>>> 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()參數(shù)必須是list，此函數(shù)將list變成堆，實(shí)時(shí)操作。從而能夠在任何情況下使用堆的函數(shù)。
>>> heap
[-4, 2, 1, 23, 7, 2, 18, 23, 42, 37, 8]
>>> heapq.heappop(heap)#如下是為了找到第3小的元素
-4
>>> heapq.heappop(heap)
1
>>> heapq.heappop(heap)
2
>>>

>>> h=[]   #定義一個(gè)list
>>> from heapq import * #引入heapq模塊
>>> h
[]
>>> heappush(h,5)  #向堆中依次增加數(shù)值
>>> heappush(h,2)
>>> heappush(h,3)
>>> heappush(h,9)
>>> h    #h的值
[2, 5, 3, 9]
>>> heappop(h)   #從h中刪除最小的，并返回該值
2
>>> h
[3, 5, 9]
>>> h.append(1)   #注意，如果不是壓入堆中，而是通過(guò)append追加一個(gè)數(shù)值
>>> h    #堆的函數(shù)并不能操作這個(gè)增加的數(shù)值，或者說(shuō)它堆對(duì)來(lái)講是不存在的
[3, 5, 9, 1]
>>> heappop(h)   #從h中能夠找到的最小值是3,而不是1
3
>>> heappush(h,2)  #這時(shí)，不僅將2壓入到堆內(nèi)，而且1也進(jìn)入了堆。
>>> h
[1, 2, 9, 5]
>>> heappop(h)   #操作對(duì)象已經(jīng)包含了1
1

>>> h
[1, 2, 9, 5]
>>> heappop(h)
1
>>> heappushpop(h,4)  #增加4同時(shí)刪除最小值2并返回該最小值，與下列操作等同：
2    #heappush(h,4),heappop(h)
>>> h
[4, 5, 9]

>>> a=[3,6,1]
>>> heapify(a)   #將a變成堆之后，可以對(duì)其操作
>>> heappop(a)
1
>>> b=[4,2,5]   #b不是堆，如果對(duì)其進(jìn)行操作，顯示結(jié)果如下
>>> heappop(b)   #按照順序，刪除第一個(gè)數(shù)值并返回,不會(huì)從中挑選出最小的
4
>>> heapify(b)   #變成堆之后，再操作
>>> heappop(b)
2

>>> a=[]
>>> heapreplace(a,3)  #如果list空，則報(bào)錯(cuò)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: index out of range
>>> heappush(a,3)
>>> a
[3]
>>> heapreplace(a,2)  #先執(zhí)行刪除（heappop(a)->3),再執(zhí)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)  #先從堆a(bǔ)中找出最小值并返回，然后加入6
2
>>> a
[4, 5, 9, 6]
>>> heapreplace(a,1)  #1是后來(lái)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]