Python includes the heapq module for min-heaps, but I need a max heap. What should I use for a max-heap implementation in Python?
This question is related to
python
data-structures
heap
recursive-datastructures
If you are inserting keys that are comparable but not int-like, you could potentially override the comparison operators on them (i.e. <= become > and > becomes <=). Otherwise, you can override heapq._siftup in the heapq module (it's all just Python code, in the end).
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]
class MaxHeap:
def __init__(self, items=[]):
self.heap = [0]
self.items = items
for item in items:
self.heap.append(item)
self.__floatUp(len(self.heap)-1)
"""
self.heap = [0] heap index stars form 1
self.heap.append(item) append items from list to heap
append items to it's proper place
"""
def push(self, data):
self.heap.append(data)
self.__floatUp(len(self.heap)-1)
"""
Append data to the end of the list the using float up put it to the proper place
"""
def peek(self):
if self.heap[1]:
return self.heap[1]
else:
return False
"""
To return the first element form list
"""
def pop(self):
if (len(self.heap)) > 2:
self.__swap(1,len(self.heap)-1)
max = self.heap.pop()
self.__bubbleDown(1)
elif (len(self.heap)) == 2:
max = self.heap.pop()
else:
return False
return max
"""
If length is greater than two then swap first and last value as only the last value can be removed from the tree
Place the new first item after removing of the item to its proper place using bubble down
"""
def __swap(self, i, j):
self.heap[i], self.heap[j] = self.heap[j], self.heap[i]
"""
Swap is used in a bubble down and bubble up for proper placing of values is list
"""
def __floatUp(self, index):
parent = index // 2
if index <= 1:
return
elif self.heap[index] > self.heap[parent]:
self.__swap(index, parent)
self.__floatUp(parent)
"""
if value and parents , where parent < value , swap where parent, value = value,parent
After swap again check parent > value == True if false then again continue
"""
def __bubbleDown(self, index):
left = index * 2
right = index * 2 + 1
largest = index
if len(self.heap) > left and self.heap[largest] < self.heap[left]:
largest = left
if len(self.heap) > right and self.heap[largest] < self.heap[right]:
largest = right
if largest != index:
self.__swap(index, largest)
self.__bubbleDown(largest)
"""
self.heap[largest] < self.heap[left] : largest = left
current value and lower left value is bigger largest = left
self.heap[largest] > self.heap[right]
current value and lower value right value if lower right value is bigger than largest value
after getting the largest value we will swap larger value and smaller value and them again check that the current value is
bigger than is down value then again swap space and check again until every node is settled at its the correct place
"""
def __str__(self):
return str(self.heap)
"""
To str representation of the list
"""
m = MaxHeap([95, 3, 21])
m.push(10)
print(m)
print(m.pop())
print(m.peek())
Multiply the values by -1
There you go. All the highest numbers are now the lowest and vice versa.
Just remember that when you pop an element to multiply it with -1 in order to get the original value again.
The solution is to negate your values when you store them in the heap, or invert your object comparison like so:
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)
Example of a max-heap:
maxh = []
heapq.heappush(maxh, MaxHeapObj(x))
x = maxh[0].val # fetch max value
x = heapq.heappop(maxh).val # pop max value
But you have to remember to wrap and unwrap your values, which requires knowing if you are dealing with a min- or max-heap.
Adding classes for MinHeap and MaxHeap objects can simplify your code:
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
Example usage:
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"
You can use
import heapq
listForTree = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
heapq.heapify(listForTree) # for a min heap
heapq._heapify_max(listForTree) # for a maxheap!!
If you then want to pop elements, use:
heapq.heappop(minheap) # pop from minheap
heapq._heappop_max(maxheap) # pop from maxheap
This is a simple MaxHeap implementation based on heapq. Though it only works with numeric values.
import heapq
from typing import List
class MaxHeap:
def __init__(self):
self.data = []
def top(self):
return -self.data[0]
def push(self, val):
heapq.heappush(self.data, -val)
def pop(self):
return -heapq.heappop(self.data)
Usage:
max_heap = MaxHeap()
max_heap.push(3)
max_heap.push(5)
max_heap.push(1)
print(max_heap.top()) # 5
The easiest way is to invert the value of the keys and use heapq. For example, turn 1000.0 into -1000.0 and 5.0 into -5.0.
I have created a heap wrapper that inverts the values to create a max-heap, as well as a wrapper class for a min-heap to make the library more OOP-like. Here is the gist. There are three classes; Heap (abstract class), HeapMin, and HeapMax.
Methods:
isempty() -> bool; obvious
getroot() -> int; returns min/max
push() -> None; equivalent to heapq.heappush
pop() -> int; equivalent to heapq.heappop
view_min()/view_max() -> int; alias for getroot()
pushpop() -> int; equivalent to heapq.pushpop
I implemented a max heap version of heapq and submitted it to PyPI. (Very slight change of heapq module CPython code.)
https://pypi.python.org/pypi/heapq_max/
https://github.com/he-zhe/heapq_max
Installation
pip install heapq_max
Usage
tl;dr: same as heapq module except adding ‘_max’ to all functions.
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
In case if you would like to get the largest K element using max heap, you can do the following trick:
nums= [3,2,1,5,6,4]
k = 2 #k being the kth largest element you want to get
heapq.heapify(nums)
temp = heapq.nlargest(k, nums)
return temp[-1]
Following up to Isaac Turner's excellent answer, I'd like put an example based on K Closest Points to the Origin using max heap.
from math import sqrt
import heapq
class MaxHeapObj(object):
def __init__(self, val):
self.val = val.distance
self.coordinates = val.coordinates
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)
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 peek(self):
return heapq.nsmallest(1, self.h)[0].val
def __getitem__(self, i):
return self.h[i].val
class Point():
def __init__(self, x, y):
self.distance = round(sqrt(x**2 + y**2), 3)
self.coordinates = (x, y)
def find_k_closest(points, k):
res = [Point(x, y) for (x, y) in points]
maxh = MaxHeap()
for i in range(k):
maxh.heappush(res[i])
for p in res[k:]:
if p.distance < maxh.peek():
maxh.heappop()
maxh.heappush(p)
res = [str(x.coordinates) for x in maxh.h]
print(f"{k} closest points from origin : {', '.join(res)}")
points = [(10, 8), (-2, 4), (0, -2), (-1, 0), (3, 5), (-2, 3), (3, 2), (0, 1)]
find_k_closest(points, 3)
Extending the int class and overriding __lt__ is one of the ways.
import queue
class MyInt(int):
def __lt__(self, other):
return self > other
def main():
q = queue.PriorityQueue()
q.put(MyInt(10))
q.put(MyInt(5))
q.put(MyInt(1))
while not q.empty():
print (q.get())
if __name__ == "__main__":
main()
To elaborate on https://stackoverflow.com/a/59311063/1328979, here is a fully documented, annotated and tested Python 3 implementation for the general case.
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
Source: Stackoverflow.com