344 lines
14 KiB
Python
344 lines
14 KiB
Python
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# -*- coding: Latin-1 -*-
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"""Heap queue algorithm (a.k.a. priority queue).
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
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all k, counting elements from 0. For the sake of comparison,
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non-existing elements are considered to be infinite. The interesting
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property of a heap is that a[0] is always its smallest element.
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Usage:
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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(x) # transforms list into a heap, in-place, in linear time
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item = heapreplace(heap, item) # pops and returns smallest item, and adds
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# new item; the heap size is unchanged
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Our API differs from textbook heap algorithms as follows:
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- We use 0-based indexing. This makes the relationship between the
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index for a node and the indexes for its children slightly less
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obvious, but is more suitable since Python uses 0-based indexing.
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- Our heappop() method returns the smallest item, not the largest.
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These two make it possible to view the heap as a regular Python list
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without surprises: heap[0] is the smallest item, and heap.sort()
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maintains the heap invariant!
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"""
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# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger
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__about__ = """Heap queues
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[explanation by Fran<EFBFBD>ois Pinard]
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
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all k, counting elements from 0. For the sake of comparison,
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non-existing elements are considered to be infinite. The interesting
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property of a heap is that a[0] is always its smallest element.
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The strange invariant above is meant to be an efficient memory
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representation for a tournament. The numbers below are `k', not a[k]:
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0
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1 2
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3 4 5 6
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7 8 9 10 11 12 13 14
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15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In
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an usual binary tournament we see in sports, each cell is the winner
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over the two cells it tops, and we can trace the winner down the tree
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to see all opponents s/he had. However, in many computer applications
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of such tournaments, we do not need to trace the history of a winner.
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To be more memory efficient, when a winner is promoted, we try to
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replace it by something else at a lower level, and the rule becomes
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that a cell and the two cells it tops contain three different items,
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but the top cell "wins" over the two topped cells.
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If this heap invariant is protected at all time, index 0 is clearly
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the overall winner. The simplest algorithmic way to remove it and
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find the "next" winner is to move some loser (let's say cell 30 in the
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diagram above) into the 0 position, and then percolate this new 0 down
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the tree, exchanging values, until the invariant is re-established.
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This is clearly logarithmic on the total number of items in the tree.
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By iterating over all items, you get an O(n ln n) sort.
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A nice feature of this sort is that you can efficiently insert new
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items while the sort is going on, provided that the inserted items are
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not "better" than the last 0'th element you extracted. This is
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especially useful in simulation contexts, where the tree holds all
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incoming events, and the "win" condition means the smallest scheduled
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time. When an event schedule other events for execution, they are
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scheduled into the future, so they can easily go into the heap. So, a
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heap is a good structure for implementing schedulers (this is what I
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used for my MIDI sequencer :-).
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Various structures for implementing schedulers have been extensively
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studied, and heaps are good for this, as they are reasonably speedy,
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the speed is almost constant, and the worst case is not much different
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than the average case. However, there are other representations which
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are more efficient overall, yet the worst cases might be terrible.
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Heaps are also very useful in big disk sorts. You most probably all
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know that a big sort implies producing "runs" (which are pre-sorted
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sequences, which size is usually related to the amount of CPU memory),
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followed by a merging passes for these runs, which merging is often
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very cleverly organised[1]. It is very important that the initial
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sort produces the longest runs possible. Tournaments are a good way
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to that. If, using all the memory available to hold a tournament, you
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replace and percolate items that happen to fit the current run, you'll
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produce runs which are twice the size of the memory for random input,
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and much better for input fuzzily ordered.
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Moreover, if you output the 0'th item on disk and get an input which
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may not fit in the current tournament (because the value "wins" over
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the last output value), it cannot fit in the heap, so the size of the
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heap decreases. The freed memory could be cleverly reused immediately
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for progressively building a second heap, which grows at exactly the
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same rate the first heap is melting. When the first heap completely
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vanishes, you switch heaps and start a new run. Clever and quite
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effective!
