AppPkg/Applications/Python/Python-2.7.10/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, count, imap, izip, tee, chain
 133 from operator import itemgetter
 134
 135 def cmp_lt(x, y):
 136     # Use __lt__ if available; otherwise, try __le__.
 137     # In Py3.x, only __lt__ will be called.
 138     return (x < y) if hasattr(x, '__lt__') else (not y <= x)
 139
 140 def heappush(heap, item):
 141     """Push item onto heap, maintaining the heap invariant."""
 142     heap.append(item)
 143     _siftdown(heap, 0, len(heap)-1)
 144
 145 def heappop(heap):
 146     """Pop the smallest item off the heap, maintaining the heap invariant."""
 147     lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
 148     if heap:
 149         returnitem = heap[0]
 150         heap[0] = lastelt
 151         _siftup(heap, 0)
 152     else:
 153         returnitem = lastelt
 154     return returnitem
 155
 156 def heapreplace(heap, item):
 157     """Pop and return the current smallest value, and add the new item.
 158
 159     This is more efficient than heappop() followed by heappush(), and can be
 160     more appropriate when using a fixed-size heap.  Note that the value
 161     returned may be larger than item!  That constrains reasonable uses of
 162     this routine unless written as part of a conditional replacement:
 163
 164         if item > heap[0]:
 165             item = heapreplace(heap, item)
 166     """
 167     returnitem = heap[0]    # raises appropriate IndexError if heap is empty
 168     heap[0] = item
 169     _siftup(heap, 0)
 170     return returnitem
 171
 172 def heappushpop(heap, item):
 173     """Fast version of a heappush followed by a heappop."""
 174     if heap and cmp_lt(heap[0], item):
 175         item, heap[0] = heap[0], item
 176         _siftup(heap, 0)
 177     return item
 178
 179 def heapify(x):
 180     """Transform list into a heap, in-place, in O(len(x)) time."""
 181     n = len(x)
 182     # Transform bottom-up.  The largest index there's any point to looking at
 183     # is the largest with a child index in-range, so must have 2*i + 1 < n,
 184     # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
 185     # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
 186     # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
 187     for i in reversed(xrange(n//2)):
 188         _siftup(x, i)
 189
 190 def _heappushpop_max(heap, item):
 191     """Maxheap version of a heappush followed by a heappop."""
 192     if heap and cmp_lt(item, heap[0]):
 193         item, heap[0] = heap[0], item
 194         _siftup_max(heap, 0)
 195     return item
 196
 197 def _heapify_max(x):
 198     """Transform list into a maxheap, in-place, in O(len(x)) time."""
 199     n = len(x)
 200     for i in reversed(range(n//2)):
 201         _siftup_max(x, i)
 202
 203 def nlargest(n, iterable):
 204     """Find the n largest elements in a dataset.
 205
 206     Equivalent to:  sorted(iterable, reverse=True)[:n]
 207     """
 208     if n < 0:
 209         return []
 210     it = iter(iterable)
 211     result = list(islice(it, n))
 212     if not result:
 213         return result
 214     heapify(result)
 215     _heappushpop = heappushpop
 216     for elem in it:
 217         _heappushpop(result, elem)
 218     result.sort(reverse=True)
 219     return result
 220
 221 def nsmallest(n, iterable):
 222     """Find the n smallest elements in a dataset.
 223
 224     Equivalent to:  sorted(iterable)[:n]
 225     """
 226     if n < 0:
 227         return []
 228     it = iter(iterable)
 229     result = list(islice(it, n))
 230     if not result:
 231         return result
 232     _heapify_max(result)
 233     _heappushpop = _heappushpop_max
 234     for elem in it:
 235         _heappushpop(result, elem)
 236     result.sort()
 237     return result
 238
 239 # 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
 240 # is the index of a leaf with a possibly out-of-order value.  Restore the
 241 # heap invariant.
 242 def _siftdown(heap, startpos, pos):
 243     newitem = heap[pos]
 244     # Follow the path to the root, moving parents down until finding a place
 245     # newitem fits.
 246     while pos > startpos:
 247         parentpos = (pos - 1) >> 1
 248         parent = heap[parentpos]
 249         if cmp_lt(newitem, parent):
 250             heap[pos] = parent
 251             pos = parentpos
 252             continue
 253         break
 254     heap[pos] = newitem
 255
 256 # The child indices of heap index pos are already heaps, and we want to make
 257 # a heap at index pos too.  We do this by bubbling the smaller child of
 258 # pos up (and so on with that child's children, etc) until hitting a leaf,
 259 # then using _siftdown to move the oddball originally at index pos into place.
