#### /Lib/heapq.py

Python | 393 lines | 359 code | 3 blank | 31 comment | 0 complexity | 33f8f0eba1662815249e87b96194fa55 MD5 | raw file

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 François 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', 'heappushpop'] 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 heappushpop(heap, item): 169 """Fast version of a heappush followed by a heappop.""" 170 if heap and heap[0] < item: 171 item, heap[0] = heap[0], item 172 _siftup(heap, 0) 173 return item 174 175def 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(xrange(n//2)): 184 _siftup(x, i) 185 186def 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 202def 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 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. 236def _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 newitem < parent: 244 heap[pos] = parent 245 pos = parentpos 246 continue 247 break 248 heap[pos] = newitem 249 250# The child indices of heap index pos are already heaps, and we want to make 251# a heap at index pos too. We do this by bubbling the smaller child of 252# pos up (and so on with that child's children, etc) until hitting a leaf, 253# then using _siftdown to move the oddball originally at index pos into place. 254# 255# We *could* break out of the loop as soon as we find a pos where newitem <= 256# both its children, but turns out that's not a good idea, and despite that 257# many books write the algorithm that way. During a heap pop, the last array 258# element is sifted in, and that tends to be large, so that comparing it 259# against values starting from the root usually doesn't pay (= usually doesn't 260# get us out of the loop early). See Knuth, Volume 3, where this is 261# explained and quantified in an exercise. 262# 263# Cutting the # of comparisons is important, since these routines have no 264# way to extract "the priority" from an array element, so that intelligence 265# is likely to be hiding in custom __cmp__ methods, or in array elements 266# storing (priority, record) tuples. Comparisons are thus potentially 267# expensive. 268# 269# On random arrays of length 1000, making this change cut the number of 270# comparisons made by heapify() a little, and those made by exhaustive 271# heappop() a lot, in accord with theory. Here are typical results from 3 272# runs (3 just to demonstrate how small the variance is): 273# 274# Compares needed by heapify Compares needed by 1000 heappops 275# -------------------------- -------------------------------- 276# 1837 cut to 1663 14996 cut to 8680 277# 1855 cut to 1659 14966 cut to 8678 278# 1847 cut to 1660 15024 cut to 8703 279# 280# Building the heap by using heappush() 1000 times instead required 281# 2198, 2148, and 2219 compares: heapify() is more efficient, when 282# you can use it. 283# 284# The total compares needed by list.sort() on the same lists were 8627, 285# 8627, and 8632 (this should be compared to the sum of heapify() and 286# heappop() compares): list.sort() is (unsurprisingly!) more efficient 287# for sorting. 288 289def _siftup(heap, pos): 290 endpos = len(heap) 291 startpos = pos 292 newitem = heap[pos] 293 # Bubble up the smaller child until hitting a leaf. 294 childpos = 2*pos + 1 # leftmost child position 295 while childpos < endpos: 296 # Set childpos to index of smaller child. 297 rightpos = childpos + 1 298 if rightpos < endpos and not heap[childpos] < heap[rightpos]: 299 childpos = rightpos 300 # Move the smaller child up. 301 heap[pos] = heap[childpos] 302 pos = childpos 303 childpos = 2*pos + 1 304 # The leaf at pos is empty now. Put newitem there, and bubble it up 305 # to its final resting place (by sifting its parents down). 306 heap[pos] = newitem 307 _siftdown(heap, startpos, pos) 308 309# If available, use C implementation 310try: 311 from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest, heappushpop 312except ImportError: 313 pass 314 315def merge(*iterables): 316 '''Merge multiple sorted inputs into a single sorted output. 317 318 Similar to sorted(itertools.chain(*iterables)) but returns a generator, 319 does not pull the data into memory all at once, and assumes that each of 320 the input streams is already sorted (smallest to largest). 321 322 >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25])) 323 [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25] 324 325 ''' 326 _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration 327 328 h = [] 329 h_append = h.append 330 for itnum, it in enumerate(map(iter, iterables)): 331 try: 332 next = it.next 333 h_append([next(), itnum, next]) 334 except _StopIteration: 335 pass 336 heapify(h) 337 338 while 1: 339 try: 340 while 1: 341 v, itnum, next = s = h[0] # raises IndexError when h is empty 342 yield v 343 s[0] = next() # raises StopIteration when exhausted 344 _heapreplace(h, s) # restore heap condition 345 except _StopIteration: 346 _heappop(h) # remove empty iterator 347 except IndexError: 348 return 349 350# Extend the implementations of nsmallest and nlargest to use a key= argument 351_nsmallest = nsmallest 352def nsmallest(n, iterable, key=None): 353 """Find the n smallest elements in a dataset. 354 355 Equivalent to: sorted(iterable, key=key)[:n] 356 """ 357 if key is None: 358 it = izip(iterable, count()) # decorate 359 result = _nsmallest(n, it) 360 return map(itemgetter(0), result) # undecorate 361 in1, in2 = tee(iterable) 362 it = izip(imap(key, in1), count(), in2) # decorate 363 result = _nsmallest(n, it) 364 return map(itemgetter(2), result) # undecorate 365 366_nlargest = nlargest 367def nlargest(n, iterable, key=None): 368 """Find the n largest elements in a dataset. 369 370 Equivalent to: sorted(iterable, key=key, reverse=True)[:n] 371 """ 372 if key is None: 373 it = izip(iterable, imap(neg, count())) # decorate 374 result = _nlargest(n, it) 375 return map(itemgetter(0), result) # undecorate 376 in1, in2 = tee(iterable) 377 it = izip(imap(key, in1), imap(neg, count()), in2) # decorate 378 result = _nlargest(n, it) 379 return map(itemgetter(2), result) # undecorate 380 381if __name__ == "__main__": 382 # Simple sanity test 383 heap = [] 384 data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] 385 for item in data: 386 heappush(heap, item) 387 sort = [] 388 while heap: 389 sort.append(heappop(heap)) 390 print sort 391 392 import doctest 393 doctest.testmod()