Python包含了用于min-堆的heapq模块,但我需要一个max堆。在Python中我应该使用什么来实现最大堆?
当前回答
我实现了一个最大堆版本的heapq,并将它提交给PyPI。(对heapq模块CPython代码的改动很小。)
https://pypi.python.org/pypi/heapq_max/
https://github.com/he-zhe/heapq_max
安装
pip install heapq_max
使用
dr:与heapq模块相同,只是所有函数都增加了' _max '。
heap_max = [] # creates an empty heap
heappush_max(heap_max, item) # pushes a new item on the heap
item = heappop_max(heap_max) # pops the largest item from the heap
item = heap_max[0] # largest item on the heap without popping it
heapify_max(x) # transforms list into a heap, in-place, in linear time
item = heapreplace_max(heap_max, item) # pops and returns largest item, and
# adds new item; the heap size is unchanged
其他回答
允许您选择任意数量的最大或最小的项目
import heapq
heap = [23, 7, -4, 18, 23, 42, 37, 2, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
heapq.heapify(heap)
print(heapq.nlargest(3, heap)) # [42, 42, 37]
print(heapq.nsmallest(3, heap)) # [-4, -4, 2]
为了详细说明https://stackoverflow.com/a/59311063/1328979,这里有一个针对一般情况的完整文档、注释和测试的Python 3实现。
from __future__ import annotations # To allow "MinHeap.push -> MinHeap:"
from typing import Generic, List, Optional, TypeVar
from heapq import heapify, heappop, heappush, heapreplace
T = TypeVar('T')
class MinHeap(Generic[T]):
'''
MinHeap provides a nicer API around heapq's functionality.
As it is a minimum heap, the first element of the heap is always the
smallest.
>>> h = MinHeap([3, 1, 4, 2])
>>> h[0]
1
>>> h.peek()
1
>>> h.push(5) # N.B.: the array isn't always fully sorted.
[1, 2, 4, 3, 5]
>>> h.pop()
1
>>> h.pop()
2
>>> h.pop()
3
>>> h.push(3).push(2)
[2, 3, 4, 5]
>>> h.replace(1)
2
>>> h
[1, 3, 4, 5]
'''
def __init__(self, array: Optional[List[T]] = None):
if array is None:
array = []
heapify(array)
self.h = array
def push(self, x: T) -> MinHeap:
heappush(self.h, x)
return self # To allow chaining operations.
def peek(self) -> T:
return self.h[0]
def pop(self) -> T:
return heappop(self.h)
def replace(self, x: T) -> T:
return heapreplace(self.h, x)
def __getitem__(self, i) -> T:
return self.h[i]
def __len__(self) -> int:
return len(self.h)
def __str__(self) -> str:
return str(self.h)
def __repr__(self) -> str:
return str(self.h)
class Reverse(Generic[T]):
'''
Wrap around the provided object, reversing the comparison operators.
>>> 1 < 2
True
>>> Reverse(1) < Reverse(2)
False
>>> Reverse(2) < Reverse(1)
True
>>> Reverse(1) <= Reverse(2)
False
>>> Reverse(2) <= Reverse(1)
True
>>> Reverse(2) <= Reverse(2)
True
>>> Reverse(1) == Reverse(1)
True
>>> Reverse(2) > Reverse(1)
False
>>> Reverse(1) > Reverse(2)
True
>>> Reverse(2) >= Reverse(1)
False
>>> Reverse(1) >= Reverse(2)
True
>>> Reverse(1)
1
'''
def __init__(self, x: T) -> None:
self.x = x
def __lt__(self, other: Reverse) -> bool:
return other.x.__lt__(self.x)
def __le__(self, other: Reverse) -> bool:
return other.x.__le__(self.x)
def __eq__(self, other) -> bool:
return self.x == other.x
def __ne__(self, other: Reverse) -> bool:
return other.x.__ne__(self.x)
def __ge__(self, other: Reverse) -> bool:
return other.x.__ge__(self.x)
def __gt__(self, other: Reverse) -> bool:
return other.x.__gt__(self.x)
def __str__(self):
return str(self.x)
def __repr__(self):
return str(self.x)
class MaxHeap(MinHeap):
'''
MaxHeap provides an implement of a maximum-heap, as heapq does not provide
it. As it is a maximum heap, the first element of the heap is always the
largest. It achieves this by wrapping around elements with Reverse,
which reverses the comparison operations used by heapq.
