Python：输入答案并随机从响应列表中获取回复

Question

我想实现的功能是，当你输入一个字母时，如果这个字母是正确的，就会自动给出一个反馈，告诉你这个字母是对的。不过，每次输入正确字母时，反馈的内容不能总是一样。

为此，我准备了一份反馈的列表，并使用了一个随机函数来从这个列表中选取反馈。

reaction=['good job','lucky guess!',you\'re on a roll]
react=random.choice(reaction)

我尝试把这个随机反馈放在

for letter in rand:
rand_list.append(letter)

之后，但这样做并不是我想要的效果，因为这样每次输入正确字母时，反馈都是一样的，只有在你猜下一个单词时才会改变。

完整的代码是：

import random

alphabeth = 'abcdefghijklmnopqrstuvwxyz'
rand_list = []
guessed_list = []

def prepWord():
    global rand, guessed_list, blank, rand_list,good,react_good
    react_good = ['Good job!', 'Lucky guess!', 'Took you a while to guess that letter!', 'You\'re on a roll!']

    words = ['note', 'pencil', 'paper','foo']
    rand = random.choice(words)
    guessed_list = []
    blank = ['_']*len(rand)
    rand_list = []
    for letter in rand:
        rand_list.append(letter)

    startPlay()

def startPlay():
    print('Welcome to Hangman. You have 8 tires to guess the secret word.')
    gameQ = input('Ready to play Hangman? y or n: ')
    if gameQ == 'y' or gameQ == 'Y':
        print('Guess the letters:')
        print(blank)
        checkAnswer()
    elif gameQ == 'n' or gameQ == 'N':
        print('goodbye')
        print('*********************')

    else:
        print('Invalid answer. Please try again')
        startPlay()


def playAgain():
    again = input('Would you like to play again? y or n --> ')
    if again == 'y':
        prepWord()
    elif again == 'n':
        print('Thanks for playing')
    else:
        print('Invalid answer. Please type y or n only')
        print(' ')
        playAgain()

def checkAnswer():
    tries = 0
    x = True
    while x:
        answer = input('').lower()
        if answer not in guessed_list:
            guessed_list.append(answer)
            if len(answer)>1:
                print('One letter at a time.')

            elif answer not in alphabeth:
                print('Invalid character, please try again.')
            else:
                if answer in rand:
                    print("The letter {} is in the word.".format(answer))

                    indices = [b for b, letter in enumerate(rand_list) if letter == answer]
                    for b in indices:
                        blank[b] = answer
                        print (blank)

                else:
                    print ("I'm sorry the letter {} is not in the word. Please try again.".format(answer))
                    tries +=1
                    if tries
                    if tries == 8:
                        print('Game over. You are out of tries')
                        playAgain()
        else:
            print('Letter {} already used. Try another.'.format(answer))

        if '_' not in blank:
            print('You guessed the secret word. You win!')
            final_word = 'The secret word is '
            for letter in blank:
                final_word += letter.upper()
            print(final_word)
            print('')
            x = False
            playAgain()



prepWord()

相关问题：

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Answer 1

每次你想要一个新的消息时，都得调用一次random。如果你只调用一次并把结果保存下来，那么你的react变量就不会改变了。下面的代码在最上面定义了反应列表，方便编辑，但在checkAnswer()函数里有一行代码，每当输入正确的字母时，就会调用random.choice来获取新的反应，代码是print(random.choice(reaction))。

import random

alphabeth = 'abcdefghijklmnopqrstuvwxyz'
rand_list = []
guessed_list = []
reaction=["good job","lucky guess!","you're on a roll"]

def prepWord():
    global rand, guessed_list, blank, rand_list,good,react_good
    react_good = ['Good job!', 'Lucky guess!', 'Took you a while to guess that letter!', 'You\'re on a roll!']

    words = ['note', 'pencil', 'paper','foo']
    rand = random.choice(words)
    guessed_list = []
    blank = ['_']*len(rand)
    rand_list = []
    for letter in rand:
        rand_list.append(letter)

    startPlay()

def startPlay():
    print('Welcome to Hangman. You have 8 tires to guess the secret word.')
    gameQ = input('Ready to play Hangman? y or n: ')
    if gameQ == 'y' or gameQ == 'Y':
        print('Guess the letters:')
        print(blank)
        checkAnswer()
    elif gameQ == 'n' or gameQ == 'N':
        print('goodbye')
        print('*********************')

    else:
        print('Invalid answer. Please try again')
        startPlay()


def playAgain():
    again = input('Would you like to play again? y or n --> ')
    if again == 'y':
        prepWord()
    elif again == 'n':
        print('Thanks for playing')
    else:
        print('Invalid answer. Please type y or n only')
        print(' ')
        playAgain()

def checkAnswer():
    tries = 0
    x = True
    while x:
        answer = input('').lower()
        if answer not in guessed_list:
            guessed_list.append(answer)
            if len(answer)>1:
                print('One letter at a time.')

