What do I use for a max-heap implementation in Python?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

MinHeap, MaxHeap classes

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"

Answer 4

The easiest and ideal solution

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.

Answer 5

The easiest way is to convert every element into negative and it will solve your problem.

import heapq
heap = []
heapq.heappush(heap, 1*(-1))
heapq.heappush(heap, 10*(-1))
heapq.heappush(heap, 20*(-1))
print(heap)

The output will look like:

[-20, -1, -10]

Answer 6

I implemented a max-heap version of heapq and submitted it to PyPI. (Very slight change of the heapq module's CPython code.)

heapq_max

Heapq_max (GitHub)

Installation

pip install heapq_max

Usage

tl;dr: The same as the 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]                      # The largest item on the heap without popping it
heapify_max(x)                          # Transforms the list into a heap, in-place, in linear time
item = heapreplace_max(heap_max, item)  # Pops and returns the largest item, and
                                        # adds a new item; the heap size is unchanged

Answer 7

I also needed to use a max-heap, and I was dealing with integers, so I just wrapped the two methods that I needed from heap as follows:

import heapq


def heappush(heap, item):
    return heapq.heappush(heap, -item)


def heappop(heap):
    return -heapq.heappop(heap)

And then I just replaced my heapq.heappush() and heapq.heappop() calls with heappush() and heappop() respectively.

Answer 8

This is a simple max-heap 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

Answer 9

The simplest way:

from heapq import *

h = [5, 7, 9, 1, 3]
h_neg = [-i for i in h]
heapify(h_neg)            # heapify
heappush(h_neg, -2)       # push
print(-heappop(h_neg))    # pop
# 9

Answer 10

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).

Answer 11

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()

Answer 12

Allowing you to chose an arbitrary amount of largest or smallest items

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]

Answer 13

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

Answer 14

To elaborate on Apoorv Patne's answer, 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

Answer 15

The heapq module has everything you need to implement a max-heap. It does only the heappush functionality of a max-heap. I've demonstrated below how to overcome that.

Add this function in the heapq module:

def _heappush_max(heap, item):
    """Push item onto heap, maintaining the heap invariant."""
    heap.append(item)
    _siftdown_max(heap, 0, len(heap)-1)

And at the end, add this:

try:
    from _heapq import _heappush_max
except ImportError:
    pass

Voila! It's done.

PS - to go to heapq function. First write "import heapq" in your editor and then right click 'heapq' and select go to definition.

Answer 16

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]

Answer 17

There's a built-in heap in Python, but here is how build it by yourself. The algorithm is working, but about the efficiency I don't know.

class Heap:

    def __init__(self):
        self.heap = []
        self.size = 0

    def add(self, heap):
        self.heap = heap
        self.size = len(self.heap)

    def heappush(self, value):
        self.heap.append(value)
        self.size += 1

    def heapify(self, heap, index=0):

        mid = int(self.size /2)
        """
            If you want to travel great value from the bottom to the top, you need to repeat swapping by the height of the tree.
            I don't know how I can get the height of the tree. That's why I use sezi/2.
            You can find the height by this formula:
            2^(x) = size+1  Why 2^x? Because the tree is growing exponentially
            xln(2) = ln(size+1)
            x = ln(size+1)/ln(2)
        """

        for i in range(mid):
            self.createTee(heap, index)

        return heap

    def createTee(self, heap, shiftindex):

        """
        """
        """

            This 'pos' variable refer to the index of the parent, only parent with children
                    (1)
                (2)      (3)           Here the size of the list is 7/2 = 3
            (4)   (5)  (6)  (7)        The number of parents is 3, but we use {2, 1, 0} in a 'while' loop.
                                       That is why a set 'pos' to -1.

        """
        pos = int(self.size /2) -1
        """
            This if you want to sort this heap list. We should swap the maximum value in the root of the tree with the last
            value in the list and if you want to repeat this until sort all list, you will need to prevent the function from
            change what we already sorted. I should decrease the size of the list. That will heapify on it.

