602 lines
22 KiB
Python
602 lines
22 KiB
Python
"""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 François 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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a 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', 'merge',
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'nlargest', 'nsmallest', 'heappushpop']
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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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return returnitem
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return lastelt
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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 heappushpop(heap, item):
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"""Fast version of a heappush followed by a heappop."""
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if heap and heap[0] < item:
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item, heap[0] = heap[0], item
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_siftup(heap, 0)
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return item
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def heapify(x):
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"""Transform list into a heap, in-place, in O(len(x)) 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(range(n//2)):
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_siftup(x, i)
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def _heappop_max(heap):
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"""Maxheap version of a heappop."""
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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_max(heap, 0)
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return returnitem
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return lastelt
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def _heapreplace_max(heap, item):
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"""Maxheap version of a heappop followed by a heappush."""
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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_max(heap, 0)
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return returnitem
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def _heapify_max(x):
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"""Transform list into a maxheap, in-place, in O(len(x)) time."""
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n = len(x)
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for i in reversed(range(n//2)):
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_siftup_max(x, i)
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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 newitem < parent:
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heap[pos] = parent
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pos = parentpos
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continue
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break
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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 comparison 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 not heap[childpos] < heap[rightpos]:
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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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def _siftdown_max(heap, startpos, pos):
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'Maxheap variant of _siftdown'
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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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heap[pos] = parent
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pos = parentpos
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continue
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break
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heap[pos] = newitem
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def _siftup_max(heap, pos):
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'Maxheap variant of _siftup'
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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 larger 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 larger child.
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rightpos = childpos + 1
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if rightpos < endpos and not heap[rightpos] < heap[childpos]:
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childpos = rightpos
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# Move the larger 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_max(heap, startpos, pos)
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def merge(*iterables, key=None, reverse=False):
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'''Merge multiple sorted inputs into a single sorted output.
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Similar to sorted(itertools.chain(*iterables)) but returns a generator,
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does not pull the data into memory all at once, and assumes that each of
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the input streams is already sorted (smallest to largest).
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>>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
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[0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
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If *key* is not None, applies a key function to each element to determine
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its sort order.
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>>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len))
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['dog', 'cat', 'fish', 'horse', 'kangaroo']
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'''
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h = []
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h_append = h.append
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if reverse:
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_heapify = _heapify_max
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_heappop = _heappop_max
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_heapreplace = _heapreplace_max
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direction = -1
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else:
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_heapify = heapify
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_heappop = heappop
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_heapreplace = heapreplace
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direction = 1
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if key is None:
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for order, it in enumerate(map(iter, iterables)):
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try:
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next = it.__next__
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h_append([next(), order * direction, next])
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except StopIteration:
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pass
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_heapify(h)
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while len(h) > 1:
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try:
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while True:
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value, order, next = s = h[0]
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yield value
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s[0] = next() # raises StopIteration when exhausted
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_heapreplace(h, s) # restore heap condition
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except StopIteration:
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_heappop(h) # remove empty iterator
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if h:
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# fast case when only a single iterator remains
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value, order, next = h[0]
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yield value
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yield from next.__self__
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return
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for order, it in enumerate(map(iter, iterables)):
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try:
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next = it.__next__
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value = next()
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h_append([key(value), order * direction, value, next])
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except StopIteration:
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pass
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_heapify(h)
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while len(h) > 1:
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try:
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while True:
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key_value, order, value, next = s = h[0]
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yield value
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value = next()
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s[0] = key(value)
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s[2] = value
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_heapreplace(h, s)
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except StopIteration:
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_heappop(h)
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if h:
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key_value, order, value, next = h[0]
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yield value
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yield from next.__self__
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# Algorithm notes for nlargest() and nsmallest()
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# ==============================================
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#
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# Make a single pass over the data while keeping the k most extreme values
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# in a heap. Memory consumption is limited to keeping k values in a list.
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#
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# Measured performance for random inputs:
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#
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# number of comparisons
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# n inputs k-extreme values (average of 5 trials) % more than min()
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# ------------- ---------------- --------------------- -----------------
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# 1,000 100 3,317 231.7%
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# 10,000 100 14,046 40.5%
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# 100,000 100 105,749 5.7%
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# 1,000,000 100 1,007,751 0.8%
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# 10,000,000 100 10,009,401 0.1%
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#
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# Theoretical number of comparisons for k smallest of n random inputs:
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#
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# Step Comparisons Action
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# ---- -------------------------- ---------------------------
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# 1 1.66 * k heapify the first k-inputs
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# 2 n - k compare remaining elements to top of heap
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# 3 k * (1 + lg2(k)) * ln(n/k) replace the topmost value on the heap
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# 4 k * lg2(k) - (k/2) final sort of the k most extreme values
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#
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# Combining and simplifying for a rough estimate gives:
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#
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# comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k))
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#
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# Computing the number of comparisons for step 3:
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# -----------------------------------------------
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# * For the i-th new value from the iterable, the probability of being in the
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# k most extreme values is k/i. For example, the probability of the 101st
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# value seen being in the 100 most extreme values is 100/101.
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# * If the value is a new extreme value, the cost of inserting it into the
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# heap is 1 + log(k, 2).
