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']
 131
 132 from itertools import islice, repeat, count, imap, izip, tee
 133 from operator import itemgetter, neg
 134 import bisect
 135
 136 def heappush(heap, item):
 137     """Push item onto heap, maintaining the heap invariant."""
 138     heap.append(item)
 139     _siftdown(heap, 0, len(heap)-1)
 140
 141 def 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
 152 def 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
 168 def heapify(x):
 169     """Transform list into a heap, in-place, in O(len(heap)) time."""
 170     n = len(x)
 171     # Transform bottom-up.  The largest index there's any point to looking at
 172     # is the largest with a child index in-range, so must have 2*i + 1 < n,
 173     # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
 174     # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
 175     # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
 176     for i in reversed(xrange(n//2)):
 177         _siftup(x, i)
 178
 179 def nlargest(n, iterable):
 180     """Find the n largest elements in a dataset.
 181
 182     Equivalent to:  sorted(iterable, reverse=True)[:n]
 183     """
 184     it = iter(iterable)
 185     result = list(islice(it, n))
 186     if not result:
 187         return result
 188     heapify(result)
 189     _heapreplace = heapreplace
 190     sol = result[0]         # sol --> smallest of the nlargest
 191     for elem in it:
 192         if elem <= sol:
 193             continue
 194         _heapreplace(result, elem)
 195         sol = result[0]
 196     result.sort(reverse=True)
 197     return result
 198
 199 def nsmallest(n, iterable):
 200     """Find the n smallest elements in a dataset.
 201
 202     Equivalent to:  sorted(iterable)[:n]
 203     """
 204     if hasattr(iterable, '__len__') and n * 10 <= len(iterable):
 205         # For smaller values of n, the bisect method is faster than a minheap.
 206         # It is also memory efficient, consuming only n elements of space.
 207         it = iter(iterable)
 208         result = sorted(islice(it, 0, n))
 209         if not result:
 210             return result
 211         insort = bisect.insort
 212         pop = result.pop
 213         los = result[-1]    # los --> Largest of the nsmallest
 214         for elem in it:
 215             if los <= elem:
 216                 continue
 217             insort(result, elem)
 218             pop()
 219             los = result[-1]
 220         return result
 221     # An alternative approach manifests the whole iterable in memory but
 222     # saves comparisons by heapifying all at once.  Also, saves time
 223     # over bisect.insort() which has O(n) data movement time for every
 224     # insertion.  Finding the n smallest of an m length iterable requires
 225     #    O(m) + O(n log m) comparisons.
 226     h = list(iterable)
 227     heapify(h)
 228     return map(heappop, repeat(h, min(n, len(h))))
 229
 230 # 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
 231 # is the index of a leaf with a possibly out-of-order value.  Restore the
 232 # heap invariant.
 233 def _siftdown(heap, startpos, pos):
 234     newitem = heap[pos]
 235     # Follow the path to the root, moving parents down until finding a place
 236     # newitem fits.
 237     while pos > startpos:
 238         parentpos = (pos - 1) >> 1
 239         parent = heap[parentpos]
 240         if parent <= newitem:
 241             break
 242         heap[pos] = parent
 243         pos = parentpos
 244     heap[pos] = newitem
 245
 246 # The child indices of heap index pos are already heaps, and we want to make
 247 # a heap at index pos too.  We do this by bubbling the smaller child of
 248 # pos up (and so on with that child's children, etc) until hitting a leaf,
 249 # then using _siftdown to move the oddball originally at index pos into place.
 250 #
 251 # We *could* break out of the loop as soon as we find a pos where newitem <=
 252 # both its children, but turns out that's not a good idea, and despite that
 253 # many books write the algorithm that way.  During a heap pop, the last array
 254 # element is sifted in, and that tends to be large, so that comparing it
 255 # against values starting from the root usually doesn't pay (= usually doesn't
 256 # get us out of the loop early).  See Knuth, Volume 3, where this is
 257 # explained and quantified in an exercise.
 258 #
 259 # Cutting the # of comparisons is important, since these routines have no
 260 # way to extract "the priority" from an array element, so that intelligence
 261 # is likely to be hiding in custom __cmp__ methods, or in array elements
 262 # storing (priority, record) tuples.  Comparisons are thus potentially
 263 # expensive.