 260 #
 261 # We *could* break out of the loop as soon as we find a pos where newitem <=
 262 # both its children, but turns out that's not a good idea, and despite that
 263 # many books write the algorithm that way.  During a heap pop, the last array
 264 # element is sifted in, and that tends to be large, so that comparing it
 265 # against values starting from the root usually doesn't pay (= usually doesn't
 266 # get us out of the loop early).  See Knuth, Volume 3, where this is
 267 # explained and quantified in an exercise.
 268 #
 269 # Cutting the # of comparisons is important, since these routines have no
 270 # way to extract "the priority" from an array element, so that intelligence
 271 # is likely to be hiding in custom __cmp__ methods, or in array elements
 272 # storing (priority, record) tuples.  Comparisons are thus potentially
 273 # expensive.
 274 #
 275 # On random arrays of length 1000, making this change cut the number of
 276 # comparisons made by heapify() a little, and those made by exhaustive
 277 # heappop() a lot, in accord with theory.  Here are typical results from 3
 278 # runs (3 just to demonstrate how small the variance is):
 279 #
 280 # Compares needed by heapify     Compares needed by 1000 heappops
 281 # --------------------------     --------------------------------
 282 # 1837 cut to 1663               14996 cut to 8680
 283 # 1855 cut to 1659               14966 cut to 8678
 284 # 1847 cut to 1660               15024 cut to 8703
 285 #
 286 # Building the heap by using heappush() 1000 times instead required
 287 # 2198, 2148, and 2219 compares:  heapify() is more efficient, when
 288 # you can use it.
 289 #
 290 # The total compares needed by list.sort() on the same lists were 8627,
 291 # 8627, and 8632 (this should be compared to the sum of heapify() and
 292 # heappop() compares):  list.sort() is (unsurprisingly!) more efficient
 293 # for sorting.
 294
 295 def _siftup(heap, pos):
 296     endpos = len(heap)
 297     startpos = pos
 298     newitem = heap[pos]
 299     # Bubble up the smaller child until hitting a leaf.
 300     childpos = 2*pos + 1    # leftmost child position
 301     while childpos < endpos:
 302         # Set childpos to index of smaller child.
 303         rightpos = childpos + 1
 304         if rightpos < endpos and not cmp_lt(heap[childpos], heap[rightpos]):
 305             childpos = rightpos
 306         # Move the smaller child up.
 307         heap[pos] = heap[childpos]
 308         pos = childpos
 309         childpos = 2*pos + 1
 310     # The leaf at pos is empty now.  Put newitem there, and bubble it up
 311     # to its final resting place (by sifting its parents down).
 312     heap[pos] = newitem
 313     _siftdown(heap, startpos, pos)
 314
 315 def _siftdown_max(heap, startpos, pos):
 316     'Maxheap variant of _siftdown'
 317     newitem = heap[pos]
 318     # Follow the path to the root, moving parents down until finding a place
 319     # newitem fits.
 320     while pos > startpos:
 321         parentpos = (pos - 1) >> 1
 322         parent = heap[parentpos]
 323         if cmp_lt(parent, newitem):
 324             heap[pos] = parent
 325             pos = parentpos
 326             continue
 327         break
 328     heap[pos] = newitem
 329
 330 def _siftup_max(heap, pos):
 331     'Maxheap variant of _siftup'
 332     endpos = len(heap)
 333     startpos = pos
 334     newitem = heap[pos]
 335     # Bubble up the larger child until hitting a leaf.
 336     childpos = 2*pos + 1    # leftmost child position
 337     while childpos < endpos:
 338         # Set childpos to index of larger child.
 339         rightpos = childpos + 1
 340         if rightpos < endpos and not cmp_lt(heap[rightpos], heap[childpos]):
 341             childpos = rightpos
 342         # Move the larger child up.
 343         heap[pos] = heap[childpos]
 344         pos = childpos
 345         childpos = 2*pos + 1
 346     # The leaf at pos is empty now.  Put newitem there, and bubble it up
 347     # to its final resting place (by sifting its parents down).
