Add missing issue number in Misc/NEWS entry.
[python.git] / Lib / heapq.py
bloba44d1beb047c2f431703d48e7111890c753cf282
1 # -*- coding: latin-1 -*-
3 """Heap queue algorithm (a.k.a. priority queue).
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.
10 Usage:
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
20 Our API differs from textbook heap algorithms as follows:
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.
26 - Our heappop() method returns the smallest item, not the largest.
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 """
33 # Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger
35 __about__ = """Heap queues
37 [explanation by François Pinard]
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.
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]:
49 1 2
51 3 4 5 6
53 7 8 9 10 11 12 13 14
55 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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.
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.
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 :-).
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.
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.
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!
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. :-)
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! :-)
129 __all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
130 'nlargest', 'nsmallest', 'heappushpop']
132 from itertools import islice, repeat, count, imap, izip, tee, chain
133 from operator import itemgetter
134 import bisect
136 def heappush(heap, item):
137 """Push item onto heap, maintaining the heap invariant."""
138 heap.append(item)
139 _siftdown(heap, 0, len(heap)-1)
141 def 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
152 def heapreplace(heap, item):
153 """Pop and return the current smallest value, and add the new item.
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:
160 if item > heap[0]:
161 item = heapreplace(heap, item)
163 returnitem = heap[0] # raises appropriate IndexError if heap is empty
164 heap[0] = item
165 _siftup(heap, 0)
166 return returnitem
168 def 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
175 def 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)
186 def nlargest(n, iterable):
187 """Find the n largest elements in a dataset.
189 Equivalent to: sorted(iterable, reverse=True)[:n]
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
202 def nsmallest(n, iterable):
203 """Find the n smallest elements in a dataset.
205 Equivalent to: sorted(iterable)[:n]
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))))
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.
236 def _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
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.
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.
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.
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):
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
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.
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.
289 def _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)
309 # If available, use C implementation
310 try:
311 from _heapq import *
312 except ImportError:
313 pass
315 def merge(*iterables):
316 '''Merge multiple sorted inputs into a single sorted output.
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).
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]
326 _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration
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)
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
350 # Extend the implementations of nsmallest and nlargest to use a key= argument
351 _nsmallest = nsmallest
352 def nsmallest(n, iterable, key=None):
353 """Find the n smallest elements in a dataset.
355 Equivalent to: sorted(iterable, key=key)[:n]
357 # Short-cut for n==1 is to use min() when len(iterable)>0
358 if n == 1:
359 it = iter(iterable)
360 head = list(islice(it, 1))
361 if not head:
362 return []
363 if key is None:
364 return [min(chain(head, it))]
365 return [min(chain(head, it), key=key)]
367 # When n>=size, it's faster to use sort()
368 try:
369 size = len(iterable)
370 except (TypeError, AttributeError):
371 pass
372 else:
373 if n >= size:
374 return sorted(iterable, key=key)[:n]
376 # When key is none, use simpler decoration
377 if key is None:
378 it = izip(iterable, count()) # decorate
379 result = _nsmallest(n, it)
380 return map(itemgetter(0), result) # undecorate
382 # General case, slowest method
383 in1, in2 = tee(iterable)
384 it = izip(imap(key, in1), count(), in2) # decorate
385 result = _nsmallest(n, it)
386 return map(itemgetter(2), result) # undecorate
388 _nlargest = nlargest
389 def nlargest(n, iterable, key=None):
390 """Find the n largest elements in a dataset.
392 Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
395 # Short-cut for n==1 is to use max() when len(iterable)>0
396 if n == 1:
397 it = iter(iterable)
398 head = list(islice(it, 1))
399 if not head:
400 return []
401 if key is None:
402 return [max(chain(head, it))]
403 return [max(chain(head, it), key=key)]
405 # When n>=size, it's faster to use sort()
406 try:
407 size = len(iterable)
408 except (TypeError, AttributeError):
409 pass
410 else:
411 if n >= size:
412 return sorted(iterable, key=key, reverse=True)[:n]
414 # When key is none, use simpler decoration
415 if key is None:
416 it = izip(iterable, count(0,-1)) # decorate
417 result = _nlargest(n, it)
418 return map(itemgetter(0), result) # undecorate
420 # General case, slowest method
421 in1, in2 = tee(iterable)
422 it = izip(imap(key, in1), count(0,-1), in2) # decorate
423 result = _nlargest(n, it)
424 return map(itemgetter(2), result) # undecorate
426 if __name__ == "__main__":
427 # Simple sanity test
428 heap = []
429 data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
430 for item in data:
431 heappush(heap, item)
432 sort = []
433 while heap:
434 sort.append(heappop(heap))
435 print sort
437 import doctest
438 doctest.testmod()