 264 #
 265 # On random arrays of length 1000, making this change cut the number of
 266 # comparisons made by heapify() a little, and those made by exhaustive
 267 # heappop() a lot, in accord with theory.  Here are typical results from 3
 268 # runs (3 just to demonstrate how small the variance is):
 269 #
 270 # Compares needed by heapify     Compares needed by 1000 heappops
 271 # --------------------------     --------------------------------
 272 # 1837 cut to 1663               14996 cut to 8680
 273 # 1855 cut to 1659               14966 cut to 8678
 274 # 1847 cut to 1660               15024 cut to 8703
 275 #
 276 # Building the heap by using heappush() 1000 times instead required
 277 # 2198, 2148, and 2219 compares:  heapify() is more efficient, when
 278 # you can use it.
 279 #
 280 # The total compares needed by list.sort() on the same lists were 8627,
 281 # 8627, and 8632 (this should be compared to the sum of heapify() and
 282 # heappop() compares):  list.sort() is (unsurprisingly!) more efficient
 283 # for sorting.
 284
 285 def _siftup(heap, pos):
 286     endpos = len(heap)
 287     startpos = pos
 288     newitem = heap[pos]
 289     # Bubble up the smaller child until hitting a leaf.
 290     childpos = 2*pos + 1    # leftmost child position
 291     while childpos < endpos:
 292         # Set childpos to index of smaller child.
 293         rightpos = childpos + 1
 294         if rightpos < endpos and heap[rightpos] <= heap[childpos]:
 295             childpos = rightpos
 296         # Move the smaller child up.
 297         heap[pos] = heap[childpos]
 298         pos = childpos
 299         childpos = 2*pos + 1
 300     # The leaf at pos is empty now.  Put newitem there, and bubble it up
 301     # to its final resting place (by sifting its parents down).
 302     heap[pos] = newitem
 303     _siftdown(heap, startpos, pos)
 304
 305 # If available, use C implementation
 306 try:
 307     from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest
 308 except ImportError:
 309     pass
 310
 311 def merge(*iterables):
 312     '''Merge multiple sorted inputs into a single sorted output.
 313
 314     Similar to sorted(itertools.chain(*iterables)) but returns a generator,
 315     does not pull the data into memory all at once, and assumes that each of
 316     the input streams is already sorted (smallest to largest).
 317
 318     >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
 319     [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
 320
 321     '''
 322     _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
 323
 324     h = []
 325     h_append = h.append
 326     for itnum, it in enumerate(map(iter, iterables)):
 327         try:
 328             next = it.next
 329             h_append([next(), itnum, next])
 330         except _StopIteration:
 331             pass
 332     heapify(h)
 333
 334     while 1:
 335         try:
 336             while 1:
 337                 v, itnum, next = s = h[0]   # raises IndexError when h is empty
 338                 yield v
 339                 s[0] = next()               # raises StopIteration when exhausted
 340                 _heapreplace(h, s)          # restore heap condition
 341         except _StopIteration:
 342             _heappop(h)                     # remove empty iterator
 343         except IndexError:
 344             return
 345
 346 # Extend the implementations of nsmallest and nlargest to use a key= argument
 347 _nsmallest = nsmallest
 348 def nsmallest(n, iterable, key=None):
 349     """Find the n smallest elements in a dataset.
 350
 351     Equivalent to:  sorted(iterable, key=key)[:n]
 352     """
 353     in1, in2 = tee(iterable)
 354     it = izip(imap(key, in1), count(), in2)                 # decorate
 355     result = _nsmallest(n, it)
 356     return map(itemgetter(2), result)                       # undecorate
 357
 358 _nlargest = nlargest
 359 def nlargest(n, iterable, key=None):
 360     """Find the n largest elements in a dataset.
 361
 362     Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
 363     """
 364     in1, in2 = tee(iterable)
 365     it = izip(imap(key, in1), imap(neg, count()), in2)      # decorate
 366     result = _nlargest(n, it)
 367     return map(itemgetter(2), result)                       # undecorate
 368
 369 if __name__ == "__main__":
 370     # Simple sanity test
 371     heap = []
 372     data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
 373     for item in data:
 374         heappush(heap, item)
 375     sort = []
 376     while heap:
 377         sort.append(heappop(heap))
 378     print sort
 379
 380     import doctest
 381     doctest.testmod()