前言
Python没有提供堆的数据类型，heapq本质上就是堆的操作方法合集。heapq是Python的一个高级模块，注释描述是：Heap queue algorithm (a.k.a. priority queue).
可以先回顾一下堆的概念，参考：堆
heapq也可以理解为堆的Python实现，由于没有用到C代码，所以直接把源代码贴最后，供读者参考。
主要接口
heapq里接口操作的对象是list
比较重大的接口：heappush、heappop、heapify、heapreplace、merge、nlargest、nsmallest、heappushpop。
其中heappush、 heappop、heapreplace、heappushpop源代码很简单，就简单描述一下怎么实现的。heapify是建堆，是对堆从C++翻译成python，也就不多赘述。 merge、nlargest、 nsmallest倒是可以说道说道。
heappush/heappop
heappush就是往列表做了append，然后再向上调整
heappop就是取堆顶，然后把堆尾放在堆顶，最后对堆顶做向下调整
heappush和heappop的时间复杂度都是log(k)
heapreplace/heappushpop
heapreplace是返回堆顶，再把新元素放堆顶，然后对堆顶做向下调整
heappushpop是快速实现同时进行heappush和heappop，做法是用新元素和堆顶做比较，如果堆顶大于新元素，则直接返回新元素，否则返回堆顶，并且把新元素放在堆顶后向下调整
heapreplace和heappushpop的时间复杂度都是log(k)
merge
merge是一个 generator（生成器）。先看一下定义：
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]
merge是对多路归并的实现，这在算法—外部排序&多路归并和堆-优先级队列中都有描述
实现上就是维护一个堆，每次取堆顶，取完堆顶后对堆顶的数据那一路数据next。这一路数据用完了，就直接pop这一路数据，直到只剩下一路数据。直接对最后一路数据遍历即可。
分析一下效率，假设有k路数据，总的数据量是n。先建了个k大小的堆，就是O(k)，然后最多对n-k个数据进行了heapreplace。所以总的时间复杂度是O(k+(n-k)logk)
不过代码实现上依赖异常来保障正常流程，笔者觉得这样实现不太好，可能有性能问题，也滥用了异常。异常用来实现正常流程，容易让人误解coder的意图，可读性不高，容易造成困扰
nlargest/nsmallest
def nsmallest(n, iterable, key=None)
def nsmallest(n, iterable, key=None):
"""Find the n smallest elements in a dataset.
Equivalent to: sorted(iterable, key=key)[:n]
"""
# some code
def nlargest(n, iterable, key=None):
"""Find the n largest elements in a dataset.
Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
"""
nlargest(n, iterable, key=None)结果上等价于sorted(iterable, key=key, reverse=True)[:n]
nsmallest(n, iterable, key=None)结果上等价于sorted(iterable, key=key)[:n]
功能上可以看出来就是从一段数据上获取最大或者最小的n个数，就是实现堆-解决海量数据topk问题。
实现上就是先建个k（参数中的n）大小的堆，再对剩下的数据用heappushpop来筛选。
比直接排序再获取效率上高了许多。排序的时间复杂度是O(nlogn)，而用堆来实现上是，建堆O(k)+每次的堆调整O((n-k)logk)，就是O(k+(n-k)logk)。用堆来实现topk问题还有一个优势就是面对大量数据可以分批拉到内存中，不用担心内存不够用。
源代码
# -*- 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
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 ]
from itertools import islice, count, imap, izip, tee, chain
from operator import itemgetter
def cmp_lt(x, y):
# Use __lt__ if available; otherwise, try __le__.
# In Py3.x, only __lt__ will be called.
return (x < y) if hasattr(x, __lt__ ) else (not y <= x)
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 cmp_lt(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(xrange(n//2)):
_siftup(x, i)
def _heappushpop_max(heap, item):
"""Maxheap version of a heappush followed by a heappop."""
if heap and cmp_lt(item, heap[0]):
item, heap[0] = heap[0], item
_siftup_max(heap, 0)
return item
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)
def nlargest(n, iterable):
"""Find the n largest elements in a dataset.
Equivalent to: sorted(iterable, reverse=True)[:n]
"""
if n < 0:
return []
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 n < 0:
return []
it = iter(iterable)
result = list(islice(it, n))
if not result:
return result
_heapify_max(result)
_heappushpop = _heappushpop_max
for elem in it:
_heappushpop(result, elem)
result.sort()
return result
# 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 cmp_lt(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 cmp_lt(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 cmp_lt(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 cmp_lt(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)
# If available, use C implementation
try:
from _heapq import *
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
_len = len
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 _len(h) > 1:
try:
while 1:
v, itnum, next = s = h[0]
yield v
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
v, itnum, next = h[0]
yield v
for v in next.__self__:
yield v
# 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]
"""
# Short-cut for n==1 is to use min() when len(iterable)>0
if n == 1:
it = iter(iterable)
head = list(islice(it, 1))
if not head:
return []
if key is None:
return [min(chain(head, it))]
return [min(chain(head, it), key=key)]
# 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 = izip(iterable, count()) # decorate
result = _nsmallest(n, it)
return map(itemgetter(0), result) # undecorate
# General case, slowest method
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]
"""
# Short-cut for n==1 is to use max() when len(iterable)>0
if n == 1:
it = iter(iterable)
head = list(islice(it, 1))
if not head:
return []
if key is None:
return [max(chain(head, it))]
return [max(chain(head, it), key=key)]
# 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 = izip(iterable, count(0,-1)) # decorate
result = _nlargest(n, it)
return map(itemgetter(0), result) # undecorate
# General case, slowest method
in1, in2 = tee(iterable)
it = izip(imap(key, in1), count(0,-1), 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()
引用
8.4. heapq — 堆队列算法
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