|  | """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 | 
|  | a 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'] | 
|  |  | 
|  | 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) | 
|  | return returnitem | 
|  | return lastelt | 
|  |  | 
|  | 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(x)) 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(range(n//2)): | 
|  | _siftup(x, i) | 
|  |  | 
|  | def _heappop_max(heap): | 
|  | """Maxheap version of a heappop.""" | 
|  | lastelt = heap.pop()    # raises appropriate IndexError if heap is empty | 
|  | if heap: | 
|  | returnitem = heap[0] | 
|  | heap[0] = lastelt | 
|  | _siftup_max(heap, 0) | 
|  | return returnitem | 
|  | return lastelt | 
|  |  | 
|  | def _heapreplace_max(heap, item): | 
|  | """Maxheap version of a heappop followed by a heappush.""" | 
|  | returnitem = heap[0]    # raises appropriate IndexError if heap is empty | 
|  | heap[0] = item | 
|  | _siftup_max(heap, 0) | 
|  | return returnitem | 
|  |  | 
|  | def _heapify_max(x): | 
|  | """Transform list into a maxheap, in-place, in O(len(x)) time.""" | 
|  | n = len(x) | 
|  | for i in reversed(range(n//2)): | 
|  | _siftup_max(x, i) | 
|  |  | 
|  | # '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 comparison 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) | 
|  |  | 
|  | def _siftdown_max(heap, startpos, pos): | 
|  | 'Maxheap variant of _siftdown' | 
|  | 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 parent < newitem: | 
|  | heap[pos] = parent | 
|  | pos = parentpos | 
|  | continue | 
|  | break | 
|  | heap[pos] = newitem | 
|  |  | 
|  | def _siftup_max(heap, pos): | 
|  | 'Maxheap variant of _siftup' | 
|  | endpos = len(heap) | 
|  | startpos = pos | 
|  | newitem = heap[pos] | 
|  | # Bubble up the larger child until hitting a leaf. | 
|  | childpos = 2*pos + 1    # leftmost child position | 
|  | while childpos < endpos: | 
|  | # Set childpos to index of larger child. | 
|  | rightpos = childpos + 1 | 
|  | if rightpos < endpos and not heap[rightpos] < heap[childpos]: | 
|  | childpos = rightpos | 
|  | # Move the larger 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_max(heap, startpos, pos) | 
|  |  | 
|  | def merge(*iterables, key=None, reverse=False): | 
|  | '''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] | 
|  |  | 
|  | If *key* is not None, applies a key function to each element to determine | 
|  | its sort order. | 
|  |  | 
|  | >>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len)) | 
|  | ['dog', 'cat', 'fish', 'horse', 'kangaroo'] | 
|  |  | 
|  | ''' | 
|  |  | 
|  | h = [] | 
|  | h_append = h.append | 
|  |  | 
|  | if reverse: | 
|  | _heapify = _heapify_max | 
|  | _heappop = _heappop_max | 
|  | _heapreplace = _heapreplace_max | 
|  | direction = -1 | 
|  | else: | 
|  | _heapify = heapify | 
|  | _heappop = heappop | 
|  | _heapreplace = heapreplace | 
|  | direction = 1 | 
|  |  | 
|  | if key is None: | 
|  | for order, it in enumerate(map(iter, iterables)): | 
|  | try: | 
|  | next = it.__next__ | 
|  | h_append([next(), order * direction, next]) | 
|  | except StopIteration: | 
|  | pass | 
|  | _heapify(h) | 
|  | while len(h) > 1: | 
|  | try: | 
|  | while True: | 
|  | value, order, next = s = h[0] | 
|  | yield value | 
|  | s[0] = next()           # raises StopIteration when exhausted | 
|  | _heapreplace(h, s)      # restore heap condition | 
|  | except StopIteration: | 
|  | _heappop(h)                 # remove empty iterator | 
|  | if h: | 
|  | # fast case when only a single iterator remains | 
|  | value, order, next = h[0] | 
|  | yield value | 
|  | yield from next.__self__ | 
|  | return | 
|  |  | 
|  | for order, it in enumerate(map(iter, iterables)): | 
|  | try: | 
|  | next = it.__next__ | 
|  | value = next() | 
|  | h_append([key(value), order * direction, value, next]) | 
|  | except StopIteration: | 
|  | pass | 
|  | _heapify(h) | 
|  | while len(h) > 1: | 
|  | try: | 
|  | while True: | 
|  | key_value, order, value, next = s = h[0] | 
|  | yield value | 
|  | value = next() | 
|  | s[0] = key(value) | 
|  | s[2] = value | 
|  | _heapreplace(h, s) | 
|  | except StopIteration: | 
|  | _heappop(h) | 
|  | if h: | 
|  | key_value, order, value, next = h[0] | 
|  | yield value | 
|  | yield from next.__self__ | 
|  |  | 
|  |  | 
|  | # Algorithm notes for nlargest() and nsmallest() | 
|  | # ============================================== | 
|  | # | 
|  | # Make a single pass over the data while keeping the k most extreme values | 
|  | # in a heap.  Memory consumption is limited to keeping k values in a list. | 
|  | # | 
|  | # Measured performance for random inputs: | 
|  | # | 
|  | #                                   number of comparisons | 
|  | #    n inputs     k-extreme values  (average of 5 trials)   % more than min() | 
|  | # -------------   ----------------  ---------------------   ----------------- | 
|  | #      1,000           100                  3,317               231.7% | 
|  | #     10,000           100                 14,046                40.5% | 
|  | #    100,000           100                105,749                 5.7% | 
|  | #  1,000,000           100              1,007,751                 0.8% | 
|  | # 10,000,000           100             10,009,401                 0.1% | 
|  | # | 
|  | # Theoretical number of comparisons for k smallest of n random inputs: | 
|  | # | 
|  | # Step   Comparisons                  Action | 
|  | # ----   --------------------------   --------------------------- | 
|  | #  1     1.66 * k                     heapify the first k-inputs | 
|  | #  2     n - k                        compare remaining elements to top of heap | 
|  | #  3     k * (1 + lg2(k)) * ln(n/k)   replace the topmost value on the heap | 
|  | #  4     k * lg2(k) - (k/2)           final sort of the k most extreme values | 
|  | # | 
|  | # Combining and simplifying for a rough estimate gives: | 
|  | # | 
|  | #        comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k)) | 
|  | # | 
|  | # Computing the number of comparisons for step 3: | 
|  | # ----------------------------------------------- | 
|  | # * For the i-th new value from the iterable, the probability of being in the | 
|  | #   k most extreme values is k/i.  For example, the probability of the 101st | 
|  | #   value seen being in the 100 most extreme values is 100/101. | 
|  | # * If the value is a new extreme value, the cost of inserting it into the | 
|  | #   heap is 1 + log(k, 2). | 
|  | # * The probability times the cost gives: | 
|  | #            (k/i) * (1 + log(k, 2)) | 
|  | # * Summing across the remaining n-k elements gives: | 
|  | #            sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1)) | 
|  | # * This reduces to: | 
|  | #            (H(n) - H(k)) * k * (1 + log(k, 2)) | 
|  | # * Where H(n) is the n-th harmonic number estimated by: | 
|  | #            gamma = 0.5772156649 | 
|  | #            H(n) = log(n, e) + gamma + 1 / (2 * n) | 
|  | #   http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence | 
|  | # * Substituting the H(n) formula: | 
|  | #            comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2) | 
|  | # | 
|  | # Worst-case for step 3: | 
|  | # ---------------------- | 
|  | # In the worst case, the input data is reversed sorted so that every new element | 
|  | # must be inserted in the heap: | 
|  | # | 
|  | #             comparisons = 1.66 * k + log(k, 2) * (n - k) | 
|  | # | 
|  | # Alternative Algorithms | 
|  | # ---------------------- | 
|  | # Other algorithms were not used because they: | 
|  | # 1) Took much more auxiliary memory, | 
|  | # 2) Made multiple passes over the data. | 
|  | # 3) Made more comparisons in common cases (small k, large n, semi-random input). | 
|  | # See the more detailed comparison of approach at: | 
|  | # http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest | 
|  |  | 
|  | def nsmallest(n, iterable, key=None): | 
|  | """Find the n smallest elements in a dataset. | 
|  |  | 
|  | Equivalent to:  sorted(iterable, key=key)[:n] | 
