1 """Heap queue algorithm (a.k.a. priority queue). 2 3 Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 4 all k, counting elements from 0. For the sake of comparison, 5 non-existing elements are considered to be infinite. The interesting 6 property of a heap is that a[0] is always its smallest element. 7 8 Usage: 9 10 heap = [] # creates an empty heap 11 heappush(heap, item) # pushes a new item on the heap 12 item = heappop(heap) # pops the smallest item from the heap 13 item = heap[0] # smallest item on the heap without popping it 14 heapify(x) # transforms list into a heap, in-place, in linear time 15 item = heapreplace(heap, item) # pops and returns smallest item, and adds 16 # new item; the heap size is unchanged 17 18 Our API differs from textbook heap algorithms as follows: 19 20 - We use 0-based indexing. This makes the relationship between the 21 index for a node and the indexes for its children slightly less 22 obvious, but is more suitable since Python uses 0-based indexing. 23 24 - Our heappop() method returns the smallest item, not the largest. 25 26 These two make it possible to view the heap as a regular Python list 27 without surprises: heap[0] is the smallest item, and heap.sort() 28 maintains the heap invariant! 29 """ 30 31 # Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger 32 33 __about__ = """Heap queues 34 35 [explanation by Franois Pinard] 36 37 Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 38 all k, counting elements from 0. For the sake of comparison, 39 non-existing elements are considered to be infinite. The interesting 40 property of a heap is that a[0] is always its smallest element. 41 42 The strange invariant above is meant to be an efficient memory 43 representation for a tournament. The numbers below are `k', not a[k]: 44 45 0 46 47 1 2 48 49 3 4 5 6 50 51 7 8 9 10 11 12 13 14 52 53 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 54 55 56 In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In 57 a usual binary tournament we see in sports, each cell is the winner 58 over the two cells it tops, and we can trace the winner down the tree 59 to see all opponents s/he had. However, in many computer applications 60 of such tournaments, we do not need to trace the history of a winner. 61 To be more memory efficient, when a winner is promoted, we try to 62 replace it by something else at a lower level, and the rule becomes 63 that a cell and the two cells it tops contain three different items, 64 but the top cell "wins" over the two topped cells. 65 66 If this heap invariant is protected at all time, index 0 is clearly 67 the overall winner. The simplest algorithmic way to remove it and 68 find the "next" winner is to move some loser (let's say cell 30 in the 69 diagram above) into the 0 position, and then percolate this new 0 down 70 the tree, exchanging values, until the invariant is re-established. 71 This is clearly logarithmic on the total number of items in the tree. 72 By iterating over all items, you get an O(n ln n) sort. 73 74 A nice feature of this sort is that you can efficiently insert new 75 items while the sort is going on, provided that the inserted items are 76 not "better" than the last 0'th element you extracted. This is 77 especially useful in simulation contexts, where the tree holds all 78 incoming events, and the "win" condition means the smallest scheduled 79 time. When an event schedule other events for execution, they are 80 scheduled into the future, so they can easily go into the heap. So, a 81 heap is a good structure for implementing schedulers (this is what I 82 used for my MIDI sequencer :-). 83 84 Various structures for implementing schedulers have been extensively 85 studied, and heaps are good for this, as they are reasonably speedy, 86 the speed is almost constant, and the worst case is not much different 87 than the average case. However, there are other representations which 88 are more efficient overall, yet the worst cases might be terrible. 