Python 巨蟒中的石头布和剪刀

我对编码还不熟悉，我创建的石头剪刀布游戏并没有按预期的那样工作。 如果有人输入单词“石头”、“布”或“剪刀”，那么程序将按预期运行。 然而，当某人输入除石头、布或剪刀以外的单词时，程序应说“我不懂，请再试一次”，并提示用户输入另一个输入，它确实输入了，但随后程序结束，而不是继续按预期工作。代码如下：

# The game of Rock Paper Scissors

import random

choices = ['rock', 'paper', 'scissors']
computer_choice = random.choice(choices)
selections = 'The computer chose ' + computer_choice + ' so'
computer_score = 0
user_score = 0

def choose_option():
    user_choice = input('Please choose Rock, Paper or Scissors. (q to quit)\n>>> ')
    while user_choice != 'q':    
        if user_choice in ['ROCK', 'Rock', 'rock', 'R', 'r']:
            user_choice = 'rock'
        elif user_choice in ['PAPER', 'Paper', 'paper', 'P', 'p']:
            user_choice = 'paper'
        elif user_choice in ['SCISSORS','Scissors', 'scissors', 'S', 's']:
            user_choice = 'scissors'
        else:
            print("I don't understand, please try again.")
            choose_option()           
        return user_choice

user_choice = choose_option()

while user_choice != 'q':
    if user_choice == computer_choice:
        print(selections + ' it\'s a tie')
    elif (user_choice == 'rock' and computer_choice == 'scissors'):
        user_score += 1
        print(selections + ' you won! :)')
    elif (user_choice == 'paper' and computer_choice == 'rock'):
        user_score += 1
        print(selections + ' you won! :)')
    elif (user_choice == 'scissors' and computer_choice == 'paper'):
        user_score += 1
        print(selections + ' you won! :)')   
    elif (user_choice == 'rock' and computer_choice == 'paper'):
        computer_score += 1
        print(selections + ' you lost :(')
    elif (user_choice == 'paper' and computer_choice == 'scissors'):
        computer_score += 1
        print(selections + ' you lost :(')
    elif (user_choice == 'scissors' and computer_choice == 'rock'):
        computer_score += 1
        print(selections + ' you lost :(')
    else: 
        break
    print('You: ' + str(user_score) + "    VS    " + "Computer: " + str(computer_score))
    computer_choice = random.choice(choices)
    selections = 'The computer chose ' + computer_choice + ' so'
    user_choice = choose_option()

而不是

else:
            print("I don't understand, please try again.")
            choose_option()

你可以用

while user_choice != 'q':    
        user_choice = input('Please choose Rock, Paper or Scissors. (q to quit)\n>>> ')
        if user_choice in ['ROCK', 'Rock', 'rock', 'R', 'r']:
            user_choice = 'rock'
        elif user_choice in ['PAPER', 'Paper', 'paper', 'P', 'p']:
            user_choice = 'paper'
        elif user_choice in ['SCISSORS','Scissors', 'scissors', 'S', 's']:
            user_choice = 'scissors'
        else:
            print("I don't understand, please try again.")
            continue          
        return user_choice

---

继续

将导致程序跳过循环的其余部分（此处跳过

返回

），并在循环时继续下一次迭代

（返回到
用户选择=…
）。还要注意的是，我认为还有一个bug，如果
用户选择
是
“q”
，结果实际上不会返回。
您的问题似乎是由于调用函数内部的函数引起的。这为我解决了问题：

def choose_option():
   user_choice = input('Please choose Rock, Paper or Scissors. (q to quit)\n>>> ')
   while user_choice != 'q':    
     if user_choice in ['ROCK', 'Rock', 'rock', 'R', 'r']:
        user_choice = 'rock'
        return user_choice
     elif user_choice in ['PAPER', 'Paper', 'paper', 'P', 'p']:
        return user_choice
     elif user_choice in ['SCISSORS','Scissors', 'scissors', 'S', 's']:
        return user_choice
     else:
        while user_choice.lower() not in ["rock", "paper", "scissors", "s","r","p","q"]:
            print("I don't understand, please try again.")
            user_choice = input('Please choose Rock, Paper or Scissors. (q to quit)\n>>> ')

这样，如果计算机不喜欢输入，它可以请求另一个输入。
我还建议使用
如果user\u choice.lower（）在[]
中，只比键入所有选项简单一点

希望这有帮助
 您需要从递归调用返回，即
else
子句中的
return choose_option（）
。作为补充说明，您可以使用['rock'，'r']中的
if user_choice.lower（）：
甚至使用类似正则表达式的-
if re.search（r'（剪刀），user_choice，flags=re.i）