|  | """ | 
|  |  | 
|  | # Short-cut for n==1 is to use min() | 
|  | if n == 1: | 
|  | it = iter(iterable) | 
|  | sentinel = object() | 
|  | result = min(it, default=sentinel, key=key) | 
|  | return [] if result is sentinel else [result] | 
|  |  | 
|  | # When n>=size, it's faster to use sorted() | 
|  | try: | 
|  | size = len(iterable) | 
|  | except (TypeError, AttributeError): | 
|  | pass | 
|  | else: | 
|  | if n >= size: | 
|  | return sorted(iterable, key=key)[:n] | 
|  |  | 
|  | # When key is none, use simpler decoration | 
|  | if key is None: | 
|  | it = iter(iterable) | 
|  | # put the range(n) first so that zip() doesn't | 
|  | # consume one too many elements from the iterator | 
|  | result = [(elem, i) for i, elem in zip(range(n), it)] | 
|  | if not result: | 
|  | return result | 
|  | _heapify_max(result) | 
|  | top = result[0][0] | 
|  | order = n | 
|  | _heapreplace = _heapreplace_max | 
|  | for elem in it: | 
|  | if elem < top: | 
|  | _heapreplace(result, (elem, order)) | 
|  | top, _order = result[0] | 
|  | order += 1 | 
|  | result.sort() | 
|  | return [elem for (elem, order) in result] | 
|  |  | 
|  | # General case, slowest method | 
|  | it = iter(iterable) | 
|  | result = [(key(elem), i, elem) for i, elem in zip(range(n), it)] | 
|  | if not result: | 
|  | return result | 
|  | _heapify_max(result) | 
|  | top = result[0][0] | 
|  | order = n | 
|  | _heapreplace = _heapreplace_max | 
|  | for elem in it: | 
|  | k = key(elem) | 
|  | if k < top: | 
|  | _heapreplace(result, (k, order, elem)) | 
|  | top, _order, _elem = result[0] | 
|  | order += 1 | 
|  | result.sort() | 
|  | return [elem for (k, order, elem) in result] | 
|  |  | 
|  | def nlargest(n, iterable, key=None): | 
|  | """Find the n largest elements in a dataset. | 
|  |  | 
|  | Equivalent to:  sorted(iterable, key=key, reverse=True)[:n] | 
|  | """ | 
|  |  | 
|  | # Short-cut for n==1 is to use max() | 
|  | if n == 1: | 
|  | it = iter(iterable) | 
|  | sentinel = object() | 
|  | result = max(it, default=sentinel, key=key) | 
|  | return [] if result is sentinel else [result] | 
|  |  | 
|  | # When n>=size, it's faster to use sorted() | 
|  | try: | 
|  | size = len(iterable) | 
|  | except (TypeError, AttributeError): | 
|  | pass | 
|  | else: | 
|  | if n >= size: | 
|  | return sorted(iterable, key=key, reverse=True)[:n] | 
|  |  | 
|  | # When key is none, use simpler decoration | 
|  | if key is None: | 
|  | it = iter(iterable) | 
|  | result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)] | 
|  | if not result: | 
|  | return result | 
|  | heapify(result) | 
|  | top = result[0][0] | 
|  | order = -n | 
|  | _heapreplace = heapreplace | 
|  | for elem in it: | 
|  | if top < elem: | 
|  | _heapreplace(result, (elem, order)) | 
|  | top, _order = result[0] | 
|  | order -= 1 | 
|  | result.sort(reverse=True) | 
|  | return [elem for (elem, order) in result] | 
|  |  | 
|  | # General case, slowest method | 
|  | it = iter(iterable) | 
|  | result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)] | 
|  | if not result: | 
|  | return result | 
|  | heapify(result) | 
|  | top = result[0][0] | 
|  | order = -n | 
|  | _heapreplace = heapreplace | 
|  | for elem in it: | 
|  | k = key(elem) | 
|  | if top < k: | 
|  | _heapreplace(result, (k, order, elem)) | 
|  | top, _order, _elem = result[0] | 
|  | order -= 1 | 
|  | result.sort(reverse=True) | 
|  | return [elem for (k, order, elem) in result] | 
|  |  | 
|  | # If available, use C implementation | 
|  | try: | 
|  | from _heapq import * | 
|  | except ImportError: | 
|  | pass | 
|  | try: | 
|  | from _heapq import _heapreplace_max | 
|  | except ImportError: | 
|  | pass | 
|  | try: | 
|  | from _heapq import _heapify_max | 
|  | except ImportError: | 
|  | pass | 
|  | try: | 
|  | from _heapq import _heappop_max | 
|  | except ImportError: | 
|  | pass | 
|  |  | 
|  |  | 
|  | if __name__ == "__main__": | 
|  |  | 
|  | import doctest # pragma: no cover | 
|  | print(doctest.testmod()) # pragma: no cover |