Minor fiddling, including a simple class to implement a heap iterator

in the test file.  I have docs for heapq.heapify ready to check in, but
Jack appears to have left behind a stale lock in the Doc/lib directory.
This commit is contained in:
Tim Peters 2002-08-03 02:11:26 +00:00
parent 0e0a479821
commit aa7d24319e
2 changed files with 23 additions and 10 deletions

View file

@ -13,7 +13,7 @@
heappush(heap, item) # pushes a new item on the heap
item = heappop(heap) # pops the smallest item from the heap
item = heap[0] # smallest item on the heap without popping it
heapify(heap) # transform list into a heap, in-place, in linear time
heapify(x) # transforms list into a heap, in-place, in linear time
Our API differs from textbook heap algorithms as follows:
@ -175,16 +175,16 @@ def heappop(heap):
returnitem = lastelt
return returnitem
def heapify(heap):
"""Transform list heap into a heap, in-place, in O(len(heap)) time."""
n = len(heap)
def heapify(x):
"""Transform list into a heap, in-place, in O(len(heap)) 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 xrange(n//2 - 1, -1, -1):
_siftdown(heap, i)
_siftdown(x, i)
if __name__ == "__main__":
# Simple sanity test

View file

@ -12,6 +12,20 @@ def check_invariant(heap):
parentpos = (pos-1) >> 1
verify(heap[parentpos] <= item)
# An iterator returning a heap's elements, smallest-first.
class heapiter(object):
def __init__(self, heap):
self.heap = heap
def next(self):
try:
return heappop(self.heap)
except IndexError:
raise StopIteration
def __iter__(self):
return self
def test_main():
# 1) Push 100 random numbers and pop them off, verifying all's OK.
heap = []
@ -47,17 +61,16 @@ def test_main():
check_invariant(heap)
# 5) Less-naive "N-best" algorithm, much faster (if len(data) is big
# enough <wink>) than sorting all of data. However, if we had a max
# heap instead of a min heap, it would go much faster still via
# heap instead of a min heap, it could go faster still via
# heapify'ing all of data (linear time), then doing 10 heappops
# (10 log-time steps).
heap = data[:10]
heapify(heap)
for item in data[10:]:
if item > heap[0]: # this gets rarer and rarer the longer we run
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)
heappop(heap)
heap.sort()
vereq(heap, data_sorted[-10:])
vereq(list(heapiter(heap)), data_sorted[-10:])
# Make user happy
if verbose:
print "All OK"