Question

这是我的代码

players = []
while len(players) >= 0:
    name = input('Enter a name: ')
    players.append(name)
    if name == '':
        players.pop()
        break
    else:
        pass             

player_dict = {name: [] for name in players}
print(player_dict)

for name in player_dict:
    answer = input(name + ', who will win the fight? ')
    player_dict[name].append(answer)

fight_winner = input('Who won the fight? ')

for name in player_dict:
    if answer == fight_winner:
        print(name + ' = Correct')
    else:
        print(name + ' = Incorrect')
print(player_dict)

这是我在运行代码时看到的内容

Enter a name: bill
Enter a name: bob
Enter a name: 
{'bill': [], 'bob': []}
bill, who will win the fight? red
bob, who will win the fight? blue
Who won the fight? red
bill = Incorrect
bob = Incorrect
{'bill': ['red'], 'bob': ['blue']}

我希望看到bill = Correct。如何访问每个indivuals值并通过if语句运行它？感谢您的帮助

Answer 1

更改

getattr(self.MyQThread, 'signal' + str(n)).emit(value)

要：

for name in player_dict:
    if answer == fight_winner:
        print(name + ' = Correct')
    else:
        print(name + ' = Incorrect')
print(player_dict)

您的for name, colors in player_dict.items(): if fight_winner in colors: print(name + ' = Correct') else: print(name + ' = Incorrect') print(player_dict)是名称以外的颜色。您将颜色存储到列表中，如下所示：

player_dict [名称] .append（回答）

键是name，值是颜色列表。因此，当您迭代字典fight_winner时，您应该找到基于颜色player_dict的{{1}}

你不能在条件语句name中使用变量答案，原因是答案的值总是由最后一次迭代分配，并且它可能没有给你正确的比较。

例如，如果订单为蓝色然后是红色，则答案为红色，并且您将fight_winner设置为蓝色，两个玩家都为if answer == fight_winner:。

以下是我的测试：

fight_winner

Answer 2

在目前的表单中，您可以通过更改以下内容来使代码正常工作：

if answer == fight_winner:

到：

if player_dict[name][0] == fight_winner:

或（甚至更好）：

answer = player_dict[name][0]
if answer == fight_winner:

但是，我不认为每个玩家的词典条目都需要列表，因为您只为每个玩家存储单个值。请考虑使用以下内容：

player_dict = {}

for name in players:
    answer = input(name + ', who will win the fight? ')
    player_dict[name] = answer

fight_winner = input('Who won the fight? ')

for name, answer in player_dict:
    if answer == fight_winner:
        print(name + ' = Correct')
    else:
        print(name + ' = Incorrect')

这种方法似乎更具可读性（至少在我看来），并且不使用player_dict中的列表，因为它们不是必需的（在当前的代码形式中）。

如何在此词典中使用if语句

2 个答案: