481 lines
18 KiB
Python
481 lines
18 KiB
Python
"""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,


nonexisting 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, inplace, 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 0based 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 0based 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




"""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,


nonexisting 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 reestablished.


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 presorted


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 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 fixedsize 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, inplace, in O(len(x)) time."""


n = len(x)


# Transform bottomup. The largest index there's any point to looking at


# is the largest with a child index inrange, so must have 2*i + 1 < n,


# or i < (n1)/2. If n is even = 2*j, this is (2*j1)/2 = j1/2 so


# j1 is the largest, which is n//2  1. If n is odd = 2*j+1, this is


# (2*j+11)/2 = j so j1 is the largest, and that's again n//21.


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, inplace, 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]


"""


from itertools import islice, count, tee, chain


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]


"""


from itertools import islice, count, tee, chain


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 outoforder 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]


"""


from itertools import islice, count, tee, chain


# Shortcut 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]


"""




from itertools import islice, count, tee, chain


# Shortcut 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()
