详解Python中heapq模块的用法
(编辑：jimmy 日期: 2024/11/29 浏览：3 次 )
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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