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/Lib/heapq.py

http://unladen-swallow.googlecode.com/
Python | 393 lines | 359 code | 3 blank | 31 comment | 0 complexity | 33f8f0eba1662815249e87b96194fa55 MD5 | raw file
  1# -*- coding: Latin-1 -*-
  2
  3"""Heap queue algorithm (a.k.a. priority queue).
  4
  5Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
  6all k, counting elements from 0.  For the sake of comparison,
  7non-existing elements are considered to be infinite.  The interesting
  8property of a heap is that a[0] is always its smallest element.
  9
 10Usage:
 11
 12heap = []            # creates an empty heap
 13heappush(heap, item) # pushes a new item on the heap
 14item = heappop(heap) # pops the smallest item from the heap
 15item = heap[0]       # smallest item on the heap without popping it
 16heapify(x)           # transforms list into a heap, in-place, in linear time
 17item = heapreplace(heap, item) # pops and returns smallest item, and adds
 18                               # new item; the heap size is unchanged
 19
 20Our API differs from textbook heap algorithms as follows:
 21
 22- We use 0-based indexing.  This makes the relationship between the
 23  index for a node and the indexes for its children slightly less
 24  obvious, but is more suitable since Python uses 0-based indexing.
 25
 26- Our heappop() method returns the smallest item, not the largest.
 27
 28These two make it possible to view the heap as a regular Python list
 29without surprises: heap[0] is the smallest item, and heap.sort()
 30maintains the heap invariant!
 31"""
 32
 33# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger
 34
 35__about__ = """Heap queues
 36
 37[explanation by Fran├žois Pinard]
 38
 39Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
 40all k, counting elements from 0.  For the sake of comparison,
 41non-existing elements are considered to be infinite.  The interesting
 42property of a heap is that a[0] is always its smallest element.
 43
 44The strange invariant above is meant to be an efficient memory
 45representation for a tournament.  The numbers below are `k', not a[k]:
 46
 47                                   0
 48
 49                  1                                 2
 50
 51          3               4                5               6
 52
 53      7       8       9       10      11      12      13      14
 54
 55    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30
 56
 57
 58In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In
 59an usual binary tournament we see in sports, each cell is the winner
 60over the two cells it tops, and we can trace the winner down the tree
 61to see all opponents s/he had.  However, in many computer applications
 62of such tournaments, we do not need to trace the history of a winner.
 63To be more memory efficient, when a winner is promoted, we try to
 64replace it by something else at a lower level, and the rule becomes
 65that a cell and the two cells it tops contain three different items,
 66but the top cell "wins" over the two topped cells.
 67
 68If this heap invariant is protected at all time, index 0 is clearly
 69the overall winner.  The simplest algorithmic way to remove it and
 70find the "next" winner is to move some loser (let's say cell 30 in the
 71diagram above) into the 0 position, and then percolate this new 0 down
 72the tree, exchanging values, until the invariant is re-established.
 73This is clearly logarithmic on the total number of items in the tree.
 74By iterating over all items, you get an O(n ln n) sort.
 75
 76A nice feature of this sort is that you can efficiently insert new
 77items while the sort is going on, provided that the inserted items are
 78not "better" than the last 0'th element you extracted.  This is
 79especially useful in simulation contexts, where the tree holds all
 80incoming events, and the "win" condition means the smallest scheduled
 81time.  When an event schedule other events for execution, they are
 82scheduled into the future, so they can easily go into the heap.  So, a
 83heap is a good structure for implementing schedulers (this is what I
 84used for my MIDI sequencer :-).
 85
 86Various structures for implementing schedulers have been extensively
 87studied, and heaps are good for this, as they are reasonably speedy,
 88the speed is almost constant, and the worst case is not much different
 89than the average case.  However, there are other representations which
 90are more efficient overall, yet the worst cases might be terrible.
 91
 92Heaps are also very useful in big disk sorts.  You most probably all
 93know that a big sort implies producing "runs" (which are pre-sorted
 94sequences, which size is usually related to the amount of CPU memory),
 95followed by a merging passes for these runs, which merging is often
 96very cleverly organised[1].  It is very important that the initial
 97sort produces the longest runs possible.  Tournaments are a good way
 98to that.  If, using all the memory available to hold a tournament, you
 99replace and percolate items that happen to fit the current run, you'll
100produce runs which are twice the size of the memory for random input,
101and much better for input fuzzily ordered.
102
103Moreover, if you output the 0'th item on disk and get an input which
104may not fit in the current tournament (because the value "wins" over
105the last output value), it cannot fit in the heap, so the size of the
106heap decreases.  The freed memory could be cleverly reused immediately
107for progressively building a second heap, which grows at exactly the
108same rate the first heap is melting.  When the first heap completely
109vanishes, you switch heaps and start a new run.  Clever and quite
110effective!
111
112In a word, heaps are useful memory structures to know.  I use them in
113a few applications, and I think it is good to keep a `heap' module
114around. :-)
115
116--------------------
117[1] The disk balancing algorithms which are current, nowadays, are
118more annoying than clever, and this is a consequence of the seeking
119capabilities of the disks.  On devices which cannot seek, like big
120tape drives, the story was quite different, and one had to be very
121clever to ensure (far in advance) that each tape movement will be the
122most effective possible (that is, will best participate at
123"progressing" the merge).  Some tapes were even able to read
124backwards, and this was also used to avoid the rewinding time.
125Believe me, real good tape sorts were quite spectacular to watch!
126From all times, sorting has always been a Great Art! :-)
127"""
128
129__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
130           'nlargest', 'nsmallest', 'heappushpop']
131
132from itertools import islice, repeat, count, imap, izip, tee
133from operator import itemgetter, neg
134import bisect
135
136def heappush(heap, item):
137    """Push item onto heap, maintaining the heap invariant."""
138    heap.append(item)
139    _siftdown(heap, 0, len(heap)-1)
140
141def heappop(heap):
142    """Pop the smallest item off the heap, maintaining the heap invariant."""
143    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
144    if heap:
145        returnitem = heap[0]
146        heap[0] = lastelt
147        _siftup(heap, 0)
148    else:
149        returnitem = lastelt
150    return returnitem
151
152def heapreplace(heap, item):
153    """Pop and return the current smallest value, and add the new item.
154
155    This is more efficient than heappop() followed by heappush(), and can be
156    more appropriate when using a fixed-size heap.  Note that the value
157    returned may be larger than item!  That constrains reasonable uses of
158    this routine unless written as part of a conditional replacement:
159
160        if item > heap[0]:
161            item = heapreplace(heap, item)
162    """
163    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
164    heap[0] = item
165    _siftup(heap, 0)
166    return returnitem
167
168def heappushpop(heap, item):
169    """Fast version of a heappush followed by a heappop."""
170    if heap and heap[0] < item:
171        item, heap[0] = heap[0], item
172        _siftup(heap, 0)
173    return item
174
175def heapify(x):
176    """Transform list into a heap, in-place, in O(len(heap)) time."""
177    n = len(x)
178    # Transform bottom-up.  The largest index there's any point to looking at
179    # is the largest with a child index in-range, so must have 2*i + 1 < n,
180    # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
181    # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
182    # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
183    for i in reversed(xrange(n//2)):
184        _siftup(x, i)
185
186def nlargest(n, iterable):
187    """Find the n largest elements in a dataset.
188
189    Equivalent to:  sorted(iterable, reverse=True)[:n]
190    """
191    it = iter(iterable)
192    result = list(islice(it, n))
193    if not result:
194        return result
195    heapify(result)
196    _heappushpop = heappushpop
197    for elem in it:
198        _heappushpop(result, elem)
199    result.sort(reverse=True)
200    return result
201
202def nsmallest(n, iterable):
203    """Find the n smallest elements in a dataset.
204
205    Equivalent to:  sorted(iterable)[:n]
206    """
207    if hasattr(iterable, '__len__') and n * 10 <= len(iterable):
208        # For smaller values of n, the bisect method is faster than a minheap.
209        # It is also memory efficient, consuming only n elements of space.
210        it = iter(iterable)
211        result = sorted(islice(it, 0, n))
212        if not result:
213            return result
214        insort = bisect.insort
215        pop = result.pop
216        los = result[-1]    # los --> Largest of the nsmallest
217        for elem in it:
218            if los <= elem:
219                continue
220            insort(result, elem)
221            pop()
222            los = result[-1]
223        return result
224    # An alternative approach manifests the whole iterable in memory but
225    # saves comparisons by heapifying all at once.  Also, saves time
226    # over bisect.insort() which has O(n) data movement time for every
227    # insertion.  Finding the n smallest of an m length iterable requires
228    #    O(m) + O(n log m) comparisons.
229    h = list(iterable)
230    heapify(h)
231    return map(heappop, repeat(h, min(n, len(h))))
232
233# 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
234# is the index of a leaf with a possibly out-of-order value.  Restore the
235# heap invariant.
236def _siftdown(heap, startpos, pos):
237    newitem = heap[pos]
238    # Follow the path to the root, moving parents down until finding a place
239    # newitem fits.
240    while pos > startpos:
241        parentpos = (pos - 1) >> 1
242        parent = heap[parentpos]
243        if newitem < parent:
244            heap[pos] = parent
245            pos = parentpos
246            continue
247        break
248    heap[pos] = newitem
249
250# The child indices of heap index pos are already heaps, and we want to make
251# a heap at index pos too.  We do this by bubbling the smaller child of
252# pos up (and so on with that child's children, etc) until hitting a leaf,
253# then using _siftdown to move the oddball originally at index pos into place.