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In a word, heaps are useful memory structures to know. I use them in
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a few applications, and I think it is good to keep a `heap' module
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around. :-)
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--------------------
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[1] The disk balancing algorithms which are current, nowadays, are
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more annoying than clever, and this is a consequence of the seeking
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capabilities of the disks. On devices which cannot seek, like big
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tape drives, the story was quite different, and one had to be very
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clever to ensure (far in advance) that each tape movement will be the
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most effective possible (that is, will best participate at
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"progressing" the merge). Some tapes were even able to read
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backwards, and this was also used to avoid the rewinding time.
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Believe me, real good tape sorts were quite spectacular to watch!
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From all times, sorting has always been a Great Art! :-)
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"""
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__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'nlargest',
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'nsmallest']
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from itertools import islice, repeat, count, imap, izip, tee
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from operator import itemgetter, neg
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import bisect
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def heappush(heap, item):
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"""Push item onto heap, maintaining the heap invariant."""
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heap.append(item)
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_siftdown(heap, 0, len(heap)-1)
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def heappop(heap):
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"""Pop the smallest item off the heap, maintaining the heap invariant."""
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lastelt = heap.pop() # raises appropriate IndexError if heap is empty
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if heap:
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returnitem = heap[0]
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heap[0] = lastelt
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_siftup(heap, 0)
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else:
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returnitem = lastelt
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return returnitem
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def heapreplace(heap, item):
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"""Pop and return the current smallest value, and add the new item.
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This is more efficient than heappop() followed by heappush(), and can be
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more appropriate when using a fixed-size heap. Note that the value
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returned may be larger than item! That constrains reasonable uses of
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this routine unless written as part of a conditional replacement:
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if item > heap[0]:
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item = heapreplace(heap, item)
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"""
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returnitem = heap[0] # raises appropriate IndexError if heap is empty
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heap[0] = item
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_siftup(heap, 0)
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return returnitem
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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 reversed(xrange(n//2)):
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_siftup(x, i)
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def nlargest(n, iterable):
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"""Find the n largest elements in a dataset.
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Equivalent to: sorted(iterable, reverse=True)[:n]
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"""
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it = iter(iterable)
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result = list(islice(it, n))
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if not result:
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return result
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heapify(result)
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_heapreplace = heapreplace
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sol = result[0] # sol --> smallest of the nlargest
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for elem in it:
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if elem <= sol:
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continue
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_heapreplace(result, elem)
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sol = result[0]
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result.sort(reverse=True)
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return result
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def nsmallest(n, iterable):
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"""Find the n smallest elements in a dataset.
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Equivalent to: sorted(iterable)[:n]
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"""
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if hasattr(iterable, '__len__') and n * 10 <= len(iterable):
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# For smaller values of n, the bisect method is faster than a minheap.
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# It is also memory efficient, consuming only n elements of space.
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it = iter(iterable)
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result = sorted(islice(it, 0, n))
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if not result:
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return result
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insort = bisect.insort
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pop = result.pop
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los = result[-1] # los --> Largest of the nsmallest
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for elem in it:
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if los <= elem:
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continue
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insort(result, elem)
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pop()
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los = result[-1]
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return result
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# An alternative approach manifests the whole iterable in memory but
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# saves comparisons by heapifying all at once. Also, saves time
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# over bisect.insort() which has O(n) data movement time for every
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# insertion. Finding the n smallest of an m length iterable requires
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# O(m) + O(n log m) comparisons.
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h = list(iterable)
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heapify(h)
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return map(heappop, repeat(h, min(n, len(h))))
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# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos
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# is the index of a leaf with a possibly out-of-order value. Restore the
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# heap invariant.
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def _siftdown(heap, startpos, pos):
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newitem = heap[pos]
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# Follow the path to the root, moving parents down until finding a place
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# newitem fits.
