Question

我正在构建一个Hangman游戏作为我的第一个python项目。当我构建它时，游戏目前正在终端中运行。我没有使用PyGame。这里有更多相关的代码。

five_letters = ['birds', 'hands', 'trees']
difficulty = 'easy'

def get_word(foo):
"""Based on difficulty generate a game_word, then turn it into a list to be returned"""
    if foo == 'easy':
        foo = random.choice(five_letters)
        foo = list(foo)

    if foo == 'medium':
        foo = random.choice(ten_letters)
        foo = list(foo)

    if foo == 'hard':
        foo = random.choice(fifteen_letters)
        foo = list(foo)

    return foo

game_word = get_word(difficulty)

我想要列出并执行以下操作：

检查玩家的猜测是否在列表中（我知道我可以迭代一个字符串）。
如果猜测在字符串中，则将其放在输出中的适当空格中。

我将字符串作为列表返回，以便我可以在列表中找到正确猜测的值，将其放在输出中的正确位置。

例如

game_word = 'birds'
player_guess = 's'

我想输出

_ _ _ _ s

但也许我说这一切都错了。我觉得我可以缩短选择随机字符串的函数部分，然后将其转换为列表。

Answer 1

您可以使用：

from random import choice
foo = [choice(a_list)]

Answer 2

我也开始使用Python 3.x开始我的旅程了，这是我刚刚制作的快速代码（如果你遇到问题，可以用它作为参考）：

from random import choice as chc

def hangman(diff):
    #Fill in the words for every difficulty type:
    easy = ['this', 'that']
    medium = ['bicycle', 'bathroom']
    hard = ['superlongword', 'othersuperlongw']
    if diff == 'easy':
        word = chc(easy)
    elif diff == 'medium':
        word = chc(medium)
    elif diff == 'hard':
        word = chc(hard)
    else:
        raise ValueError('Bad difficulty choosen. Terminating.')
    shadow = ['_' for item in word] #list which is showing the progress
    shad_str = ''
    for item in shadow:
        shad_str += item
    print('Choosen word is {} characters long and looks like this:\n{}'.format(len(shad_str), shad_str))
    while '_' in shadow:
        letter = input('Type in next char: ')
        if len(letter) == 1:    #anti-cheat - always have to give 1 char
            if letter in word:
                for i in range(len(word)):   #makes it work correctly even when the letter shows up more than one time
                    if(letter == word[i]):
                        shadow[i] = letter
                temp = ''
                for item in shadow:
                    temp += item
                print(temp)
            else:
                print("This letter isn't in the choosen word! Try again.")
        else:
                print('No cheating! Only one character allowed!')
    else:
        shad_str = ''
        for item in shadow:
            shad_str += item
        print('The game has ended. The word was "{}".'.format(shad_str))

我确定检查的全部内容都可以在一个功能中完成（这就是为什么只是为了简单的模式），所以你可以调用＆＃39;打＆＃39;功能3次，具体取决于您选择的难度模式编辑：那是不需要的，我只是注意到你可以用3 ifs来决定差异。如果用户选择正确的难度，现在游戏每次都有效。

如何随机化并将字符串转换为python中的列表

2 个答案: