AppPkg/Applications/Python/Python-2.7.2/Lib/heapq.py

   1 # -*- coding: latin-1 -*-
   2
   3 """Heap queue algorithm (a.k.a. priority queue).
   4
   5 Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
   6 all k, counting elements from 0.  For the sake of comparison,
   7 non-existing elements are considered to be infinite.  The interesting
   8 property of a heap is that a[0] is always its smallest element.
   9
  10 Usage:
  11
  12 heap = []            # creates an empty heap
  13 heappush(heap, item) # pushes a new item on the heap
  14 item = heappop(heap) # pops the smallest item from the heap
  15 item = heap[0]       # smallest item on the heap without popping it
  16 heapify(x)           # transforms list into a heap, in-place, in linear time
  17 item = heapreplace(heap, item) # pops and returns smallest item, and adds
  18                                # new item; the heap size is unchanged
  19
  20 Our 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
  28 These two make it possible to view the heap as a regular Python list
  29 without surprises: heap[0] is the smallest item, and heap.sort()
  30 maintains 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 François Pinard]
  38
  39 Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
  40 all k, counting elements from 0.  For the sake of comparison,
  41 non-existing elements are considered to be infinite.  The interesting
  42 property of a heap is that a[0] is always its smallest element.
  43
  44 The strange invariant above is meant to be an efficient memory
  45 representation 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
  58 In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In
  59 an usual binary tournament we see in sports, each cell is the winner
  60 over the two cells it tops, and we can trace the winner down the tree
  61 to see all opponents s/he had.  However, in many computer applications
  62 of such tournaments, we do not need to trace the history of a winner.
  63 To be more memory efficient, when a winner is promoted, we try to
  64 replace it by something else at a lower level, and the rule becomes
  65 that a cell and the two cells it tops contain three different items,
  66 but the top cell "wins" over the two topped cells.
  67
  68 If this heap invariant is protected at all time, index 0 is clearly
  69 the overall winner.  The simplest algorithmic way to remove it and
  70 find the "next" winner is to move some loser (let's say cell 30 in the
  71 diagram above) into the 0 position, and then percolate this new 0 down
  72 the tree, exchanging values, until the invariant is re-established.
  73 This is clearly logarithmic on the total number of items in the tree.
  74 By iterating over all items, you get an O(n ln n) sort.
  75
  76 A nice feature of this sort is that you can efficiently insert new
  77 items while the sort is going on, provided that the inserted items are
  78 not "better" than the last 0'th element you extracted.  This is
  79 especially useful in simulation contexts, where the tree holds all
  80 incoming events, and the "win" condition means the smallest scheduled
  81 time.  When an event schedule other events for execution, they are
  82 scheduled into the future, so they can easily go into the heap.  So, a
  83 heap is a good structure for implementing schedulers (this is what I
  84 used for my MIDI sequencer :-).
  85
  86 Various structures for implementing schedulers have been extensively
  87 studied, and heaps are good for this, as they are reasonably speedy,
  88 the speed is almost constant, and the worst case is not much different
  89 than the average case.  However, there are other representations which
  90 are more efficient overall, yet the worst cases might be terrible.
  91
  92 Heaps are also very useful in big disk sorts.  You most probably all
  93 know that a big sort implies producing "runs" (which are pre-sorted
  94 sequences, which size is usually related to the amount of CPU memory),
  95 followed by a merging passes for these runs, which merging is often
  96 very cleverly organised[1].  It is very important that the initial
  97 sort produces the longest runs possible.  Tournaments are a good way
  98 to that.  If, using all the memory available to hold a tournament, you
  99 replace and percolate items that happen to fit the current run, you'll
 100 produce runs which are twice the size of the memory for random input,
 101 and much better for input fuzzily ordered.
 102
 103 Moreover, if you output the 0'th item on disk and get an input which
 104 may not fit in the current tournament (because the value "wins" over
 105 the last output value), it cannot fit in the heap, so the size of the
 106 heap decreases.  The freed memory could be cleverly reused immediately
 107 for progressively building a second heap, which grows at exactly the
 108 same rate the first heap is melting.  When the first heap completely
 109 vanishes, you switch heaps and start a new run.  Clever and quite
 110 effective!
 111
 112 In a word, heaps are useful memory structures to know.  I use them in
 113 a few applications, and I think it is good to keep a `heap' module
 114 around. :-)
 115
 116 --------------------
 117 [1] The disk balancing algorithms which are current, nowadays, are
 118 more annoying than clever, and this is a consequence of the seeking
 119 capabilities of the disks.  On devices which cannot seek, like big
 120 tape drives, the story was quite different, and one had to be very
 121 clever to ensure (far in advance) that each tape movement will be the
 122 most effective possible (that is, will best participate at
 123 "progressing" the merge).  Some tapes were even able to read
 124 backwards, and this was also used to avoid the rewinding time.
 125 Believe me, real good tape sorts were quite spectacular to watch!
 126 From all times, sorting has always been a Great Art! :-)
 127 """
 128
 129 __all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
 130            'nlargest', 'nsmallest', 'heappushpop']
 131
 132 from itertools import islice, repeat, count, imap, izip, tee, chain
 133 from operator import itemgetter
 134 import bisect
 135
 136 def cmp_lt(x, y):
 137     # Use __lt__ if available; otherwise, try __le__.
 138     # In Py3.x, only __lt__ will be called.
 139     return (x < y) if hasattr(x, '__lt__') else (not y <= x)
 140
 141 def heappush(heap, item):
 142     """Push item onto heap, maintaining the heap invariant."""
 143     heap.append(item)
 144     _siftdown(heap, 0, len(heap)-1)
 145
 146 def heappop(heap):
 147     """Pop the smallest item off the heap, maintaining the heap invariant."""
 148     lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
 149     if heap:
 150         returnitem = heap[0]
 151         heap[0] = lastelt
 152         _siftup(heap, 0)
 153     else:
 154         returnitem = lastelt
 155     return returnitem
 156
 157 def heapreplace(heap, item):
 158     """Pop and return the current smallest value, and add the new item.
 159
 160     This is more efficient than heappop() followed by heappush(), and can be
 161     more appropriate when using a fixed-size heap.  Note that the value
 162     returned may be larger than item!  That constrains reasonable uses of
 163     this routine unless written as part of a conditional replacement:
 164
 165         if item > heap[0]:
 166             item = heapreplace(heap, item)
 167     """
 168     returnitem = heap[0]    # raises appropriate IndexError if heap is empty
 169     heap[0] = item
 170     _siftup(heap, 0)
 171     return returnitem
 172
 173 def heappushpop(heap, item):
 174     """Fast version of a heappush followed by a heappop."""
 175     if heap and cmp_lt(heap[0], item):
 176         item, heap[0] = heap[0], item
 177         _siftup(heap, 0)
 178     return item
 179
 180 def heapify(x):
 181     """Transform list into a heap, in-place, in O(len(x)) time."""
 182     n = len(x)
 183     # Transform bottom-up.  The largest index there's any point to looking at
 184     # is the largest with a child index in-range, so must have 2*i + 1 < n,
 185     # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
 186     # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
 187     # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
 188     for i in reversed(xrange(n//2)):
 189         _siftup(x, i)
 190
 191 def nlargest(n, iterable):
 192     """Find the n largest elements in a dataset.
 193
 194     Equivalent to:  sorted(iterable, reverse=True)[:n]
 195     """
 196     it = iter(iterable)
 197     result = list(islice(it, n))
 198     if not result:
 199         return result
 200     heapify(result)
 201     _heappushpop = heappushpop
 202     for elem in it:
 203         _heappushpop(result, elem)
 204     result.sort(reverse=True)
 205     return result
 206
 207 def nsmallest(n, iterable):
 208     """Find the n smallest elements in a dataset.
 209
 210     Equivalent to:  sorted(iterable)[:n]
 211     """
 212     if hasattr(iterable, '__len__') and n * 10 <= len(iterable):
 213         # For smaller values of n, the bisect method is faster than a minheap.
 214         # It is also memory efficient, consuming only n elements of space.
