Python cookbook（数据结构与算法）找到最大或最小的N个元素实现方法示例

>>> import heapq
>>> nums=[1,8,2,23,7,-4,18,23,42,37,2]
>>> print(heapq.nlargest(3,nums))
[42, 37, 23]
>>> print(heapq.nsmallest(3,nums))
[-4, 1, 2]
>>>

# example.py
#
# Example of using heapq to find the N smallest or largest items
import heapq
portfolio = [
 {'name': 'IBM', 'shares': 100, 'price': 91.1},
 {'name': 'AAPL', 'shares': 50, 'price': 543.22},
 {'name': 'FB', 'shares': 200, 'price': 21.09},
 {'name': 'HPQ', 'shares': 35, 'price': 31.75},
 {'name': 'YHOO', 'shares': 45, 'price': 16.35},
 {'name': 'ACME', 'shares': 75, 'price': 115.65}
]
cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])
print(cheap)
print(expensive)

Python 3.4.0 (v3.4.0:04f714765c13, Mar 16 2014, 19:24:06) [MSC v.1600 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> ================================ RESTART ================================
>>>
[{'name': 'YHOO', 'price': 16.35, 'shares': 45}, {'name': 'FB', 'price': 21.09, 'shares': 200}, {'name': 'HPQ', 'price': 31.75, 'shares': 35}]
[{'name': 'AAPL', 'price': 543.22, 'shares': 50}, {'name': 'ACME', 'price': 115.65, 'shares': 75}, {'name': 'IBM', 'price': 91.1, 'shares': 100}]
>>>

>>> import heapq
>>> nums=[1,8,2,23,7,-4,18,23,42,37,2]
>>> heap=list(nums)
>>> heap
[1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
>>> heapq.heapify(heap) #heapify()参数必须是list，此函数将list变成堆，实时操作。从而能够在任何情况下使用堆的函数。
>>> heap
[-4, 2, 1, 23, 7, 2, 18, 23, 42, 37, 8]
>>> heapq.heappop(heap)#如下是为了找到第3小的元素
-4
>>> heapq.heappop(heap)
1
>>> heapq.heappop(heap)
2
>>>

>>> h=[]   #定义一个list
>>> from heapq import * #引入heapq模块
>>> h
[]
>>> heappush(h,5)  #向堆中依次增加数值
>>> heappush(h,2)
>>> heappush(h,3)
>>> heappush(h,9)
>>> h    #h的值
[2, 5, 3, 9]
>>> heappop(h)   #从h中删除最小的，并返回该值
2
>>> h
[3, 5, 9]
>>> h.append(1)   #注意，如果不是压入堆中，而是通过append追加一个数值
>>> h    #堆的函数并不能操作这个增加的数值，或者说它堆对来讲是不存在的
[3, 5, 9, 1]
>>> heappop(h)   #从h中能够找到的最小值是3,而不是1
3
>>> heappush(h,2)  #这时，不仅将2压入到堆内，而且1也进入了堆。
>>> h
[1, 2, 9, 5]
>>> heappop(h)   #操作对象已经包含了1
1

>>> h
[1, 2, 9, 5]
>>> heappop(h)
1
>>> heappushpop(h,4)  #增加4同时删除最小值2并返回该最小值，与下列操作等同：
2    #heappush(h,4),heappop(h)
>>> h
[4, 5, 9]

>>> a=[3,6,1]
>>> heapify(a)   #将a变成堆之后，可以对其操作
>>> heappop(a)
1
>>> b=[4,2,5]   #b不是堆，如果对其进行操作，显示结果如下
>>> heappop(b)   #按照顺序，删除第一个数值并返回,不会从中挑选出最小的
4
>>> heapify(b)   #变成堆之后，再操作
>>> heappop(b)
2

>>> a=[]
>>> heapreplace(a,3)  #如果list空，则报错
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: index out of range
>>> heappush(a,3)
>>> a
[3]
>>> heapreplace(a,2)  #先执行删除（heappop(a)->3),再执行加入（heappush(a,2))
3
>>> a
[2]
>>> heappush(a,5)
>>> heappush(a,9)
>>> heappush(a,4)
>>> a
[2, 4, 9, 5]
>>> heapreplace(a,6)  #先从堆a中找出最小值并返回，然后加入6
2
>>> a
[4, 5, 9, 6]
>>> heapreplace(a,1)  #1是后来加入的，在1加入之前，a中的最小值是4
4
>>> a
[1, 5, 9, 6]

>>> a=[2,4,6]
>>> b=[1,3,5]
>>> c=merge(a,b)
>>> list(c)
[1, 2, 3, 4, 5, 6]