diff -r ffa851df0825 -r 2fb8b9db1c86 symbian-qemu-0.9.1-12/python-2.6.1/Lib/heapq.py --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/symbian-qemu-0.9.1-12/python-2.6.1/Lib/heapq.py Fri Jul 31 15:01:17 2009 +0100 @@ -0,0 +1,385 @@ +# -*- coding: Latin-1 -*- + +"""Heap queue algorithm (a.k.a. priority queue). + +Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for +all k, counting elements from 0. For the sake of comparison, +non-existing elements are considered to be infinite. The interesting +property of a heap is that a[0] is always its smallest element. + +Usage: + +heap = [] # creates an empty heap +heappush(heap, item) # pushes a new item on the heap +item = heappop(heap) # pops the smallest item from the heap +item = heap[0] # smallest item on the heap without popping it +heapify(x) # transforms list into a heap, in-place, in linear time +item = heapreplace(heap, item) # pops and returns smallest item, and adds + # new item; the heap size is unchanged + +Our API differs from textbook heap algorithms as follows: + +- We use 0-based indexing. This makes the relationship between the + index for a node and the indexes for its children slightly less + obvious, but is more suitable since Python uses 0-based indexing. + +- Our heappop() method returns the smallest item, not the largest. + +These two make it possible to view the heap as a regular Python list +without surprises: heap[0] is the smallest item, and heap.sort() +maintains the heap invariant! +""" + +# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger + +__about__ = """Heap queues + +[explanation by François Pinard] + +Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for +all k, counting elements from 0. For the sake of comparison, +non-existing elements are considered to be infinite. The interesting +property of a heap is that a[0] is always its smallest element. + +The strange invariant above is meant to be an efficient memory +representation for a tournament. The numbers below are `k', not a[k]: + + 0 + + 1 2 + + 3 4 5 6 + + 7 8 9 10 11 12 13 14 + + 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 + + +In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In +an usual binary tournament we see in sports, each cell is the winner +over the two cells it tops, and we can trace the winner down the tree +to see all opponents s/he had. However, in many computer applications +of such tournaments, we do not need to trace the history of a winner. +To be more memory efficient, when a winner is promoted, we try to +replace it by something else at a lower level, and the rule becomes +that a cell and the two cells it tops contain three different items, +but the top cell "wins" over the two topped cells. + +If this heap invariant is protected at all time, index 0 is clearly +the overall winner. The simplest algorithmic way to remove it and +find the "next" winner is to move some loser (let's say cell 30 in the +diagram above) into the 0 position, and then percolate this new 0 down +the tree, exchanging values, until the invariant is re-established. +This is clearly logarithmic on the total number of items in the tree. +By iterating over all items, you get an O(n ln n) sort. + +A nice feature of this sort is that you can efficiently insert new +items while the sort is going on, provided that the inserted items are +not "better" than the last 0'th element you extracted. This is +especially useful in simulation contexts, where the tree holds all +incoming events, and the "win" condition means the smallest scheduled +time. When an event schedule other events for execution, they are +scheduled into the future, so they can easily go into the heap. So, a +heap is a good structure for implementing schedulers (this is what I +used for my MIDI sequencer :-). + +Various structures for implementing schedulers have been extensively +studied, and heaps are good for this, as they are reasonably speedy, +the speed is almost constant, and the worst case is not much different +than the average case. However, there are other representations which +are more efficient overall, yet the worst cases might be terrible. + +Heaps are also very useful in big disk sorts. You most probably all +know that a big sort implies producing "runs" (which are pre-sorted +sequences, which size is usually related to the amount of CPU memory), +followed by a merging passes for these runs, which merging is often +very cleverly organised[1]. It is very important that the initial +sort produces the longest runs possible. Tournaments are a good way +to that. If, using all the memory available to hold a tournament, you +replace and percolate items that happen to fit the current run, you'll +produce runs which are twice the size of the memory for random input, +and much better for input fuzzily ordered. + +Moreover, if you output the 0'th item on disk and get an input which +may not fit in the current tournament (because