        """

        newsize = self.size - shiftindex
        while pos >= 0:
            left_child = pos * 2 + 1
            right_child = pos * 2 + 2
            # This means that left child is exist
            if left_child < newsize:
                if right_child < newsize:

                    # If the right child exits, we want to check if the left
                    # child > rightchild.
                    #
                    # If the right child doesn't exist, we can check that
                    # we will get error out of range.
                    if heap[pos] < heap[left_child] and heap[left_child]  > heap[right_child]:
                        heap[left_child], heap[pos] = heap[pos], heap[left_child]
                # Here if the right child doesn't exist
                else:
                    if heap[pos] < heap[left_child]:
                        heap[left_child], heap[pos] = heap[pos], heap[left_child]
            # If the right child exists
            if right_child < newsize:
                if heap[pos] < heap[right_child]:
                    heap[right_child], heap[pos] = heap[pos], heap[right_child]
            pos -= 1

        return heap

    def sort(self):
        k = 1
        for i in range(self.size -1, 0, -1):
            """
            Because this is max-heap, we swap root with last element in the list

            """
            self.heap [0], self.heap[i] = self.heap[i], self.heap[0]
            self.heapify(self.heap, k)
            k += 1

        return self.heap


h = Heap()
h.add([5, 7, 0, 8, 9, 10, 20, 30, 50, -1])
h.heappush(-2)
print(" before heapify ")
print(h.heap)
print(" after heapify ")
print(h.heapify(h.heap, 0))
print(" after sort ")
print(h.sort())

Output

Before heapify

[5, 7, 0, 8, 9, 10, 20, 30, 50, -1, -2]

After heapify

[50, 30, 20, 8, 9, 10, 0, 7, 5, -1, -2]

After sort

[-2, -1, 0, 5, 7, 8, 9, 10, 20, 30, 50]

Answer 18

arr = [3, 4, 5, 1, 2, 3, 0, 7, 8, 90, 67, 31, 2, 5, 567]
# max-heap sort will lead the array to ascending order
def maxheap(arr, p):

    for i in range(len(arr)-p):
        if i > 0:
            child = i
            parent = (i + 1)//2 - 1

            while arr[child]> arr[parent] and child != 0:
                arr[child], arr[parent] = arr[parent], arr[child]
                child = parent
                parent = (parent + 1)//2 -1


def heapsort(arr):
    for i in range(len(arr)):
        maxheap(arr, i)
        arr[0], arr[len(arr)-i-1] = arr[len(arr)-i-1], arr[0]

    return arr


print(heapsort(arr))

Try this.

Answer 19

I've created a package called heap_class that implements max-heaps, and also wraps the various heap functions into a list-compatible environment.

>>> from heap_class import Heap
>>> h = Heap([3, 1, 9, 20], max=True)
>>> h.pop()
20

>>> h.peek()  # The same as h[0]
9

>>> h.push(17)  # Or h.append(17)
>>> h[0]  # The same as h.peek()
17

>>> h[1]  # Inefficient, but it works
9

Get a min-heap from a max-heap.

>>> y = reversed(h)
>>> y.peek()
1

>>> y  # The representation is inefficient, but correct
Heap([1, 3, 9, 17], max=False)

>>> 9 in y
True

>>> y.raw()  # Underlying heap structure
[1, 3, 17, 9]

As others have mentioned, working with strings and complex objects in a max-heap is rather hard in heapq because of the different forms of negation. It is easy with the heap_class implementation:

>>> h = Heap(('aa', 4), ('aa', 5), ('zz', 2), ('zz', 1), max=True)
>>> h.pop()
('zz', 2)

Custom keys are supported and work with subsequent pushes/appends and pops:

>>> 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')

(The implementation is built on heapq functions, so it is all in C or with C-wrappers, except heappush and heapreplace on max-heap which is in Python.)