 215         it = iter(iterable)
 216         result = sorted(islice(it, 0, n))
 217         if not result:
 218             return result
 219         insort = bisect.insort
 220         pop = result.pop
 221         los = result[-1]    # los --> Largest of the nsmallest
 222         for elem in it:
 223             if cmp_lt(elem, los):
 224                 insort(result, elem)
 225                 pop()
 226                 los = result[-1]
 227         return result
 228     # An alternative approach manifests the whole iterable in memory but
 229     # saves comparisons by heapifying all at once.  Also, saves time
 230     # over bisect.insort() which has O(n) data movement time for every
 231     # insertion.  Finding the n smallest of an m length iterable requires
 232     #    O(m) + O(n log m) comparisons.
 233     h = list(iterable)
 234     heapify(h)
 235     return map(heappop, repeat(h, min(n, len(h))))
 236
 237 # 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
 238 # is the index of a leaf with a possibly out-of-order value.  Restore the
 239 # heap invariant.
 240 def _siftdown(heap, startpos, pos):
 241     newitem = heap[pos]
 242     # Follow the path to the root, moving parents down until finding a place
 243     # newitem fits.
 244     while pos > startpos:
 245         parentpos = (pos - 1) >> 1
 246         parent = heap[parentpos]
 247         if cmp_lt(newitem, parent):
 248             heap[pos] = parent
 249             pos = parentpos
 250             continue
 251         break
 252     heap[pos] = newitem
 253
 254 # The child indices of heap index pos are already heaps, and we want to make
 255 # a heap at index pos too.  We do this by bubbling the smaller child of
 256 # pos up (and so on with that child's children, etc) until hitting a leaf,
 257 # then using _siftdown to move the oddball originally at index pos into place.
 258 #
 259 # We *could* break out of the loop as soon as we find a pos where newitem <=
 260 # both its children, but turns out that's not a good idea, and despite that
 261 # many books write the algorithm that way.  During a heap pop, the last array
 262 # element is sifted in, and that tends to be large, so that comparing it
 263 # against values starting from the root usually doesn't pay (= usually doesn't
 264 # get us out of the loop early).  See Knuth, Volume 3, where this is
 265 # explained and quantified in an exercise.
 266 #
 267 # Cutting the # of comparisons is important, since these routines have no
 268 # way to extract "the priority" from an array element, so that intelligence
 269 # is likely to be hiding in custom __cmp__ methods, or in array elements
 270 # storing (priority, record) tuples.  Comparisons are thus potentially
 271 # expensive.
 272 #
 273 # On random arrays of length 1000, making this change cut the number of
 274 # comparisons made by heapify() a little, and those made by exhaustive
 275 # heappop() a lot, in accord with theory.  Here are typical results from 3
 276 # runs (3 just to demonstrate how small the variance is):
 277 #
 278 # Compares needed by heapify     Compares needed by 1000 heappops
 279 # --------------------------     --------------------------------
 280 # 1837 cut to 1663               14996 cut to 8680
 281 # 1855 cut to 1659               14966 cut to 8678
 282 # 1847 cut to 1660               15024 cut to 8703
 283 #
 284 # Building the heap by using heappush() 1000 times instead required
 285 # 2198, 2148, and 2219 compares:  heapify() is more efficient, when
 286 # you can use it.
 287 #
 288 # The total compares needed by list.sort() on the same lists were 8627,
 289 # 8627, and 8632 (this should be compared to the sum of heapify() and
 290 # heappop() compares):  list.sort() is (unsurprisingly!) more efficient
 291 # for sorting.
 292
 293 def _siftup(heap, pos):
 294     endpos = len(heap)
 295     startpos = pos
 296     newitem = heap[pos]
 297     # Bubble up the smaller child until hitting a leaf.
 298     childpos = 2*pos + 1    # leftmost child position
 299     while childpos < endpos:
 300         # Set childpos to index of smaller child.
 301         rightpos = childpos + 1
 302         if rightpos < endpos and not cmp_lt(heap[childpos], heap[rightpos]):
 303             childpos = rightpos
 304         # Move the smaller child up.
 305         heap[pos] = heap[childpos]
 306         pos = childpos
 307         childpos = 2*pos + 1
 308     # The leaf at pos is empty now.  Put newitem there, and bubble it up
 309     # to its final resting place (by sifting its parents down).
 310     heap[pos] = newitem
 311     _siftdown(heap, startpos, pos)
 312
 313 # If available, use C implementation
 314 try:
 315     from _heapq import *
 316 except ImportError:
 317     pass
 318
 319 def merge(*iterables):
 320     '''Merge multiple sorted inputs into a single sorted output.
 321
 322     Similar to sorted(itertools.chain(*iterables)) but returns a generator,
 323     does not pull the data into memory all at once, and assumes that each of
 324     the input streams is already sorted (smallest to largest).
 325
 326     >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
 327     [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
 328
 329     '''
 330     _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
 331
 332     h = []
 333     h_append = h.append
 334     for itnum, it in enumerate(map(iter, iterables)):
 335         try:
 336             next = it.next
 337             h_append([next(), itnum, next])
 338         except _StopIteration:
 339             pass
 340     heapify(h)
 341
 342     while 1:
 343         try:
 344             while 1:
 345                 v, itnum, next = s = h[0]   # raises IndexError when h is empty
 346                 yield v
 347                 s[0] = next()               # raises StopIteration when exhausted
 348                 _heapreplace(h, s)          # restore heap condition
 349         except _StopIteration:
 350             _heappop(h)                     # remove empty iterator
 351         except IndexError:
 352             return
 353
 354 # Extend the implementations of nsmallest and nlargest to use a key= argument
 355 _nsmallest = nsmallest
 356 def nsmallest(n, iterable, key=None):
 357     """Find the n smallest elements in a dataset.
 358
 359     Equivalent to:  sorted(iterable, key=key)[:n]
 360     """
 361     # Short-cut for n==1 is to use min() when len(iterable)>0
 362     if n == 1:
 363         it = iter(iterable)
 364         head = list(islice(it, 1))
 365         if not head:
 366             return []
 367         if key is None:
 368             return [min(chain(head, it))]
 369         return [min(chain(head, it), key=key)]
 370
 371     # When n>=size, it's faster to use sorted()
 372     try:
 373         size = len(iterable)
 374     except (TypeError, AttributeError):
 375         pass
 376     else:
 377         if n >= size:
 378             return sorted(iterable, key=key)[:n]
 379
 380     # When key is none, use simpler decoration
 381     if key is None:
 382         it = izip(iterable, count())                        # decorate
 383         result = _nsmallest(n, it)
 384         return map(itemgetter(0), result)                   # undecorate
 385
 386     # General case, slowest method
 387     in1, in2 = tee(iterable)
 388     it = izip(imap(key, in1), count(), in2)                 # decorate
 389     result = _nsmallest(n, it)
 390     return map(itemgetter(2), result)                       # undecorate
 391
 392 _nlargest = nlargest
 393 def nlargest(n, iterable, key=None):
 394     """Find the n largest elements in a dataset.
 395
 396     Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
 397     """
 398
 399     # Short-cut for n==1 is to use max() when len(iterable)>0
 400     if n == 1:
 401         it = iter(iterable)
 402         head = list(islice(it, 1))
 403         if not head:
 404             return []
 405         if key is None:
 406             return [max(chain(head, it))]
 407         return [max(chain(head, it), key=key)]
 408
 409     # When n>=size, it's faster to use sorted()
 410     try:
 411         size = len(iterable)
 412     except (TypeError, AttributeError):
 413         pass
 414     else:
 415         if n >= size:
 416             return sorted(iterable, key=key, reverse=True)[:n]
 417
 418     # When key is none, use simpler decoration
 419     if key is None:
 420         it = izip(iterable, count(0,-1))                    # decorate
 421         result = _nlargest(n, it)
 422         return map(itemgetter(0), result)                   # undecorate
 423
 424     # General case, slowest method
 425     in1, in2 = tee(iterable)
 426     it = izip(imap(key, in1), count(0,-1), in2)             # decorate
 427     result = _nlargest(n, it)
 428     return map(itemgetter(2), result)                       # undecorate
 429
 430 if __name__ == "__main__":
 431     # Simple sanity test
 432     heap = []
 433     data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
 434     for item in data:
 435         heappush(heap, item)
 436     sort = []
 437     while heap:
 438         sort.append(heappop(heap))
 439     print sort
 440
 441     import doctest
 442     doctest.testmod()