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while pos > startpos:
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parentpos = (pos - 1) >> 1
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parent = heap[parentpos]
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if parent <= newitem:
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break
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heap[pos] = parent
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pos = parentpos
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heap[pos] = newitem
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# The child indices of heap index pos are already heaps, and we want to make
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# a heap at index pos too. We do this by bubbling the smaller child of
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# pos up (and so on with that child's children, etc) until hitting a leaf,
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# then using _siftdown to move the oddball originally at index pos into place.
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#
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# We *could* break out of the loop as soon as we find a pos where newitem <=
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# both its children, but turns out that's not a good idea, and despite that
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# many books write the algorithm that way. During a heap pop, the last array
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# element is sifted in, and that tends to be large, so that comparing it
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# against values starting from the root usually doesn't pay (= usually doesn't
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# get us out of the loop early). See Knuth, Volume 3, where this is
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# explained and quantified in an exercise.
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#
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# Cutting the # of comparisons is important, since these routines have no
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# way to extract "the priority" from an array element, so that intelligence
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# is likely to be hiding in custom __cmp__ methods, or in array elements
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# storing (priority, record) tuples. Comparisons are thus potentially
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# expensive.
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#
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# On random arrays of length 1000, making this change cut the number of
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# comparisons made by heapify() a little, and those made by exhaustive
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# heappop() a lot, in accord with theory. Here are typical results from 3
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# runs (3 just to demonstrate how small the variance is):
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#
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# Compares needed by heapify Compares needed by 1000 heappops
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# -------------------------- --------------------------------
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# 1837 cut to 1663 14996 cut to 8680
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# 1855 cut to 1659 14966 cut to 8678
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# 1847 cut to 1660 15024 cut to 8703
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#
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# Building the heap by using heappush() 1000 times instead required
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# 2198, 2148, and 2219 compares: heapify() is more efficient, when
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# you can use it.
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#
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# The total compares needed by list.sort() on the same lists were 8627,
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# 8627, and 8632 (this should be compared to the sum of heapify() and
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# heappop() compares): list.sort() is (unsurprisingly!) more efficient
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# for sorting.
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def _siftup(heap, pos):
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endpos = len(heap)
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startpos = pos
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newitem = heap[pos]
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# Bubble up the smaller child until hitting a leaf.
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childpos = 2*pos + 1 # leftmost child position
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while childpos < endpos:
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# Set childpos to index of smaller child.
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rightpos = childpos + 1
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if rightpos < endpos and heap[rightpos] <= heap[childpos]:
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childpos = rightpos
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# Move the smaller child up.
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heap[pos] = heap[childpos]
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pos = childpos
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childpos = 2*pos + 1
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# The leaf at pos is empty now. Put newitem there, and bubble it up
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# to its final resting place (by sifting its parents down).
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heap[pos] = newitem
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_siftdown(heap, startpos, pos)
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# If available, use C implementation
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try:
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from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest
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except ImportError:
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pass
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# Extend the implementations of nsmallest and nlargest to use a key= argument
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_nsmallest = nsmallest
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def nsmallest(n, iterable, key=None):
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"""Find the n smallest elements in a dataset.
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Equivalent to: sorted(iterable, key=key)[:n]
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"""
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in1, in2 = tee(iterable)
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it = izip(imap(key, in1), count(), in2) # decorate
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result = _nsmallest(n, it)
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return map(itemgetter(2), result) # undecorate
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_nlargest = nlargest
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def nlargest(n, iterable, key=None):
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"""Find the n largest elements in a dataset.
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Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
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"""
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in1, in2 = tee(iterable)
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it = izip(imap(key, in1), imap(neg, count()), in2) # decorate
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result = _nlargest(n, it)
|
|||
|
|
return map(itemgetter(2), result) # undecorate
|
|||
|
|
|
|||
|
|
if __name__ == "__main__":
|
|||
|
|
# Simple sanity test
|
|||
|
|
heap = []
|
|||
|
|
data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
|
|||
|
|
for item in data:
|
|||
|
|
heappush(heap, item)
|
|||
|
|
sort = []
|
|||
|
|
while heap:
|
|||
|
|
sort.append(heappop(heap))
|
|||
|
|
print sort
|