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# * The probability times the cost gives:
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# (k/i) * (1 + log(k, 2))
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# * Summing across the remaining n-k elements gives:
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# sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1))
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# * This reduces to:
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# (H(n) - H(k)) * k * (1 + log(k, 2))
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# * Where H(n) is the n-th harmonic number estimated by:
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# gamma = 0.5772156649
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# H(n) = log(n, e) + gamma + 1 / (2 * n)
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# http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence
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# * Substituting the H(n) formula:
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# comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2)
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#
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# Worst-case for step 3:
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# ----------------------
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# In the worst case, the input data is reversed sorted so that every new element
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# must be inserted in the heap:
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#
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# comparisons = 1.66 * k + log(k, 2) * (n - k)
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#
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# Alternative Algorithms
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# ----------------------
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# Other algorithms were not used because they:
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# 1) Took much more auxiliary memory,
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# 2) Made multiple passes over the data.
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# 3) Made more comparisons in common cases (small k, large n, semi-random input).
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# See the more detailed comparison of approach at:
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# http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest
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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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# Short-cut for n==1 is to use min()
|
|
if n == 1:
|
|
it = iter(iterable)
|
|
sentinel = object()
|
|
result = min(it, default=sentinel, key=key)
|
|
return [] if result is sentinel else [result]
|
|
|
|
# When n>=size, it's faster to use sorted()
|
|
try:
|
|
size = len(iterable)
|
|
except (TypeError, AttributeError):
|
|
pass
|
|
else:
|
|
if n >= size:
|
|
return sorted(iterable, key=key)[:n]
|
|
|
|
# When key is none, use simpler decoration
|
|
if key is None:
|
|
it = iter(iterable)
|
|
# put the range(n) first so that zip() doesn't
|
|
# consume one too many elements from the iterator
|
|
result = [(elem, i) for i, elem in zip(range(n), it)]
|
|
if not result:
|
|
return result
|
|
_heapify_max(result)
|
|
top = result[0][0]
|
|
order = n
|
|
_heapreplace = _heapreplace_max
|
|
for elem in it:
|
|
if elem < top:
|
|
_heapreplace(result, (elem, order))
|
|
top, _order = result[0]
|
|
order += 1
|
|
result.sort()
|
|
return [elem for (elem, order) in result]
|
|
|
|
# General case, slowest method
|
|
it = iter(iterable)
|
|
result = [(key(elem), i, elem) for i, elem in zip(range(n), it)]
|
|
if not result:
|
|
return result
|
|
_heapify_max(result)
|
|
top = result[0][0]
|
|
order = n
|
|
_heapreplace = _heapreplace_max
|
|
for elem in it:
|
|
k = key(elem)
|
|
if k < top:
|
|
_heapreplace(result, (k, order, elem))
|
|
top, _order, _elem = result[0]
|
|
order += 1
|
|
result.sort()
|
|
return [elem for (k, order, elem) in result]
|
|
|
|
def nlargest(n, iterable, key=None):
|
|
"""Find the n largest elements in a dataset.
|
|
|
|
Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
|
|
"""
|
|
|
|
# Short-cut for n==1 is to use max()
|
|
if n == 1:
|
|
it = iter(iterable)
|
|
sentinel = object()
|
|
result = max(it, default=sentinel, key=key)
|
|
return [] if result is sentinel else [result]
|
|
|
|
# When n>=size, it's faster to use sorted()
|
|
try:
|
|
size = len(iterable)
|
|
except (TypeError, AttributeError):
|
|
pass
|
|
else:
|
|
if n >= size:
|
|
return sorted(iterable, key=key, reverse=True)[:n]
|
|
|
|
# When key is none, use simpler decoration
|
|
if key is None:
|
|
it = iter(iterable)
|
|
result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)]
|
|
if not result:
|
|
return result
|
|
heapify(result)
|
|
top = result[0][0]
|
|
order = -n
|
|
_heapreplace = heapreplace
|
|
for elem in it:
|
|
if top < elem:
|
|
_heapreplace(result, (elem, order))
|
|
top, _order = result[0]
|
|
order -= 1
|
|
result.sort(reverse=True)
|
|
return [elem for (elem, order) in result]
|
|
|
|
# General case, slowest method
|
|
it = iter(iterable)
|
|
result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)]
|
|
if not result:
|
|
return result
|
|
heapify(result)
|
|
top = result[0][0]
|
|
order = -n
|
|
_heapreplace = heapreplace
|
|
for elem in it:
|
|
k = key(elem)
|
|
if top < k:
|
|
_heapreplace(result, (k, order, elem))
|
|
top, _order, _elem = result[0]
|
|
order -= 1
|
|
result.sort(reverse=True)
|
|
return [elem for (k, order, elem) in result]
|
|
|
|
# If available, use C implementation
|
|
try:
|
|
from _heapq import *
|
|
except ImportError:
|
|
pass
|
|
try:
|
|
from _heapq import _heapreplace_max
|
|
except ImportError:
|
|
pass
|
|
try:
|
|
from _heapq import _heapify_max
|
|
except ImportError:
|
|
pass
|
|
try:
|
|
from _heapq import _heappop_max
|
|
except ImportError:
|
|
pass
|
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
import doctest # pragma: no cover
|
|
print(doctest.testmod()) # pragma: no cover
|