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.
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10
Lib/heapq.py
10
Lib/heapq.py
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@ -13,7 +13,7 @@ heap = [] # creates an empty heap
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heappush(heap, item) # pushes a new item on the heap
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item = heappop(heap) # pops the smallest item from the heap
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item = heap[0] # smallest item on the heap without popping it
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heapify(heap) # transform list into a heap, in-place, in linear time
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heapify(x) # transforms list into a heap, in-place, in linear time
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Our API differs from textbook heap algorithms as follows:
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@ -175,16 +175,16 @@ def heappop(heap):
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returnitem = lastelt
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return returnitem
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def heapify(heap):
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"""Transform list heap into a heap, in-place, in O(len(heap)) time."""
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n = len(heap)
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def heapify(x):
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"""Transform list into a heap, in-place, in O(len(heap)) time."""
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n = len(x)
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# Transform bottom-up. The largest index there's any point to looking at
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# is the largest with a child index in-range, so must have 2*i + 1 < n,
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# or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
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# j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is
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# (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
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for i in xrange(n//2 - 1, -1, -1):
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_siftdown(heap, i)
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_siftdown(x, i)
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if __name__ == "__main__":
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# Simple sanity test
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@ -12,6 +12,20 @@ def check_invariant(heap):
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parentpos = (pos-1) >> 1
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verify(heap[parentpos] <= item)
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# An iterator returning a heap's elements, smallest-first.
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class heapiter(object):
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def __init__(self, heap):
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self.heap = heap
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def next(self):
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try:
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return heappop(self.heap)
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except IndexError:
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raise StopIteration
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def __iter__(self):
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return self
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def test_main():
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# 1) Push 100 random numbers and pop them off, verifying all's OK.
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heap = []
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@ -47,17 +61,16 @@ def test_main():
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check_invariant(heap)
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# 5) Less-naive "N-best" algorithm, much faster (if len(data) is big
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# enough <wink>) than sorting all of data. However, if we had a max
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# heap instead of a min heap, it would go much faster still via
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# heap instead of a min heap, it could go faster still via
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# heapify'ing all of data (linear time), then doing 10 heappops
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# (10 log-time steps).
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heap = data[:10]
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heapify(heap)
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for item in data[10:]:
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if item > heap[0]: # this gets rarer and rarer the longer we run
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if item > heap[0]: # this gets rarer the longer we run
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heappop(heap) # we know heap[0] isn't in best 10 anymore
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heappush(heap, item)
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heappop(heap)
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heap.sort()
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vereq(heap, data_sorted[-10:])
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vereq(list(heapiter(heap)), data_sorted[-10:])
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# Make user happy
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if verbose:
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print "All OK"
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