89 90 Heaps are also very useful in big disk sorts. You most probably all 91 know that a big sort implies producing "runs" (which are pre-sorted 92 sequences, which size is usually related to the amount of CPU memory), 93 followed by a merging passes for these runs, which merging is often 94 very cleverly organised[1]. It is very important that the initial 95 sort produces the longest runs possible. Tournaments are a good way 96 to that. If, using all the memory available to hold a tournament, you 97 replace and percolate items that happen to fit the current run, you'll 98 produce runs which are twice the size of the memory for random input, 99 and much better for input fuzzily ordered. 100 101 Moreover, if you output the 0'th item on disk and get an input which 102 may not fit in the current tournament (because the value "wins" over 103 the last output value), it cannot fit in the heap, so the size of the 104 heap decreases. The freed memory could be cleverly reused immediately 105 for progressively building a second heap, which grows at exactly the 106 same rate the first heap is melting. When the first heap completely 107 vanishes, you switch heaps and start a new run. Clever and quite 108 effective! 109 110 In a word, heaps are useful memory structures to know. I use them in 111 a few applications, and I think it is good to keep a `heap' module 112 around. :-) 113 114 -------------------- 115 [1] The disk balancing algorithms which are current, nowadays, are 116 more annoying than clever, and this is a consequence of the seeking 117 capabilities of the disks. On devices which cannot seek, like big 118 tape drives, the story was quite different, and one had to be very 119 clever to ensure (far in advance) that each tape movement will be the 120 most effective possible (that is, will best participate at 121 "progressing" the merge). Some tapes were even able to read 122 backwards, and this was also used to avoid the rewinding time. 123 Believe me, real good tape sorts were quite spectacular to watch! 124 From all times, sorting has always been a Great Art! :-) 125 """ 126 127 __all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge', 128 'nlargest', 'nsmallest', 'heappushpop'] 129 130 def heappush(heap, item): 131 """Push item onto heap, maintaining the heap invariant.""" 132 heap.append(item) 133 _siftdown(heap, 0, len(heap)-1) 134 135 def heappop(heap): 136 """Pop the smallest item off the heap, maintaining the heap invariant.""" 137 lastelt = heap.pop() # raises appropriate IndexError if heap is empty 138 if heap: 139 returnitem = heap[0] 140 heap[0] = lastelt 141 _siftup(heap, 0) 142 return returnitem 143 return lastelt 144 145 def heapreplace(heap, item): 146 """Pop and return the current smallest value, and add the new item. 147 148 This is more efficient than heappop() followed by heappush(), and can be 149 more appropriate when using a fixed-size heap. Note that the value 150 returned may be larger than item! That constrains reasonable uses of 151 this routine unless written as part of a conditional replacement: 152 153 if item > heap[0]: 154 item = heapreplace(heap, item) 155 """ 156 returnitem = heap[0] # raises appropriate IndexError if heap is empty 157 heap[0] = item 158 _siftup(heap, 0) 159 return returnitem 160 161 def heappushpop(heap, item): 162 """Fast version of a heappush followed by a heappop.""" 163 if heap and heap[0] < item: 164 item, heap[0] = heap[0], item 165 _siftup(heap, 0) 166 return item 167 168 def heapify(x): 169 """Transform list into a heap, in-place, in O(len(x)) 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(range(n//2)): 177 _siftup(x, i) 178 179 def _heappop_max(heap): 180 """Maxheap version of a heappop.""" 181 lastelt = heap.pop() # raises appropriate IndexError if heap is empty 182 if heap: 183 returnitem = heap[0] 184 heap[0] = lastelt 185 _siftup_max(heap, 0) 186 return returnitem 187 return lastelt 188 189 def _heapreplace_max(heap, item): 190 """Maxheap version of a heappop followed by a heappush.""" 191 returnitem = heap[0] # raises appropriate IndexError if heap is empty 192 heap[0] = item 193 _siftup_max(heap, 0) 194 return returnitem 195 196 def _heapify_max(x): 197 """Transform list into a maxheap, in-place, in O(len(x)) time.""" 198 n = len(x) 199 for i in reversed(range(n//2)): 200 _siftup_max(x, i) 201 202 # 'heap' is a heap at all indices >= startpos, except possibly for pos. pos 203 # is the index of a leaf with a possibly out-of-order value. Restore the 204 # heap invariant. 