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