heapq with custom compare predicate
Solution 1:
According to the heapq documentation, the way to customize the heap order is to have each element on the heap to be a tuple, with the first tuple element being one that accepts normal Python comparisons.
The functions in the heapq module are a bit cumbersome (since they are not object-oriented), and always require our heap object (a heapified list) to be explicitly passed as the first parameter. We can kill two birds with one stone by creating a very simple wrapper class that will allow us to specify a key
function, and present the heap as an object.
The class below keeps an internal list, where each element is a tuple, the first member of which is a key, calculated at element insertion time using the key
parameter, passed at Heap instantiation:
# -*- coding: utf-8 -*-
import heapq
class MyHeap(object):
def __init__(self, initial=None, key=lambda x:x):
self.key = key
self.index = 0
if initial:
self._data = [(key(item), i, item) for i, item in enumerate(initial)]
self.index = len(self._data)
heapq.heapify(self._data)
else:
self._data = []
def push(self, item):
heapq.heappush(self._data, (self.key(item), self.index, item))
self.index += 1
def pop(self):
return heapq.heappop(self._data)[2]
(The extra self.index
part is to avoid clashes when the evaluated key value is a draw and the stored value is not directly comparable - otherwise heapq could fail with TypeError)
Solution 2:
Define a class, in which override the __lt__()
function. See example below (works in Python 3.7):
import heapq
class Node(object):
def __init__(self, val: int):
self.val = val
def __repr__(self):
return f'Node value: {self.val}'
def __lt__(self, other):
return self.val < other.val
heap = [Node(2), Node(0), Node(1), Node(4), Node(2)]
heapq.heapify(heap)
print(heap) # output: [Node value: 0, Node value: 2, Node value: 1, Node value: 4, Node value: 2]
heapq.heappop(heap)
print(heap) # output: [Node value: 1, Node value: 2, Node value: 2, Node value: 4]