205 def _siftdown(heap, startpos, pos): 206 newitem = heap[pos] 207 # Follow the path to the root, moving parents down until finding a place 208 # newitem fits. 209 while pos > startpos: 210 parentpos = (pos - 1) >> 1 211 parent = heap[parentpos] 212 if newitem < parent: 213 heap[pos] = parent 214 pos = parentpos 215 continue 216 break 217 heap[pos] = newitem 218 219 # The child indices of heap index pos are already heaps, and we want to make 220 # a heap at index pos too. We do this by bubbling the smaller child of 221 # pos up (and so on with that child's children, etc) until hitting a leaf, 222 # then using _siftdown to move the oddball originally at index pos into place. 223 # 224 # We *could* break out of the loop as soon as we find a pos where newitem <= 225 # both its children, but turns out that's not a good idea, and despite that 226 # many books write the algorithm that way. During a heap pop, the last array 227 # element is sifted in, and that tends to be large, so that comparing it 228 # against values starting from the root usually doesn't pay (= usually doesn't 229 # get us out of the loop early). See Knuth, Volume 3, where this is 230 # explained and quantified in an exercise. 231 # 232 # Cutting the # of comparisons is important, since these routines have no 233 # way to extract "the priority" from an array element, so that intelligence 234 # is likely to be hiding in custom comparison methods, or in array elements 235 # storing (priority, record) tuples. Comparisons are thus potentially 236 # expensive. 237 # 238 # On random arrays of length 1000, making this change cut the number of 239 # comparisons made by heapify() a little, and those made by exhaustive 240 # heappop() a lot, in accord with theory. Here are typical results from 3 241 # runs (3 just to demonstrate how small the variance is): 242 # 243 # Compares needed by heapify Compares needed by 1000 heappops 244 # -------------------------- -------------------------------- 245 # 1837 cut to 1663 14996 cut to 8680 246 # 1855 cut to 1659 14966 cut to 8678 247 # 1847 cut to 1660 15024 cut to 8703 248 # 249 # Building the heap by using heappush() 1000 times instead required 250 # 2198, 2148, and 2219 compares: heapify() is more efficient, when 251 # you can use it. 252 # 253 # The total compares needed by list.sort() on the same lists were 8627, 254 # 8627, and 8632 (this should be compared to the sum of heapify() and 255 # heappop() compares): list.sort() is (unsurprisingly!) more efficient 256 # for sorting. 257 258 def _siftup(heap, pos): 259 endpos = len(heap) 260 startpos = pos 261 newitem = heap[pos] 262 # Bubble up the smaller child until hitting a leaf. 263 childpos = 2*pos + 1 # leftmost child position 264 while childpos < endpos: 265 # Set childpos to index of smaller child. 266 rightpos = childpos + 1 267 if rightpos < endpos and not heap[childpos] < heap[rightpos]: 268 childpos = rightpos 269 # Move the smaller child up. 270 heap[pos] = heap[childpos] 271 pos = childpos 272 childpos = 2*pos + 1 273 # The leaf at pos is empty now. Put newitem there, and bubble it up 274 # to its final resting place (by sifting its parents down). 275 heap[pos] = newitem 276 _siftdown(heap, startpos, pos) 277 278 def _siftdown_max(heap, startpos, pos): 279 'Maxheap variant of _siftdown' 280 newitem = heap[pos] 281 # Follow the path to the root, moving parents down until finding a place 282 # newitem fits. 283 while pos > startpos: 284 parentpos = (pos - 1) >> 1 285 parent = heap[parentpos] 286 if parent < newitem: 287 heap[pos] = parent 288 pos = parentpos 289 continue 290 break 291 heap[pos] = newitem 292 293 def _siftup_max(heap, pos): 294 'Maxheap variant of _siftup' 295 endpos = len(heap) 296 startpos = pos 297 newitem = heap[pos] 298 # Bubble up the larger child until hitting a leaf. 299 childpos = 2*pos + 1 # leftmost child position 300 while childpos < endpos: 301 # Set childpos to index of larger child. 302 rightpos = childpos + 1 303 if rightpos < endpos and not heap[rightpos] < heap[childpos]: 304 childpos = rightpos 305 # Move the larger child up. 306 heap[pos] = heap[childpos] 307 pos = childpos 308 childpos = 2*pos + 1 309 # The leaf at pos is empty now. Put newitem there, and bubble it up 310 # to its final resting place (by sifting its parents down). 