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 Franois 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 369 370 h = [] 371 h_append = h.append 372 for itnum, it in enumerate(map(iter, iterables)): 373 try: 374 next = it.next 375 h_append([next(), itnum, next]) 376 except _StopIteration: 377 pass 378 heapify(h) 379 380 while 1: 381 try: 382 while 1: 383 v, itnum, next = s = h[0] # raises IndexError when h is empty 384 yield v 385 s[0] = next() # raises StopIteration when exhausted 386 _heapreplace(h, s) # restore heap condition 387 except _StopIteration: 388 _heappop(h) # remove empty iterator 389 except IndexError: 390 return 391 392 # Extend the implementations of nsmallest and nlargest to use a key= argument 393 _nsmallest = nsmallest 394 def nsmallest(n, iterable, key=None): 395 """Find the n smallest elements in a dataset. 396 397 Equivalent to: sorted(iterable, key=key)[:n] 398 """ 399 # Short-cut for n==1 is to use min() 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 [min(chain(head, it))] 407 return [min(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)[:n] 417 418 # When key is none, use simpler decoration 419 if key is None: 420 it = izip(iterable, count()) # decorate 421 result = _nsmallest(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(), in2) # decorate 427 result = _nsmallest(n, it) 428 return map(itemgetter(2), result) # undecorate 429 430 _nlargest = nlargest 431 def nlargest(n, iterable, key=None): 432 """Find the n largest elements in a dataset. 433 434 Equivalent to: sorted(iterable, key=key, reverse=True)[:n] 435 """ 436 437 # Short-cut for n==1 is to use max() when len(iterable)>0 438 if n == 1: 439 it = iter(iterable) 440 head = list(islice(it, 1)) 441 if not head: 442 return [] 443 if key is None: 444 return [max(chain(head, it))] 445 return [max(chain(head, it), key=key)] 446 447 # When n>=size, it's faster to use sorted() 448 try: 449 size = len(iterable) 450 except (TypeError, AttributeError): 451 pass 452 else: 453 if n >= size: 454 return sorted(iterable, key=key, reverse=True)[:n] 455 456 # When key is none, use simpler decoration 457 if key is None: 458 it = izip(iterable, count(0,-1)) # decorate 459 result = _nlargest(n, it) 460 return map(itemgetter(0), result) # undecorate 461 462 # General case, slowest method 463 in1, in2 = tee(iterable) 464 it = izip(imap(key, in1), count(0,-1), in2) # decorate 465 result = _nlargest(n, it) 466 return map(itemgetter(2), result) # undecorate 467 468 if __name__ == "__main__": 469 # Simple sanity test 470 heap = [] 471 data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] 472 for item in data: 473 heappush(heap, item) 474 sort = [] 475 while heap: 476 sort.append(heappop(heap)) 477 print sort 478 479 import doctest 480 doctest.testmod() 481
