How do you remove an item from a dictionary by its index value?-CodePudding

I am trying to create a high score system for my game, but only want to display the top 5 high scores. I used a dictionary to store the scores and the names of the players. I want the program to remove the first score once there are more than 5 items. How do I remove items from a dictionary based on their order?

I tried to use .pop(index) like so:

highscores = {"player1":"54", "player2":"56", "player3":"63", "player4":"72", "player5":"81", "player6":"94"}
if len(highscores) > 5:
    highscores.pop(0)

However I get an error:

Traceback (most recent call last):
  File "c:\Users\-----\Documents\Python projects\Python NEA coursework\test.py", line 3, in <module>
    highscores.pop(0)
KeyError: 0

Anyone know why this happens?

CodePudding user response：

What you can do is turn your dict into a list of tuples (the items), truncate that, then turn back into a dict. For example, to always keep only the last 5 values inserted:

highscores = dict(list(highscores.items())[-5:])

(Note that it is idempotent if there were fewer than 5 items to start with).

CodePudding user response：

dict is not in ordered way. So first create ordered dict with the order you want.

You can try:

>>> import collections
>>> highscores = {"player1":"54", "player2":"56", "player3":"63", "player4":"72", "player5":"81", "player6":"94"}
>>> highscores = collections.OrderedDict(highscores)
>>> highscores.pop(list(new_dict.keys())[0])
'54'
>>> highscores
OrderedDict([('player2', '56'), ('player3', '63'), ('player4', '72'), ('player5', '81'), ('player6', '94')])