311 heap[pos] = newitem 312 _siftdown_max(heap, startpos, pos) 313 314 def merge(*iterables, key=None, reverse=False): 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 If *key* is not None, applies a key function to each element to determine 325 its sort order. 326 327 >>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len)) 328 ['dog', 'cat', 'fish', 'horse', 'kangaroo'] 329 330 ''' 331 332 h = [] 333 h_append = h.append 334 335 if reverse: 336 _heapify = _heapify_max 337 _heappop = _heappop_max 338 _heapreplace = _heapreplace_max 339 direction = -1 340 else: 341 _heapify = heapify 342 _heappop = heappop 343 _heapreplace = heapreplace 344 direction = 1 345 346 if key is None: 347 for order, it in enumerate(map(iter, iterables)): 348 try: 349 next = it.__next__ 350 h_append([next(), order * direction, next]) 351 except StopIteration: 352 pass 353 _heapify(h) 354 while len(h) > 1: 355 try: 356 while True: 357 value, order, next = s = h[0] 358 yield value 359 s[0] = next() # raises StopIteration when exhausted 360 _heapreplace(h, s) # restore heap condition 361 except StopIteration: 362 _heappop(h) # remove empty iterator 363 if h: 364 # fast case when only a single iterator remains 365 value, order, next = h[0] 366 yield value 367 yield from next.__self__ 368 return 369 370 for order, it in enumerate(map(iter, iterables)): 371 try: 372 next = it.__next__ 373 value = next() 374 h_append([key(value), order * direction, value, next]) 375 except StopIteration: 376 pass 377 _heapify(h) 378 while len(h) > 1: 379 try: 380 while True: 381 key_value, order, value, next = s = h[0] 382 yield value 383 value = next() 384 s[0] = key(value) 385 s[2] = value 386 _heapreplace(h, s) 387 except StopIteration: 388 _heappop(h) 389 if h: 390 key_value, order, value, next = h[0] 391 yield value 392 yield from next.__self__ 393 394 395 # Algorithm notes for nlargest() and nsmallest() 396 # ============================================== 397 # 398 # Make a single pass over the data while keeping the k most extreme values 399 # in a heap. Memory consumption is limited to keeping k values in a list. 400 # 401 # Measured performance for random inputs: 402 # 403 # number of comparisons 404 # n inputs k-extreme values (average of 5 trials) % more than min() 405 # ------------- ---------------- --------------------- ----------------- 406 # 1,000 100 3,317 231.7% 407 # 10,000 100 14,046 40.5% 408 # 100,000 100 105,749 5.7% 409 # 1,000,000 100 1,007,751 0.8% 410 # 10,000,000 100 10,009,401 0.1% 411 # 412 # Theoretical number of comparisons for k smallest of n random inputs: 413 # 414 # Step Comparisons Action 415 # ---- -------------------------- --------------------------- 416 # 1 1.66 * k heapify the first k-inputs 417 # 2 n - k compare remaining elements to top of heap 418 # 3 k * (1 + lg2(k)) * ln(n/k) replace the topmost value on the heap 419 # 4 k * lg2(k) - (k/2) final sort of the k most extreme values 420 # 421 # Combining and simplifying for a rough estimate gives: 422 # 423 # comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k)) 424 # 425 # Computing the number of comparisons for step 3: 426 # ----------------------------------------------- 427 # * For the i-th new value from the iterable, the probability of being in the 428 # k most extreme values is k/i. For example, the probability of the 101st 429 # value seen being in the 100 most extreme values is 100/101. 430 # * If the value is a new extreme value, the cost of inserting it into the 431 # heap is 1 + log(k, 2). 432 # * The probability times the cost gives: 433 # (k/i) * (1 + log(k, 2)) 434 # * Summing across the remaining n-k elements gives: 435 # sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1)) 436 # * This reduces to: 437 # (H(n) - H(k)) * k * (1 + log(k, 2)) 438 # * Where H(n) is the n-th harmonic number estimated by: 439 # gamma = 0.5772156649 440 # H(n) = log(n, e) + gamma + 1 / (2 * n) 441 # http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence 442 # * Substituting the H(n) formula: 443 # comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2) 444 # 445 # Worst-case for step 3: 446 # ---------------------- 447 # In the worst case, the input data is reversed sorted so that every new element 448 # must be inserted in the heap: 449 # 450 # comparisons = 1.66 * k + log(k, 2) * (n - k) 451 # 452 # Alternative Algorithms 453 # ---------------------- 454 # Other algorithms were not used because they: 455 # 1) Took much more auxiliary memory, 456 # 2) Made multiple passes over the data. 457 # 3) Made more comparisons in common cases (small k, large n, semi-random input). 