Python 3难度选择代码输入hard后重复提示输入问题求助

问题根源是while循环的逻辑条件写错了，受Python运算符优先级影响，你写的not diff == 'easy' or diff == 'hard'实际等价于(not (diff == 'easy')) or (diff == 'hard')：

• 当输入为easy时，not diff == 'easy'是False，diff == 'hard'也是False，整体条件为False，循环正常终止
• 当输入为hard时，not diff == 'easy'是True，or判断只要有一个为真整体就为真，所以循环会继续执行，不会退出，自然会重复弹出输入提示，不会执行后续的hard分支后续逻辑

推荐直接用更简洁的成员判断写法修改循环条件，可读性更高也不容易出错：

import random
# 假设words、wordseasy是你已经导入的自定义模块
diff = ''
# 仅当diff不在合法选项里时才继续循环
while diff not in ('easy', 'hard'):
    diff = input('What difficulty? [easy/hard] ')
    diff = diff.lower()

    if diff == "hard":
        word = random.choice(words.hangman_words)
        word = word.upper()
    elif diff == "easy":
        word = random.choice(wordseasy.hangman_words_easy)
        word = word.upper()
    else: 
        print('Invalid answer. Try again.')

如果要保留原有的判断写法，需要补充括号修正逻辑优先级，或者替换为and运算符：

# 写法1：加括号修正逻辑范围
while not (diff == 'easy' or diff == 'hard'):

# 写法2：用and判断两个都不满足的情况
while diff != 'easy' and diff != 'hard':

内容的提问来源于stack exchange，提问作者Kizuo