            elif answer not in alphabeth:
                print('Invalid character, please try again.')
            else:
                if answer in rand:
                    print("The letter {} is in the word.".format(answer))
                    print(random.choice(reaction))
                    indices = [b for b, letter in enumerate(rand_list) if letter == answer]
                    for b in indices:
                        blank[b] = answer
                        print (blank)

                else:
                    print ("I'm sorry the letter {} is not in the word. Please try again.".format(answer))
                    tries +=1
                    if tries == 8:
                        print('Game over. You are out of tries')
                        playAgain()
        else:
            print('Letter {} already used. Try another.'.format(answer))

        if '_' not in blank:
            print('You guessed the secret word. You win!')
            final_word = 'The secret word is '
            for letter in blank:
                final_word += letter.upper()
            print(final_word)
            print('')
            x = False
            playAgain()



prepWord()

Answer 2

我对你的代码做了一些修改。把这些改动放在这里，希望你能通过对比看到不同之处，从而对你有帮助。以下是一些改动：

1. 没有使用递归：虽然递归很强大，但通常使用循环会更安全。比如在 startPlay() 函数中，每当用户输入一个无效字符时，你就会打开另一个嵌套的 startPlay() 函数。这可能会导致同时加载很多嵌套的函数。更糟糕的是，startPlay() 可能会调用 checkAnswer()，而 checkAnswer() 又可以调用 playAgain()，然后 playAgain() 可能会调用 prepWord()，最后又回到 startPlay()。用户玩得越久，你的程序占用的内存就越多，最终可能会崩溃。

2. 没有可变的全局变量。虽然在脚本顶部定义全局变量很有用，但在函数中调用 globals 并修改它们是有风险的，这会让你的代码更难调试和重用。最好是把需要的东西从一个函数传递到另一个函数。有些人可能会反对在顶部定义任何变量，但我觉得这很有用。特别是当其他人需要自定义细节，但不需要理解代码是如何工作的时。

3. 处理列表和字符串的一些细节是不同的。你不需要自己定义字母表，因为它已经存在于 string.ascii_lowercase 中。你可以直接用 if 'p' in 'python': 来检查字符串中的字符，所以你不需要把选定的单词转换成列表。你可以用 join 把列表转换回字符串，比如 " ".join(blank)，这会把 blank 从列表转换成字符串，并在每个元素之间加一个空格。如果在 join 前使用不同的字符串，会改变每个元素之间插入的字符。

希望这些对你有帮助。你可以随意使用这段代码：

import random
import string

# Global and on top for easy editing, not changed in any function
words = ['note', 'pencil', 'paper','foo']
react_good = ["Good job!", 
              "Lucky guess!", 
              "Took you a while to guess that letter!", 
              "You're on a roll!"]


def game():
    games = 0
    while start(games):  # Checks if they want to play.
        games += 1
        play_game()


def play_game():  # returns True if won and returns False if lost.
    rand_word = random.choice(words)  # Choose the word to use.
    blank = ['_']*len(rand_word)
    guessed = []
    tries = 0
    while True:  # Not infinite, ends when a return statement is called

        print("")
        print(" ".join(blank))  # Converting to string first looks better
        answer = input('Guess a letter: ').strip().lower()  # remove whitespace

        if answer == 'exit' or answer == 'quit':
            return False  # Exit Game
        elif len(answer) != 1:
            print('One letter at a time.')
        elif answer not in string.ascii_lowercase:  # same as alphabet string
            print('Invalid character, please try again.')
        elif answer in guessed:
            print('Letter {} already used. Try another.'.format(answer))
        elif answer in rand_word:  # Correct Guess
            # Update progress
            indices = [i for i, x in enumerate(rand_word) if x == answer]
            for i in indices:
                blank[i] = answer
            # Check if they have won
            if blank == list(rand_word):  # Or could convert blank to a string
                print('You guessed the secret word. You win!')
                print('The secret word is ' + rand_word.upper())
                print('')
                return True  # Exit Game
            guessed.append(answer)
            print("The letter {} is in the word.".format(answer))
            print(random.choice(react_good))
        else:  # Incorrect Guess
            tries += 1
            # Check if they have lost.
            if tries >= 8:
                print('Game over. You are out of tries')
                return False  # Exit Game
            print ("I'm sorry the letter {} is not in the word. Please try again.".format(answer))
            guessed.append(answer)


def start(games = 0):  # Gives different messages for first time
    if games == 0:  # If first time
        print('Welcome to Hangman. You have 8 tires to guess the secret word.')

    # Loops until they give a valid answer and a return statement is called
    while True:  
        # Get answer
        if games > 0:  # or just 'if games' works too
            game_q = input('Would you like to play again? y or n --> ')
        else:
            game_q = input('Ready to play Hangman? y or n: ')

        # Check answer
        if game_q.lower() == 'y':
            return True
        elif game_q.lower() == 'n':
            if games > 0:
                print('Thanks for playing')
            else:
                print('goodbye')
                print('*********************')
            return False
        else:
            print('Invalid answer. Please try again')


game()