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()