 348     heap[pos] = newitem
 349     _siftdown_max(heap, startpos, pos)
 350
 351 # If available, use C implementation
 352 try:
 353     from _heapq import *
 354 except ImportError:
 355     pass
 356
 357 def merge(*iterables):
 358     '''Merge multiple sorted inputs into a single sorted output.
 359
 360     Similar to sorted(itertools.chain(*iterables)) but returns a generator,
 361     does not pull the data into memory all at once, and assumes that each of
 362     the input streams is already sorted (smallest to largest).
 363
 364     >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
 365     [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
 366
 367     '''
 368     _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
 369     _len = len
 370
 371     h = []
 372     h_append = h.append
 373     for itnum, it in enumerate(map(iter, iterables)):
 374         try:
 375             next = it.next
 376             h_append([next(), itnum, next])
 377         except _StopIteration:
 378             pass
 379     heapify(h)
 380
 381     while _len(h) > 1:
 382         try:
 383             while 1:
 384                 v, itnum, next = s = h[0]
 385                 yield v
 386                 s[0] = next()               # raises StopIteration when exhausted
 387                 _heapreplace(h, s)          # restore heap condition
 388         except _StopIteration:
 389             _heappop(h)                     # remove empty iterator
 390     if h:
 391         # fast case when only a single iterator remains
 392         v, itnum, next = h[0]
 393         yield v
 394         for v in next.__self__:
 395             yield v
 396
 397 # Extend the implementations of nsmallest and nlargest to use a key= argument
 398 _nsmallest = nsmallest
 399 def nsmallest(n, iterable, key=None):
 400     """Find the n smallest elements in a dataset.
 401
 402     Equivalent to:  sorted(iterable, key=key)[:n]
 403     """
 404     # Short-cut for n==1 is to use min() when len(iterable)>0
 405     if n == 1:
 406         it = iter(iterable)
 407         head = list(islice(it, 1))
 408         if not head:
 409             return []
 410         if key is None:
 411             return [min(chain(head, it))]
 412         return [min(chain(head, it), key=key)]
 413
 414     # When n>=size, it's faster to use sorted()
 415     try:
 416         size = len(iterable)
 417     except (TypeError, AttributeError):
 418         pass
 419     else:
 420         if n >= size:
 421             return sorted(iterable, key=key)[:n]
 422
 423     # When key is none, use simpler decoration
 424     if key is None:
 425         it = izip(iterable, count())                        # decorate
 426         result = _nsmallest(n, it)
 427         return map(itemgetter(0), result)                   # undecorate
 428
 429     # General case, slowest method
 430     in1, in2 = tee(iterable)
 431     it = izip(imap(key, in1), count(), in2)                 # decorate
 432     result = _nsmallest(n, it)
 433     return map(itemgetter(2), result)                       # undecorate
 434
 435 _nlargest = nlargest
 436 def nlargest(n, iterable, key=None):
 437     """Find the n largest elements in a dataset.
 438
 439     Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
 440     """
 441
 442     # Short-cut for n==1 is to use max() when len(iterable)>0
 443     if n == 1:
 444         it = iter(iterable)
 445         head = list(islice(it, 1))
 446         if not head:
 447             return []
 448         if key is None:
 449             return [max(chain(head, it))]
 450         return [max(chain(head, it), key=key)]
 451
 452     # When n>=size, it's faster to use sorted()
 453     try:
 454         size = len(iterable)
 455     except (TypeError, AttributeError):
 456         pass
 457     else:
 458         if n >= size:
 459             return sorted(iterable, key=key, reverse=True)[:n]
 460
 461     # When key is none, use simpler decoration
 462     if key is None:
 463         it = izip(iterable, count(0,-1))                    # decorate
 464         result = _nlargest(n, it)
 465         return map(itemgetter(0), result)                   # undecorate
 466
 467     # General case, slowest method
 468     in1, in2 = tee(iterable)
 469     it = izip(imap(key, in1), count(0,-1), in2)             # decorate
 470     result = _nlargest(n, it)
 471     return map(itemgetter(2), result)                       # undecorate
 472
 473 if __name__ == "__main__":
 474     # Simple sanity test
 475     heap = []
 476     data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
 477     for item in data:
 478         heappush(heap, item)
 479     sort = []
 480     while heap:
 481         sort.append(heappop(heap))
 482     print sort
 483
 484     import doctest
 485     doctest.testmod()