It is difficult to run the code because there is no way out. However, I tried several things:
In the heapq module heap[0] always indicates the smallest element. In your case, 1 is the smallest item. Therefore, changing this value from 1 to 5 should theoretically be simple. I tried heapq.heappop(heap) , which should return the smallest value. Thus, as you said in your question: โI want to update the val of the element, I donโt know, in turn, thisโ, this method automatically gets the smallest value (I assume from your question that you want to replace 1 since this is the smallest value , since it is associated with the name First ). However, when I try to run my code, I got <__main__.Element object at 0x103c15dd0> , so you should try to fix your code so that you can print the output, the same thing happens for print heap[0] , the same type of error . Then, as soon as you no longer get this error, at the end of your code block try:
s = heapq.heappop(heap) print heapq.heapreplace(5, s)
Using this approach, I get the following error: TypeError: heap argument must be a list Therefore, if you can understand how to convert s to a list, then this should work. Maybe someone can edit my answer to add this code.
Hope this helps.
SELF-EDIT:
Add this to the end of your code block, [] include it in the list that you want to receive as heapq input.
s = heapq.heappop(heap) print heapq.heapreplace([5], [s])
This returns a value of 5 on the output.
Returning to the output problem, if you indicate that you want the result to look, I can try to help you.