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, 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 heappushpop(heap, item):
 169     """Fast version of a heappush followed by a heappop."""
 170     if heap and item > heap[0]:
 171         item, heap[0] = heap[0], item
 172         _siftup(heap, 0)
 173     return item
 174
 175 def 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(range(n//2)):
 184         _siftup(x, i)
 185
 186 def 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
 202 def 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 list(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.
 236 def _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 parent <= newitem:
 244             break
 245         heap[pos] = parent
 246         pos = parentpos
 247     heap[pos] = newitem
 248
 249 # The child indices of heap index pos are already heaps, and we want to make
 250 # a heap at index pos too.  We do this by bubbling the smaller child of
 251 # pos up (and so on with that child's children, etc) until hitting a leaf,
 252 # then using _siftdown to move the oddball originally at index pos into place.
 253 #
 254 # We *could* break out of the loop as soon as we find a pos where newitem <=
 255 # both its children, but turns out that's not a good idea, and despite that
 256 # many books write the algorithm that way.  During a heap pop, the last array
 257 # element is sifted in, and that tends to be large, so that comparing it
 258 # against values starting from the root usually doesn't pay (= usually doesn't
 259 # get us out of the loop early).  See Knuth, Volume 3, where this is
 260 # explained and quantified in an exercise.
 261 #
 262 # Cutting the # of comparisons is important, since these routines have no
 263 # way to extract "the priority" from an array element, so that intelligence
 264 # is likely to be hiding in custom __cmp__ methods, or in array elements
 265 # storing (priority, record) tuples.  Comparisons are thus potentially
 266 # expensive.
 267 #
 268 # On random arrays of length 1000, making this change cut the number of
 269 # comparisons made by heapify() a little, and those made by exhaustive
 270 # heappop() a lot, in accord with theory.  Here are typical results from 3
 271 # runs (3 just to demonstrate how small the variance is):
 272 #
 273 # Compares needed by heapify     Compares needed by 1000 heappops
 274 # --------------------------     --------------------------------
 275 # 1837 cut to 1663               14996 cut to 8680
 276 # 1855 cut to 1659               14966 cut to 8678
 277 # 1847 cut to 1660               15024 cut to 8703
 278 #
 279 # Building the heap by using heappush() 1000 times instead required
 280 # 2198, 2148, and 2219 compares:  heapify() is more efficient, when
 281 # you can use it.
 282 #
 283 # The total compares needed by list.sort() on the same lists were 8627,
 284 # 8627, and 8632 (this should be compared to the sum of heapify() and
 285 # heappop() compares):  list.sort() is (unsurprisingly!) more efficient
 286 # for sorting.
 287
 288 def _siftup(heap, pos):
 289     endpos = len(heap)
 290     startpos = pos
 291     newitem = heap[pos]
 292     # Bubble up the smaller child until hitting a leaf.
 293     childpos = 2*pos + 1    # leftmost child position
 294     while childpos < endpos:
 295         # Set childpos to index of smaller child.
 296         rightpos = childpos + 1
 297         if rightpos < endpos and heap[rightpos] <= heap[childpos]:
 298             childpos = rightpos
 299         # Move the smaller child up.
 300         heap[pos] = heap[childpos]
 301         pos = childpos
 302         childpos = 2*pos + 1
 303     # The leaf at pos is empty now.  Put newitem there, and bubble it up
 304     # to its final resting place (by sifting its parents down).
 305     heap[pos] = newitem
 306     _siftdown(heap, startpos, pos)
 307
 308 # If available, use C implementation
 309 try:
 310     from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest, heappushpop
 311 except ImportError:
 312     pass
 313
 314 def merge(*iterables):
 315     '''Merge multiple sorted inputs into a single sorted output.
 316
 317     Similar to sorted(itertools.chain(*iterables)) but returns a generator,
 318     does not pull the data into memory all at once, and assumes that each of
 319     the input streams is already sorted (smallest to largest).
 320
 321     >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
 322     [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
 323
 324     '''
 325     _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
 326
 327     h = []
 328     h_append = h.append
 329     for itnum, it in enumerate(map(iter, iterables)):
 330         try:
 331             next = it.__next__
 332             h_append([next(), itnum, next])
 333         except _StopIteration:
 334             pass
 335     heapify(h)
 336
 337     while 1:
 338         try:
 339             while 1:
 340                 v, itnum, next = s = h[0]   # raises IndexError when h is empty
 341                 yield v
 342                 s[0] = next()               # raises StopIteration when exhausted
 343                 _heapreplace(h, s)          # restore heap condition
 344         except _StopIteration:
 345             _heappop(h)                     # remove empty iterator
 346         except IndexError:
 347             return
 348
 349 # Extend the implementations of nsmallest and nlargest to use a key= argument
 350 _nsmallest = nsmallest
 351 def nsmallest(n, iterable, key=None):
 352     """Find the n smallest elements in a dataset.
 353
 354     Equivalent to:  sorted(iterable, key=key)[:n]
 355     """
 356     in1, in2 = tee(iterable)
 357     keys = in1 if key is None else map(key, in1)
 358     it = zip(keys, count(), in2)                           # decorate
 359     result = _nsmallest(n, it)
 360     return list(map(itemgetter(2), result))                 # undecorate
 361
 362 _nlargest = nlargest
 363 def nlargest(n, iterable, key=None):
 364     """Find the n largest elements in a dataset.
 365
 366     Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
 367     """
 368     in1, in2 = tee(iterable)
 369     keys = in1 if key is None else map(key, in1)
 370     it = zip(keys, map(neg, count()), in2)                 # decorate
 371     result = _nlargest(n, it)
 372     return list(map(itemgetter(2), result))                 # undecorate
 373
 374 if __name__ == "__main__":
 375     # Simple sanity test
 376     heap = []
 377     data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
 378     for item in data:
 379         heappush(heap, item)
 380     sort = []
 381     while heap:
 382         sort.append(heappop(heap))
 383     print(sort)
 384
 385     import doctest
 386     doctest.testmod()