heapq.py revision e1defa4175426594be53c1bc6c3d2f02a0952bae
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', 'nlargest', 130 'nsmallest'] 131 132from itertools import islice, repeat 133import bisect 134 135def heappush(heap, item): 136 """Push item onto heap, maintaining the heap invariant.""" 137 heap.append(item) 138 _siftdown(heap, 0, len(heap)-1) 139 140def heappop(heap): 141 """Pop the smallest item off the heap, maintaining the heap invariant.""" 142 lastelt = heap.pop() # raises appropriate IndexError if heap is empty 143 if heap: 144 returnitem = heap[0] 145 heap[0] = lastelt 146 _siftup(heap, 0) 147 else: 148 returnitem = lastelt 149 return returnitem 150 151def heapreplace(heap, item): 152 """Pop and return the current smallest value, and add the new item. 153 154 This is more efficient than heappop() followed by heappush(), and can be 155 more appropriate when using a fixed-size heap. Note that the value 156 returned may be larger than item! That constrains reasonable uses of 157 this routine unless written as part of a conditional replacement: 158 159 if item > heap[0]: 160 item = heapreplace(heap, item) 161 """ 162 returnitem = heap[0] # raises appropriate IndexError if heap is empty 163 heap[0] = item 164 _siftup(heap, 0) 165 return returnitem 166 167def heapify(x): 168 """Transform list into a heap, in-place, in O(len(heap)) time.""" 169 n = len(x) 170 # Transform bottom-up. The largest index there's any point to looking at 171 # is the largest with a child index in-range, so must have 2*i + 1 < n, 172 # or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so 173 # j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is 174 # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1. 175 for i in reversed(xrange(n//2)): 176 _siftup(x, i) 177 178def nlargest(n, iterable): 179 """Find the n largest elements in a dataset. 180 181 Equivalent to: sorted(iterable, reverse=True)[:n] 182 """ 183 it = iter(iterable) 184 result = list(islice(it, n)) 185 if not result: 186 return result 187 heapify(result) 188 _heapreplace = heapreplace 189 sol = result[0] # sol --> smallest of the nlargest 190 for elem in it: 191 if elem <= sol: 192 continue 193 _heapreplace(result, elem) 194 sol = result[0] 195 result.sort(reverse=True) 196 return result 197 198def nsmallest(n, iterable): 199 """Find the n smallest elements in a dataset. 200 201 Equivalent to: sorted(iterable)[:n] 202 """ 203 if hasattr(iterable, '__len__') and n * 10 <= len(iterable): 204 # For smaller values of n, the bisect method is faster than a minheap. 205 # It is also memory efficient, consuming only n elements of space. 206 it = iter(iterable) 207 result = sorted(islice(it, 0, n)) 208 if not result: 209 return result 210 insort = bisect.insort 211 pop = result.pop 212 los = result[-1] # los --> Largest of the nsmallest 213 for elem in it: 214 if los <= elem: 215 continue 216 insort(result, elem) 217 pop() 218 los = result[-1] 219 return result 220 # An alternative approach manifests the whole iterable in memory but 221 # saves comparisons by heapifying all at once. Also, saves time 222 # over bisect.insort() which has O(n) data movement time for every 223 # insertion. Finding the n smallest of an m length iterable requires 224 # O(m) + O(n log m) comparisons. 225 h = list(iterable) 226 heapify(h) 227 return map(heappop, repeat(h, min(n, len(h)))) 228 229# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos 230# is the index of a leaf with a possibly out-of-order value. Restore the 231# heap invariant. 232def _siftdown(heap, startpos, pos): 233 newitem = heap[pos] 234 # Follow the path to the root, moving parents down until finding a place 235 # newitem fits. 236 while pos > startpos: 237 parentpos = (pos - 1) >> 1 238 parent = heap[parentpos] 239 if parent <= newitem: 240 break 241 heap[pos] = parent 242 pos = parentpos 243 heap[pos] = newitem 244 245# The child indices of heap index pos are already heaps, and we want to make 246# a heap at index pos too. We do this by bubbling the smaller child of 247# pos up (and so on with that child's children, etc) until hitting a leaf, 248# then using _siftdown to move the oddball originally at index pos into place. 249# 250# We *could* break out of the loop as soon as we find a pos where newitem <= 251# both its children, but turns out that's not a good idea, and despite that 252# many books write the algorithm that way. During a heap pop, the last array 253# element is sifted in, and that tends to be large, so that comparing it 254# against values starting from the root usually doesn't pay (= usually doesn't 255# get us out of the loop early). See Knuth, Volume 3, where this is 256# explained and quantified in an exercise. 257# 258# Cutting the # of comparisons is important, since these routines have no 259# way to extract "the priority" from an array element, so that intelligence 260# is likely to be hiding in custom __cmp__ methods, or in array elements 261# storing (priority, record) tuples. Comparisons are thus potentially 262# expensive. 263# 264# On random arrays of length 1000, making this change cut the number of 265# comparisons made by heapify() a little, and those made by exhaustive 266# heappop() a lot, in accord with theory. Here are typical results from 3 267# runs (3 just to demonstrate how small the variance is): 268# 269# Compares needed by heapify Compares needed by 1000 heappops 270# -------------------------- -------------------------------- 271# 1837 cut to 1663 14996 cut to 8680 272# 1855 cut to 1659 14966 cut to 8678 273# 1847 cut to 1660 15024 cut to 8703 274# 275# Building the heap by using heappush() 1000 times instead required 276# 2198, 2148, and 2219 compares: heapify() is more efficient, when 277# you can use it. 278# 279# The total compares needed by list.sort() on the same lists were 8627, 280# 8627, and 8632 (this should be compared to the sum of heapify() and 281# heappop() compares): list.sort() is (unsurprisingly!) more efficient 282# for sorting. 283 284def _siftup(heap, pos): 285 endpos = len(heap) 286 startpos = pos 287 newitem = heap[pos] 288 # Bubble up the smaller child until hitting a leaf. 289 childpos = 2*pos + 1 # leftmost child position 290 while childpos < endpos: 291 # Set childpos to index of smaller child. 292 rightpos = childpos + 1 293 if rightpos < endpos and heap[rightpos] <= heap[childpos]: 294 childpos = rightpos 295 # Move the smaller child up. 296 heap[pos] = heap[childpos] 297 pos = childpos 298 childpos = 2*pos + 1 299 # The leaf at pos is empty now. Put newitem there, and bubble it up 300 # to its final resting place (by sifting its parents down). 301 heap[pos] = newitem 302 _siftdown(heap, startpos, pos) 303 304# If available, use C implementation 305try: 306 from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest 307except ImportError: 308 pass 309 310if __name__ == "__main__": 311 # Simple sanity test 312 heap = [] 313 data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] 314 for item in data: 315 heappush(heap, item) 316 sort = [] 317 while heap: 318 sort.append(heappop(heap)) 319 print sort 320