heapq.py revision 00166c5532fc7562c07383a0ae2985b3ffaf253a
1# -*- coding: Latin-1 -*- 2 3"""Heap queue algorithm (a.k.a. priority queue). 4 5Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 6all k, counting elements from 0. For the sake of comparison, 7non-existing elements are considered to be infinite. The interesting 8property of a heap is that a[0] is always its smallest element. 9 10Usage: 11 12heap = [] # creates an empty heap 13heappush(heap, item) # pushes a new item on the heap 14item = heappop(heap) # pops the smallest item from the heap 15item = heap[0] # smallest item on the heap without popping it 16heapify(x) # transforms list into a heap, in-place, in linear time 17item = heapreplace(heap, item) # pops and returns smallest item, and adds 18 # new item; the heap size is unchanged 19 20Our 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 28These two make it possible to view the heap as a regular Python list 29without surprises: heap[0] is the smallest item, and heap.sort() 30maintains 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 Fran�ois Pinard] 38 39Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 40all k, counting elements from 0. For the sake of comparison, 41non-existing elements are considered to be infinite. The interesting 42property of a heap is that a[0] is always its smallest element. 43 44The strange invariant above is meant to be an efficient memory 45representation 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 58In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In 59an usual binary tournament we see in sports, each cell is the winner 60over the two cells it tops, and we can trace the winner down the tree 61to see all opponents s/he had. However, in many computer applications 62of such tournaments, we do not need to trace the history of a winner. 63To be more memory efficient, when a winner is promoted, we try to 64replace it by something else at a lower level, and the rule becomes 65that a cell and the two cells it tops contain three different items, 66but the top cell "wins" over the two topped cells. 67 68If this heap invariant is protected at all time, index 0 is clearly 69the overall winner. The simplest algorithmic way to remove it and 70find the "next" winner is to move some loser (let's say cell 30 in the 71diagram above) into the 0 position, and then percolate this new 0 down 72the tree, exchanging values, until the invariant is re-established. 73This is clearly logarithmic on the total number of items in the tree. 74By iterating over all items, you get an O(n ln n) sort. 75 76A nice feature of this sort is that you can efficiently insert new 77items while the sort is going on, provided that the inserted items are 78not "better" than the last 0'th element you extracted. This is 79especially useful in simulation contexts, where the tree holds all 80incoming events, and the "win" condition means the smallest scheduled 81time. When an event schedule other events for execution, they are 82scheduled into the future, so they can easily go into the heap. So, a 83heap is a good structure for implementing schedulers (this is what I 84used for my MIDI sequencer :-). 85 86Various structures for implementing schedulers have been extensively 87studied, and heaps are good for this, as they are reasonably speedy, 88the speed is almost constant, and the worst case is not much different 89than the average case. However, there are other representations which 90are more efficient overall, yet the worst cases might be terrible. 91 92Heaps are also very useful in big disk sorts. You most probably all 93know that a big sort implies producing "runs" (which are pre-sorted 94sequences, which size is usually related to the amount of CPU memory), 95followed by a merging passes for these runs, which merging is often 96very cleverly organised[1]. It is very important that the initial 97sort produces the longest runs possible. Tournaments are a good way 98to that. If, using all the memory available to hold a tournament, you 99replace and percolate items that happen to fit the current run, you'll 100produce runs which are twice the size of the memory for random input, 101and much better for input fuzzily ordered. 102 103Moreover, if you output the 0'th item on disk and get an input which 104may not fit in the current tournament (because the value "wins" over 105the last output value), it cannot fit in the heap, so the size of the 106heap decreases. The freed memory could be cleverly reused immediately 107for progressively building a second heap, which grows at exactly the 108same rate the first heap is melting. When the first heap completely 109vanishes, you switch heaps and start a new run. Clever and quite 110effective! 111 112In a word, heaps are useful memory structures to know. I use them in 113a few applications, and I think it is good to keep a `heap' module 114around. :-) 115 116-------------------- 117[1] The disk balancing algorithms which are current, nowadays, are 118more annoying than clever, and this is a consequence of the seeking 119capabilities of the disks. On devices which cannot seek, like big 120tape drives, the story was quite different, and one had to be very 121clever to ensure (far in advance) that each tape movement will be the 122most effective possible (that is, will best participate at 123"progressing" the merge). Some tapes were even able to read 124backwards, and this was also used to avoid the rewinding time. 125Believe me, real good tape sorts were quite spectacular to watch! 126From all times, sorting has always been a Great Art! :-) 127""" 128 129__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge', 130 'nlargest', 'nsmallest'] 131 132from itertools import islice, repeat, count, imap, izip, tee 133from operator import itemgetter, neg 134import bisect 135 136def heappush(heap, item): 137 """Push item onto heap, maintaining the heap invariant.""" 138 heap.append(item) 139 _siftdown(heap, 0, len(heap)-1) 140 141def 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 152def 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 168def 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 179def 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 199def 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. 233def _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 285def _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 306try: 307 from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest 308except ImportError: 309 pass 310 311def merge(*iterables): 312 '''Merge multiple sorted inputs into a single sorted output. 313 314 Similar to sorted(itertools.chain(*iterables)) but returns an iterable, 315 does not pull the data into memory all at once, and reduces the number 316 of comparisons by assuming that each of the input streams is already sorted. 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, siftup, _StopIteration = heappop, _siftup, StopIteration 323 324 h = [] 325 h_append = h.append 326 for it in map(iter, iterables): 327 try: 328 next = it.next 329 h_append([next(), next]) 330 except _StopIteration: 331 pass 332 heapify(h) 333 334 while 1: 335 try: 336 while 1: 337 v, next = s = h[0] # raises IndexError when h is empty 338 yield v 339 s[0] = next() # raises StopIteration when exhausted 340 siftup(h, 0) # 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 348def 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 359def 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 369if __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() 382