254#
255# We *could* break out of the loop as soon as we find a pos where newitem <=
256# both its children, but turns out that's not a good idea, and despite that
257# many books write the algorithm that way.  During a heap pop, the last array
258# element is sifted in, and that tends to be large, so that comparing it
259# against values starting from the root usually doesn't pay (= usually doesn't
260# get us out of the loop early).  See Knuth, Volume 3, where this is
261# explained and quantified in an exercise.
262#
263# Cutting the # of comparisons is important, since these routines have no
264# way to extract "the priority" from an array element, so that intelligence
265# is likely to be hiding in custom __cmp__ methods, or in array elements
266# storing (priority, record) tuples.  Comparisons are thus potentially
267# expensive.
268#
269# On random arrays of length 1000, making this change cut the number of
270# comparisons made by heapify() a little, and those made by exhaustive
271# heappop() a lot, in accord with theory.  Here are typical results from 3
272# runs (3 just to demonstrate how small the variance is):
273#
274# Compares needed by heapify     Compares needed by 1000 heappops
275# --------------------------     --------------------------------
276# 1837 cut to 1663               14996 cut to 8680
277# 1855 cut to 1659               14966 cut to 8678
278# 1847 cut to 1660               15024 cut to 8703
279#
280# Building the heap by using heappush() 1000 times instead required
281# 2198, 2148, and 2219 compares:  heapify() is more efficient, when
282# you can use it.
283#
284# The total compares needed by list.sort() on the same lists were 8627,
285# 8627, and 8632 (this should be compared to the sum of heapify() and
286# heappop() compares):  list.sort() is (unsurprisingly!) more efficient
287# for sorting.
288
289def _siftup(heap, pos):
290    endpos = len(heap)
291    startpos = pos
292    newitem = heap[pos]
293    # Bubble up the smaller child until hitting a leaf.
294    childpos = 2*pos + 1    # leftmost child position
295    while childpos < endpos:
296        # Set childpos to index of smaller child.
297        rightpos = childpos + 1
298        if rightpos < endpos and not heap[childpos] < heap[rightpos]:
299            childpos = rightpos
300        # Move the smaller child up.
301        heap[pos] = heap[childpos]
302        pos = childpos
303        childpos = 2*pos + 1
304    # The leaf at pos is empty now.  Put newitem there, and bubble it up
305    # to its final resting place (by sifting its parents down).
306    heap[pos] = newitem
307    _siftdown(heap, startpos, pos)
308
309# If available, use C implementation
310try:
311    from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest, heappushpop
312except ImportError:
313    pass
314
315def merge(*iterables):
316    '''Merge multiple sorted inputs into a single sorted output.
317
318    Similar to sorted(itertools.chain(*iterables)) but returns a generator,
319    does not pull the data into memory all at once, and assumes that each of
320    the input streams is already sorted (smallest to largest).
321
322    >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
323    [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
324
325    '''
326    _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
327
328    h = []
329    h_append = h.append
330    for itnum, it in enumerate(map(iter, iterables)):
331        try:
332            next = it.next
333            h_append([next(), itnum, next])
334        except _StopIteration:
335            pass
336    heapify(h)
337
338    while 1:
339        try:
340            while 1:
341                v, itnum, next = s = h[0]   # raises IndexError when h is empty
342                yield v
343                s[0] = next()               # raises StopIteration when exhausted
344                _heapreplace(h, s)          # restore heap condition
345        except _StopIteration:
346            _heappop(h)                     # remove empty iterator
347        except IndexError:
348            return
349
350# Extend the implementations of nsmallest and nlargest to use a key= argument
351_nsmallest = nsmallest
352def nsmallest(n, iterable, key=None):
353    """Find the n smallest elements in a dataset.
354
355    Equivalent to:  sorted(iterable, key=key)[:n]
356    """
357    if key is None:
358        it = izip(iterable, count())                        # decorate
359        result = _nsmallest(n, it)
360        return map(itemgetter(0), result)                   # undecorate
361    in1, in2 = tee(iterable)
362    it = izip(imap(key, in1), count(), in2)                 # decorate
363    result = _nsmallest(n, it)
364    return map(itemgetter(2), result)                       # undecorate
365
366_nlargest = nlargest
367def nlargest(n, iterable, key=None):
368    """Find the n largest elements in a dataset.
369
370    Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
371    """
372    if key is None:
373        it = izip(iterable, imap(neg, count()))             # decorate
374        result = _nlargest(n, it)
375        return map(itemgetter(0), result)                   # undecorate
376    in1, in2 = tee(iterable)
377    it = izip(imap(key, in1), imap(neg, count()), in2)      # decorate
378    result = _nlargest(n, it)
379    return map(itemgetter(2), result)                       # undecorate
380
381if __name__ == "__main__":
382    # Simple sanity test
383    heap = []
384    data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
385    for item in data:
386        heappush(heap, item)
387    sort = []
388    while heap:
389        sort.append(heappop(heap))
390    print sort
391
392    import doctest
393    doctest.testmod()