>>> h = MaxHeap([3, 1, 4, 2])
>>> h[0]
4
>>> h.peek()
4
>>> h.push(5) # N.B.: the array isn't always fully sorted.
[5, 4, 3, 1, 2]
>>> h.pop()
5
>>> h.pop()
4
>>> h.pop()
3
>>> h.pop()
2
>>> h.push(3).push(2).push(4)
[4, 3, 2, 1]
>>> h.replace(1)
4
>>> h
[3, 1, 2, 1]
'''
def __init__(self, array: Optional[List[T]] = None):
if array is not None:
array = [Reverse(x) for x in array] # Wrap with Reverse.
super().__init__(array)
def push(self, x: T) -> MaxHeap:
super().push(Reverse(x))
return self
def peek(self) -> T:
return super().peek().x
def pop(self) -> T:
return super().pop().x
def replace(self, x: T) -> T:
return super().replace(Reverse(x)).x
if __name__ == '__main__':
import doctest
doctest.testmod()
https://gist.github.com/marccarre/577a55850998da02af3d4b7b98152cf4
最简单的方法是反转键的值并使用heapq。例如,将1000.0转换为-1000.0,将5.0转换为-5.0。
我创建了一个名为heap_class的包,它实现了最大堆,还将各种堆函数包装到一个与列表兼容的环境中。
>>> from heap_class import Heap
>>> h = Heap([3, 1, 9, 20], max=True)
>>> h.pop()
20
>>> h.peek() # same as h[0]
9
>>> h.push(17) # or h.append(17)
>>> h[0] # same as h.peek()
17
>>> h[1] # inefficient, but works
9
从最大堆中获得最小堆。
>>> y = reversed(h)
>>> y.peek()
1
>>> y # repr is inefficient, but correct
Heap([1, 3, 9, 17], max=False)
>>> 9 in y
True
>>> y.raw() # underlying heap structure
[1, 3, 17, 9]
正如其他人所提到的,在max堆中处理字符串和复杂对象在heapq中是相当困难的,因为它们不同 否定的形式。heap_class实现简单:
>>> h = Heap(('aa', 4), ('aa', 5), ('zz', 2), ('zz', 1), max=True)
>>> h.pop()
('zz', 2)
支持自定义键,并与后续的推/追加和弹出一起工作:
>>> vals = [('Adam', 'Smith'), ('Zeta', 'Jones')]
>>> h = Heap(vals, key=lambda name: name[1])
>>> h.peek() # Jones comes before Smith
('Zeta', 'Jones')
>>> h.push(('Aaron', 'Allen'))
>>> h.peek()
('Aaron', 'Allen')
(实现是建立在heapq函数上的,所以它都是用C语言或C语言包装的,除了Python中max heap上的heappush和heapreplace)
解决方案是当你在堆中存储你的值时对其求反,或者像这样反转你的对象比较:
import heapq
class MaxHeapObj(object):
def __init__(self, val): self.val = val
def __lt__(self, other): return self.val > other.val
def __eq__(self, other): return self.val == other.val
def __str__(self): return str(self.val)
max-heap的例子:
maxh = []
heapq.heappush(maxh, MaxHeapObj(x))
x = maxh[0].val # fetch max value
x = heapq.heappop(maxh).val # pop max value
但是您必须记住包装和打开您的值,这需要知道您正在处理的是最小堆还是最大堆。
MinHeap, MaxHeap类
为MinHeap和MaxHeap对象添加类可以简化代码:
class MinHeap(object):
def __init__(self): self.h = []
def heappush(self, x): heapq.heappush(self.h, x)
def heappop(self): return heapq.heappop(self.h)
def __getitem__(self, i): return self.h[i]
def __len__(self): return len(self.h)
class MaxHeap(MinHeap):
def heappush(self, x): heapq.heappush(self.h, MaxHeapObj(x))
def heappop(self): return heapq.heappop(self.h).val
def __getitem__(self, i): return self.h[i].val
使用示例:
minh = MinHeap()
maxh = MaxHeap()
# add some values
minh.heappush(12)
maxh.heappush(12)
minh.heappush(4)
maxh.heappush(4)
# fetch "top" values
print(minh[0], maxh[0]) # "4 12"
# fetch and remove "top" values
print(minh.heappop(), maxh.heappop()) # "4 12"