the value "wins" over +the last output value), it cannot fit in the heap, so the size of the +heap decreases. The freed memory could be cleverly reused immediately +for progressively building a second heap, which grows at exactly the +same rate the first heap is melting. When the first heap completely +vanishes, you switch heaps and start a new run. Clever and quite +effective! + +In a word, heaps are useful memory structures to know. I use them in +a few applications, and I think it is good to keep a `heap' module +around. :-) + +-------------------- +[1] The disk balancing algorithms which are current, nowadays, are +more annoying than clever, and this is a consequence of the seeking +capabilities of the disks. On devices which cannot seek, like big +tape drives, the story was quite different, and one had to be very +clever to ensure (far in advance) that each tape movement will be the +most effective possible (that is, will best participate at +"progressing" the merge). Some tapes were even able to read +backwards, and this was also used to avoid the rewinding time. +Believe me, real good tape sorts were quite spectacular to watch! +From all times, sorting has always been a Great Art! :-) +""" + +__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge', + 'nlargest', 'nsmallest', 'heappushpop'] + +from itertools import islice, repeat, count, imap, izip, tee +from operator import itemgetter, neg +import bisect + +def heappush(heap, item): + """Push item onto heap, maintaining the heap invariant.""" + heap.append(item) + _siftdown(heap, 0, len(heap)-1) + +def heappop(heap): + """Pop the smallest item off the heap, maintaining the heap invariant.""" + lastelt = heap.pop() # raises appropriate IndexError if heap is empty + if heap: + returnitem = heap[0] + heap[0] = lastelt + _siftup(heap, 0) + else: + returnitem = lastelt + return returnitem + +def heapreplace(heap, item): + """Pop and return the current smallest value, and add the new item. + + This is more efficient than heappop() followed by heappush(), and can be + more appropriate when using a fixed-size heap. Note that the value + returned may be larger than item! That constrains reasonable uses of + this routine unless written as part of a conditional replacement: + + if item > heap[0]: + item = heapreplace(heap, item) + """ + returnitem = heap[0] # raises appropriate IndexError if heap is empty + heap[0] = item + _siftup(heap, 0) + return returnitem + +def heappushpop(heap, item): + """Fast version of a heappush followed by a heappop.""" + if heap and heap[0] < item: + item, heap[0] = heap[0], item + _siftup(heap, 0) + return item + +def heapify(x): + """Transform list into a heap, in-place, in O(len(heap)) time.""" + n = len(x) + # Transform bottom-up. The largest index there's any point to looking at + # is the largest with a child index in-range, so must have 2*i + 1 < n, + # or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so + # j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is + # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1. + for i in reversed(xrange(n//2)): + _siftup(x, i) + +def nlargest(n, iterable): + """Find the n largest elements in a dataset. + + Equivalent to: sorted(iterable, reverse=True)[:n] + """ + it = iter(iterable) + result = list(islice(it, n)) + if not result: + return result + heapify(result) + _heappushpop = heappushpop + for elem in it: + heappushpop(result, elem) + result.sort(reverse=True) + return result + +def nsmallest(n, iterable): + """Find the n smallest elements in a dataset. + + Equivalent to: sorted(iterable)[:n] + """ + if hasattr(iterable, '__len__') and n * 10 <= len(iterable): + # For smaller values of n, the bisect method is faster than a minheap. + # It is also memory efficient, consuming only n elements of space. + it = iter(iterable) + result = sorted(islice(it, 0, n)) + if not result: + return result + insort = bisect.insort + pop = result.pop + los = result[-1] # los --> Largest of the nsmallest + for elem in it: + if los <= elem: + continue + insort(result, elem) + pop() + los = result[-1] + return result + # An alternative approach manifests the whole iterable in memory but + # saves comparisons by heapifying all at once. Also, saves time + # over bisect.insort() which has O(n) data movement time for every + # insertion. Finding the n smallest of an m length iterable requires + # O(m) + O(n log m) comparisons. + h = list(iterable) + heapify(h) + return map(heappop, repeat(h, min(n, len(h)))) + +# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos +# is the index of a leaf with a possibly out-of-order value. Restore the +# heap invariant. +def _siftdown(heap, startpos, pos): + newitem = heap[pos] + # Follow the path to the root, moving parents down until finding a place + # newitem fits. + while pos > startpos: + parentpos = (pos - 1) >> 1 + parent = heap[parentpos] + if newitem < parent: + heap[pos] = parent + pos = parentpos + continue + break + heap[pos] = newitem + +# The child indices of heap index pos are already heaps, and we want to make +# a heap at index pos too. We do this by bubbling the smaller child of +# pos up (and so on with that child's children, etc) until hitting a leaf, +# then using _siftdown to move the oddball originally at index pos into place. +# +# We *could* break out of the loop as soon as we find a pos where newitem <= +# both its children, but turns out that's not a good idea, and despite that +# many books write the algorithm that way. During a heap pop, the last array +# element is sifted in, and that tends to be large, so that comparing it +# against values starting from the root usually doesn't pay (= usually doesn't +# get us out of the loop early). See Knuth, Volume 3, where this is +# explained and quantified in an exercise. +# +# Cutting the # of comparisons is important, since these routines have no +# way to extract "the priority" from an array element, so that intelligence +# is likely to be hiding in custom __cmp__ methods, or in array elements +# storing (priority, record) tuples. Comparisons are thus potentially +# expensive. +# +# On random arrays of length 1000, making this change cut the number of +# comparisons made by heapify() a little, and those made by exhaustive +# heappop() a lot, in accord with theory. Here are typical results from 3 +# runs (3 just to demonstrate how small the variance is): +# +# Compares needed by heapify Compares needed by 1000 heappops +# -------------------------- -------------------------------- +# 1837 cut to 1663 14996 cut to 8680 +# 1855 cut to 1659 14966 cut to 8678 +# 1847 cut to 1660 15024 cut to 8703 +# +# Building the heap by using heappush() 1000 times instead required +# 2198, 2148, and 2219 compares: heapify() is more efficient, when +# you can use it. +# +# The total compares needed by list.sort() on the same lists were 8627, +# 8627, and 8632 (this should be compared to the sum of heapify() and +# heappop() compares): list.sort() is (unsurprisingly!) more efficient +# for sorting. + +def _siftup(heap, pos): + endpos = len(heap) + startpos = pos + newitem = heap[pos] + # Bubble up the smaller child until hitting a leaf. + childpos = 2*pos + 1 # leftmost child position + while childpos < endpos: + # Set childpos to index of smaller child. + rightpos = childpos + 1 + if rightpos < endpos and not heap[childpos] < heap[rightpos]: + childpos = rightpos + # Move the smaller child up. + heap[pos] = heap[childpos] + pos = childpos + childpos = 2*pos + 1 + # The leaf at pos is empty now. Put newitem there, and bubble it up + # to its final resting place (by sifting its parents down). + heap[pos] = newitem + _siftdown(heap, startpos, pos) + +# If available, use C implementation +try: + from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest, heappushpop +except ImportError: + pass + +def merge(*iterables): + '''Merge multiple sorted inputs into a single sorted output. + + Similar to sorted(itertools.chain(*iterables)) but returns a generator, + does not pull the data into memory all at once, and assumes that each of + the input streams is already sorted (smallest to largest). + + >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25])) + [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25] + + ''' + _heappop, _heapreplace, _StopIteration = heappop, heapreplace, StopIteration + + h = [] + h_append = h.append + for itnum, it in enumerate(map(iter, iterables)): + try: + next = it.next + h_append([next(), itnum, next]) + except _StopIteration: + pass + heapify(h) + + while 1: + try: + while 1: + v, itnum, next = s = h[0] # raises IndexError when h is empty + yield v + s[0] = next() # raises StopIteration when exhausted + _heapreplace(h, s) # restore heap condition + except _StopIteration: + _heappop(h) # remove empty iterator + except IndexError: + return + +# Extend the implementations of nsmallest and nlargest to use a key= argument +_nsmallest = nsmallest +def nsmallest(n, iterable, key=None): + """Find the n smallest elements in a dataset. + + Equivalent to: sorted(iterable, key=key)[:n] + """ + in1, in2 = tee(iterable) + it = izip(imap(key, in1), count(), in2) # decorate + result = _nsmallest(n, it) + return map(itemgetter(2), result) # undecorate + +_nlargest = nlargest +def nlargest(n, iterable, key=None): + """Find the n largest elements in a dataset. + + Equivalent to: sorted(iterable, key=key, reverse=True)[:n] + """ + in1, in2 = tee(iterable) + it = izip(imap(key, in1), imap(neg, count()), in2) # decorate + result = _nlargest(n, it) + return map(itemgetter(2), result) # undecorate + +if __name__ == "__main__": + # Simple sanity test + heap = [] + data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] + for item in data: + heappush(heap, item) + sort = [] + while heap: + sort.append(heappop(heap)) + print sort + + import doctest + doctest.testmod()