适合初学者的 Python 3 刽子手

Question

print("Welcome to hangman. Are you ready to have some fun?")

def play():
 import random
 List = ["random", "words", "list"]

 word = str(random.choice(List))
 mistake = 7
 alreadySaid = set()
 board = "_" * len(word)

 print(" ".join(board))
 while mistake > 0:
     while True:
         guess = input("Please guess a letter: ")
         if len(guess) <= 1:
             break
         else:
             print("Too long. Enter only one letter.")
     if guess in word:
         alreadySaid.add(guess)
         print("Correct!",guess, " was in the word!")
         board = "".join([guess if guess in word else "_" for str in word])
         if board == word:
             print("Congratulations! You´re correct!!!")
     elif guess not in word:
         mistake -= 1
         print("Wrong!", mistake," mistakes remaining.")
         if mistake <= 0:
            print("Game Over")
      print(" ".join(board))

play()

我正在尝试用 python 3 制作刽子手，但每当我输入一个正确的字母时，它就会作为仅包含该字母的单词出现。例如，对于随机输入 r，当我想要 r _ _ _ _ _ 时，输出为 r r r r r r。你觉得哪里不对？还有其他问题吗？

Answer 1

我是否可以建议您退后一步，尝试一种不同的、更简洁的方法？除此之外，我建议您将隐藏的单词保留为一个列表，因为字符串是不可变的并且不支持项目分配，这意味着如果用户的猜测是正确的，您将无法显示字符(然后您可以 join() 当你需要它时将它作为字符串显示给用户):

import random

word_list = [
    'spam',
    'eggs',
    'foo',
    'bar'
]

word = random.choice(word_list)

guess = ['_'] * len(word)
chances = 7

while '_' in guess and chances > 0:

    print(' '.join(guess))

    char = input('Enter char: ')

    if len(char) > 1:
        print('Please enter only one char.')
        continue

    if char not in word:
        chances -= 1

        if chances == 0:
            print('Game over!')
            break
        else:
            print('You have', chances, 'chances left.')
    else:
        for i, x in enumerate(word):
            if x == char:
                guess[i] = char
else:
    print(''.join(guess))
    print('Congratulations!')