详解Python中heapq模块的用法

heapq 模块提供了堆算法。heapq是一种子节点和父节点排序的树形数据结构。这个模块提供heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2]。为了比较不存在的元素被人为是无限大的。heap最小的元素总是[0]。

打印 heapq 类型

import math

import random

from cStringIO import StringIO

def show_tree(tree, total_width=36, fill=' '):

output = StringIO()

last_row = -1

for i, n in enumerate(tree):

if i:

row = int(math.floor(math.log(i+1, 2)))

else:

row = 0

if row != last_row:

output.write('\n')

columns = 2**row

col_width = int(math.floor((total_width * 1.0) / columns))

output.write(str(n).center(col_width, fill))

last_row = row

print output.getvalue()

print '-' * total_width

print

return

data = random.sample(range(1,8), 7)

print 'data: ', data

show_tree(data)

打印结果

data: [3, 2, 6, 5, 4, 7, 1]

3

2 6

5 4 7 1

-------------------------

heapq.heappush(heap, item)

push一个元素到heap里, 修改上面的代码

heap = []

data = random.sample(range(1,8), 7)

print 'data: ', data

for i in data:

print 'add %3d:' % i

heapq.heappush(heap, i)

show_tree(heap)

打印结果

data: [6, 1, 5, 4, 3, 7, 2]

add 6:

6

------------------------------------

add 1:

1

6

------------------------------------

add 5:

1

6 5

------------------------------------

add 4:

1

4 5

6

------------------------------------

add 3:

1

3 5

6 4

------------------------------------

add 7:

1

3 5

6 4 7

------------------------------------

add 2:

1

3 2

6 4 7 5

------------------------------------

根据结果可以了解,子节点的元素大于父节点元素。而兄弟节点则不会排序。

heapq.heapify(list)

将list类型转化为heap, 在线性时间内, 重新排列列表。

print 'data: ', data

heapq.heapify(data)

print 'data: ', data

show_tree(data)

打印结果

data: [2, 7, 4, 3, 6, 5, 1]

data: [1, 3, 2, 7, 6, 5, 4]

1

3 2

7 6 5 4

------------------------------------

heapq.heappop(heap)

删除并返回堆中最小的元素, 通过heapify() 和heappop()来排序。

data = random.sample(range(1, 8), 7)

print 'data: ', data

heapq.heapify(data)

show_tree(data)

heap = []

while data:

i = heapq.heappop(data)

print 'pop %3d:' % i

show_tree(data)

heap.append(i)

print 'heap: ', heap

打印结果

data: [4, 1, 3, 7, 5, 6, 2]

1

4 2

7 5 6 3

------------------------------------

pop 1:

2

4 3

7 5 6

------------------------------------

pop 2:

3

4 6

7 5

------------------------------------

pop 3:

4

5 6

7

------------------------------------

pop 4:

5

7 6

------------------------------------

pop 5:

6

7

------------------------------------

pop 6:

7

------------------------------------

pop 7:

------------------------------------

heap: [1, 2, 3, 4, 5, 6, 7]

可以看到已排好序的heap。

heapq.heapreplace(iterable, n)

删除现有元素并将其替换为一个新值。

data = random.sample(range(1, 8), 7)

print 'data: ', data

heapq.heapify(data)

show_tree(data)

for n in [8, 9, 10]:

smallest = heapq.heapreplace(data, n)

print 'replace %2d with %2d:' % (smallest, n)

show_tree(data)

打印结果

data: [7, 5, 4, 2, 6, 3, 1]

1

2 3

5 6 7 4

------------------------------------

replace 1 with 8:

2

5 3

8 6 7 4

------------------------------------

replace 2 with 9:

3

5 4

8 6 7 9

------------------------------------

replace 3 with 10:

4

5 7

8 6 10 9

------------------------------------

heapq.nlargest(n, iterable) 和 heapq.nsmallest(n, iterable)

返回列表中的n个最大值和最小值

data = range(1,6)

l = heapq.nlargest(3, data)

print l # [5, 4, 3]

s = heapq.nsmallest(3, data)

print s # [1, 2, 3]

PS:一个计算题

构建元素个数为 K=5 的最小堆代码实例:

#!/usr/bin/env python

# -*- encoding: utf-8 -*-

# Author: kentzhan

#

import heapq

import random

heap = []

heapq.heapify(heap)

for i in range(15):

item = random.randint(10, 100)

print "comeing ", item,

if len(heap) >= 5:

top_item = heap[0] # smallest in heap

if top_item < item: # min heap

top_item = heapq.heappop(heap)

print "pop", top_item,

heapq.heappush(heap, item)

print "push", item,

else:

heapq.heappush(heap, item)

print "push", item,

pass

print heap

pass

print heap

print "sort"

heap.sort()

print heap

结果:

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