Descending order using heapq

I think you are looking instead for a sorted linked list in this case, I modify someone I found here so it will insert with ascending order (I added the pop function, for some reason it wasn't in the code, but I think you may need it):

# Python program to insert in sorted list

# Node class 
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    def sortedInsert(self, new_node):

        # Special case for the empty linked list 
        if self.head is None:
            new_node.next = self.head
            self.head = new_node

        # Special case for head at end
        elif self.head.data <= new_node.data:
            new_node.next = self.head
            self.head = new_node

        else :

            # Locate the node before the point of insertion
            current = self.head
            while(current.next is not None and
                 current.next.data > new_node.data):
                current = current.next

            new_node.next = current.next
            current.next = new_node

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to prit the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print(temp.data),
            temp = temp.next

    def pop(self):
        val = self.head.data
        self.head = self.head.next
        return val


# Driver program
llist = LinkedList()
new_node = Node(5)
llist.sortedInsert(new_node)
new_node = Node(10)
llist.sortedInsert(new_node)
new_node = Node(7)
llist.sortedInsert(new_node)
new_node = Node(3)
llist.sortedInsert(new_node)
new_node = Node(1)
llist.sortedInsert(new_node)
new_node = Node(9)
llist.sortedInsert(new_node)
print("Create Linked List")
llist.printList()

As you can see, it was just change the >= to <=, it does the job perfectly

As we discussed in the comments, your concerns about copying data when using negated values to flip a min-heap into a max-heap don't matter when you're starting with an empty heap and adding the values as you go. Since that's the use case when finding the running median of a stream of values, negating the values as you add them should work just fine.

Here's a running median generator I wrote just to double check that it works the way I expected:

def running_median(iterable):
    left_q = [] # heap of smaller-than-median elements, stored negated
    right_q = [] # heap of larger-than-median elements

    for value in iterable:
        if len(left_q) == len(right_q): # push to left_q when they're equal size
            if len(right_q) > 0 and value > right_q[0]:
                value = heapq.heapreplace(right_q, value)
            heapq.heappush(left_q, -value)
        else: # push to right_q only when it's (strictly) smaller
            if value < -left_q[0]:
                value = -heapq.heapreplace(left_q, -value)
            heapq.heappush(right_q, value)

        # len(left_q) is always >= len(right_q) so we never yield right_q[0]
        if len(left_q) > len(right_q):
            yield -left_q[0]
        else:
            yield (-left_q[0] + right_q[0]) / 2

The left_q heap stores the less-than-or-equal-to-median values. Each value is negated when it's pushed, so using the normal min-heap operations on it makes it work like a max-heap. We just need to remember to re-negate any value we take out of it, to get back to the original sign.