如何从 Python 列表中删除字典索引？

Question

我正在创建一个程序，将玩家的高分存储在街机中。 我为此使用的列表称为 PlayerID，因为它包含唯一 ID 和其他信息，例如他们在每场比赛中的高分。每当我尝试从玩家列表中成功删除字典时，它都无法正常工作，会删除多个配置文件。

这是我目前使用的代码。 Pickle 被用于数据存储。

with open("playerdata.dat",'rb') as f:
    PlayerID = pickle.load(f)    
while True:
    try:
        SearchID= int(input("Enter the ID of the profile you are removing")) # used to check if a wanted user actually exists in the program
    except ValueError:
        print("You have not provided an integer input, please try again.") #performs type check to ensure a valid input is provided
        continue
    else:
        break    

index= 0
position = -1
for Player in PlayerID:
    if Player['ID'] == SearchID:
        position = index
    else:
        index = index + 1
    try:
        PlayerID.pop(position)
    except IndexError:
        print("The ID provided does not exist.")
print("The user with ID", searchID,", has been deleted")    
with open('playerdata.dat','wb') as f:
    pickle.dump(playerID,f,pickle.HIGHEST_PROTOCOL)

此外，即使输入的整数 ID 实际上并不存在于 PlayerID 列表中，即使我有 IndexError 代码，它仍会删除多个配置文件。

Answer 1

这可能是实现此目的的更简单方法：

for index, Player in enumerate(PlayerID):
    if Player['ID'] == SearchID:
        PlayerID.pop(index)
        print("The user with ID", SearchID, ", has been deleted")
        break
else:
    print("The ID provided does not exist.")

Answer 2

问题是-1在Python中是一个有效的列表索引；它会弹出列表中的最后一个元素。

只要遇到正确的 id，就更容易弹出。此外，您可以使用enumerate 来计算索引：

for index, player in enumerate(players):
    if player['ID'] == search_id:
        players.pop(index)
        # we expect that the ID is truly unique, there is
        # only 1 occurrence of the ID.
        break

现在当然有人可能会问，为什么不使用 id->player 的字典来存储玩家 - 然后你可以这样做：

if search_id in players:
    players.pop(search_id)