_heapq_ heapsort
This module provides an implementation of heap sort algorithm. Binary heap is the maximum stack parent child node is greater than or equal to two, the minimum heap parent child nodes is less than or equal to two.
Creating the heap
There heapq stack created in two ways, one is to use an empty list, then heapq.heappush () function is added to the value of the stack, another is used heap.heapify (list) becomes stack structure conversion list
import heapq
# 第一种
"""
函数定义：
heapq.heappush(heap, item)
- Push the value item onto the heap, maintaining the heap invariant.
heapq.heappop(heap)
- Pop and return the smallest item from the heap, maintaining the heap invariant.
If the heap is empty, IndexError is raised. To access the smallest item without popping it, use heap[0].
"""
nums = [2, 3, 5, 1, 54, 23, 132]
heap = []
for num in nums:
heapq.heappush(heap, num) # 加入堆
print(heap[0]) # 如果只是想获取最小值而不是弹出，使用heap[0]
print([heapq.heappop(heap) for _ in range(len(nums))]) # 堆排序结果
# out: [1, 2, 3, 5, 23, 54, 132]
# 第二种
nums = [2, 3, 5, 1, 54, 23, 132]
heapq.heapify(nums)
print([heapq.heappop(heap) for _ in range(len(nums))]) # 堆排序结果
# out: [1, 2, 3, 5, 23, 54, 132]heapq module and a heapq.merge(*iterables)method for combining a plurality of the sorted sequence into a sequence of ordered, iterator returns a sorted value.
It is similar sorted(itertools.chain(*iterables)), but returns may be iterative.
"""
函数定义：
heapq.merge(*iterables)
- Merge multiple sorted inputs into a single sorted output (for example, merge timestamped entries from multiple log files). Returns an iterator over the sorted values.
- Similar to sorted(itertools.chain(*iterables)) but returns an iterable, does not pull the data into memory all at once, and assumes that each of the input streams is already sorted (smallest to largest).
"""
import heapq
num1 = [32, 3, 5, 34, 54, 23, 132]
num2 = [23, 2, 12, 656, 324, 23, 54]
num1 = sorted(num1)
num2 = sorted(num2)
res = heapq.merge(num1, num2)
print(list(res))Access heap contents
After the heap is created, you can () function pops up a minimum heap by `heapq.heappop.
import heapq
nums = [2, 43, 45, 23, 12]
heapq.heapify(nums)
print(heapq.heappop(nums))
# out: 2
# 如果需要所有堆排序后的元素
result = [heapq.heappop(nums) for _ in range(len(nums))]
print(result)
# out: [12, 23, 43, 45]If you need to remove the smallest element of the heap and add an element, you can use heapq.heaprepalce()the function
import heapq
nums = [1, 2, 4, 5, 3]
heapq.heapify(nums)
heapq.heapreplace(nums, 23)
print([heapq.heappop(nums) for _ in range(len(nums))])
# out: [2, 3, 4, 5, 23]Get the maximum or minimum heap
If you need to get the maximum or minimum heap range of values, it can be used heapq.nlargest()or heapq.nsmallest()function
"""
函数定义：
heapq.nlargest(n, iterable[, key])¶
- Return a list with the n largest elements from the dataset defined by iterable.
- key if provided, specifies a function of one argument that is used to extract a comparison key from each element in the iterable: key=str.lower
- Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
"""
import heapq
nums = [1, 3, 4, 5, 2]
print(heapq.nlargest(3, nums))
print(heapq.nsmallest(3, nums))
"""
输出：
[5, 4, 3]
[1, 2, 3]
"""This function takes two parameters a key, or other data structure for dict types using
import heapq
from pprint import pprint
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'])
pprint(cheap)
pprint(expensive)
"""
输出：
[{'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}]
"""heapq application
Heap achieve heap sorting algorithm
>>> def heapsort(iterable):
... h = []
... for value in iterable:
... heappush(h, value)
... return [heappop(h) for i in range(len(h))]
...
>>> heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]The algorithm sorted(iterable)is similar, but it is unstable.
The value of the stack may be a tuple, the elements may be implemented with weights sorted.
>>> h = []
>>> heappush(h, (5, 'write code'))
>>> heappush(h, (7, 'release product'))
>>> heappush(h, (1, 'write spec'))
>>> heappush(h, (3, 'create tests'))
>>> heappop(h)
(1, 'write spec')