458 # See the more detailed comparison of approach at: 459 # http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest 460 461 def nsmallest(n, iterable, key=None): 462 """Find the n smallest elements in a dataset. 463 464 Equivalent to: sorted(iterable, key=key)[:n] 465 """ 466 467 # Short-cut for n==1 is to use min() 468 if n == 1: 469 it = iter(iterable) 470 sentinel = object() 471 if key is None: 472 result = min(it, default=sentinel) 473 else: 474 result = min(it, default=sentinel, key=key) 475 return [] if result is sentinel else [result] 476 477 # When n>=size, it's faster to use sorted() 478 try: 479 size = len(iterable) 480 except (TypeError, AttributeError): 481 pass 482 else: 483 if n >= size: 484 return sorted(iterable, key=key)[:n] 485 486 # When key is none, use simpler decoration 487 if key is None: 488 it = iter(iterable) 489 # put the range(n) first so that zip() doesn't 490 # consume one too many elements from the iterator 491 result = [(elem, i) for i, elem in zip(range(n), it)] 492 if not result: 493 return result 494 _heapify_max(result) 495 top = result[0][0] 496 order = n 497 _heapreplace = _heapreplace_max 498 for elem in it: 499 if elem < top: 500 _heapreplace(result, (elem, order)) 501 top = result[0][0] 502 order += 1 503 result.sort() 504 return [r[0] for r in result] 505 506 # General case, slowest method 507 it = iter(iterable) 508 result = [(key(elem), i, elem) for i, elem in zip(range(n), it)] 509 if not result: 510 return result 511 _heapify_max(result) 512 top = result[0][0] 513 order = n 514 _heapreplace = _heapreplace_max 515 for elem in it: 516 k = key(elem) 517 if k < top: 518 _heapreplace(result, (k, order, elem)) 519 top = result[0][0] 520 order += 1 521 result.sort() 522 return [r[2] for r in result] 523 524 def nlargest(n, iterable, key=None): 525 """Find the n largest elements in a dataset. 526 527 Equivalent to: sorted(iterable, key=key, reverse=True)[:n] 528 """ 529 530 # Short-cut for n==1 is to use max() 531 if n == 1: 532 it = iter(iterable) 533 sentinel = object() 534 if key is None: 535 result = max(it, default=sentinel) 536 else: 537 result = max(it, default=sentinel, key=key) 538 return [] if result is sentinel else [result] 539 540 # When n>=size, it's faster to use sorted() 541 try: 542 size = len(iterable) 543 except (TypeError, AttributeError): 544 pass 545 else: 546 if n >= size: 547 return sorted(iterable, key=key, reverse=True)[:n] 548 549 # When key is none, use simpler decoration 550 if key is None: 551 it = iter(iterable) 552 result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)] 553 if not result: 554 return result 555 heapify(result) 556 top = result[0][0] 557 order = -n 558 _heapreplace = heapreplace 559 for elem in it: 560 if top < elem: 561 _heapreplace(result, (elem, order)) 562 top = result[0][0] 563 order -= 1 564 result.sort(reverse=True) 565 return [r[0] for r in result] 566 567 # General case, slowest method 568 it = iter(iterable) 569 result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)] 570 if not result: 571 return result 572 heapify(result) 573 top = result[0][0] 574 order = -n 575 _heapreplace = heapreplace 576 for elem in it: 577 k = key(elem) 578 if top < k: 579 _heapreplace(result, (k, order, elem)) 580 top = result[0][0] 581 order -= 1 582 result.sort(reverse=True) 583 return [r[2] for r in result] 584 585 # If available, use C implementation 586 try: 587 from _heapq import * 588 except ImportError: 589 pass 590 try: 591 from _heapq import _heapreplace_max 592 except ImportError: 593 pass 594 try: 595 from _heapq import _heapify_max 596 except ImportError: 597 pass 598 try: 599 from _heapq import _heappop_max 600 except ImportError: 601 pass 602 603 604 if __name__ == "__main__": 605 606 import doctest 607 print(doctest.testmod()) 608
