.heapify(x)
Transform list x
into a heap, in-place, in linear time.
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heapq.heappushpop(heap, item)
Push item
on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush()
followed by a separate heappop()
.
#
heapq.heapreplace(heap, item)
Pop and return the smallest item from the heap, and also push the new item. The heap size doesn’t change. If the heap is empty, IndexError is raised.
#
Source code
# Heap queue algorithm (a.k.a. priority queue)
def heappush(heap, item):
"""Push item onto heap, maintaining the heap invariant."""
heap.append(item)
_siftdown(heap, 0, len(heap)-1)
def heappop(heap):
"""Pop the smallest item off the heap, maintaining the heap invariant."""
lastelt = heap.pop() # raises appropriate IndexError if heap is empty
if heap:
returnitem = heap[0]
heap[0] = lastelt
_siftup(heap, 0)
return returnitem
return lastelt
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 unless written as part of a conditional replacement:
if item > heap[0]:
item = heapreplace(heap, item)
"""
returnitem = heap[0] # raises appropriate IndexError if heap is empty
heap[0] = item
_siftup(heap, 0)
return returnitem
def heappushpop(heap, item):
"""Fast version of a heappush followed by a heappop."""
if heap and heap[0] < item:
item, heap[0] = heap[0], item
_siftup(heap, 0)
return item
def heapify(x):
"""Transform list into a heap, in-place, in O(len(x)) time."""
n = len(x)
# Transform bottom-up. The largest index there's any point to looking at
# is the largest with a child index in-range, so must have 2*i + 1 < n,
# or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
# j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is
# (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
for i in reversed(range(n//2)):
_siftup(x, i)
# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos
# is the index of a leaf with a possibly out-of-order value. Restore the
# heap invariant.
def _siftdown(heap, startpos, pos):
newitem = heap[pos]
# Follow the path to the root, moving parents down until finding a place
# newitem fits.
while pos > startpos:
parentpos = (pos - 1) >> 1
parent = heap[parentpos]
if newitem < parent:
heap[pos] = parent
pos = parentpos
continue
break
heap[pos] = newitem
def _siftup(heap, pos):
endpos = len(heap)
startpos = pos
newitem = heap[pos]
# Bubble up the smaller child until hitting a leaf.
childpos = 2*pos + 1 # leftmost child position
while childpos < endpos:
# Set childpos to index of smaller child.
rightpos = childpos + 1
if rightpos < endpos and not heap[childpos] < heap[rightpos]:
childpos = rightpos
# Move the smaller child up.
heap[pos] = heap[childpos]
pos = childpos
childpos = 2*pos + 1
# The leaf at pos is empty now. Put newitem there, and bubble it up
# to its final resting place (by sifting its parents down).
heap[pos] = newitem
_siftdown(heap, startpos, pos)