Question

我今天一直在写一个程序（一个刽子手游戏），并遇到了一些障碍，我知道你们不喜欢用这个作为家庭作业答案的人，但我有搜索了很长一段时间，虽然我在开发早期发布了这个程序，但我已经开辟了新的领域。我正在处理列表，元组，词典等等，这个程序一直很棒。对不起文字墙！

问题：

使用列表时有没有办法引用字符串？

上下文 - 定义一个函数（最后一行是测试是否有效）

def click_1 (text):
    key_1 = word.index [letter]
    hidden_word[key_1] = letter
    print (hidden_word)

这引发了一个很长的错误，我不会发布，这里是重要的＆＃39;在我眼中：

File "/Users/foo/Desktop/Hangman.py", line 19, in click_1
    key_1 = word.index [letter]
TypeError: 'builtin_function_or_method' object is not subscriptable

如果有人可以帮助我，那就太好了。我尝试过使用＆＃39;＆＃39;声明，但我无法使其工作 - 目前，虽然我怀疑它可能是解决方案的一部分。我试过了：

key_1 = word.index [letter in word]

和

key_1 = word.index([letter] in word)

这两个都不起作用。

提前为一位年轻的程序员干杯！

Answer 1

我不知道你的功能是支持做什么的，但这是你遇到的问题：

def click_1(text):
    key_1 = word.index [letter]
    # word.index(letter) calls the function,
    # word.index[letter] tries to treat the function like a dictionary
    # and pull up its value attached to the key `letter`
    hidden_word[key_1] = letter
    print (hidden_word)

或者我建议看一下不同的设计模式。请尝试改为：

the_word = generate_random_word_from(word_list)
# which is probably as simple as random.choice(word_list)

guessed_letters = set()

def display_word():
    global the_word
    global guessed_letters
    # globals are bad in practice, but this is actually a great use case
    # for a global if you haven't wrapped this all up in OOP yet!

    masked_list = [letter if letter in guessed_letters else "_" for
                     letter in the_word]
    # this is literally:
    # # masked_list = []
    # # for letter in the_word:
    # #     if letter in guessed_letters:
    # #         masked_list.append(letter)
    # #     else:
    # #         masked_list.append("_")
    masked_word = ''.join(masked_list)
    # join the list on an empty string, e.g.:
    # ['s','_','m','e','w','_','r','d'] becomes 's_mew_rd'
    print(masked_word)
    # this wouldn't be a bad place to trigger your win condition, too.
    # Could be as simple as
    # # if "_" not in masked_word:
    # #     you_win()

def guess_letter():
    global the_word
    global guessed_letters

    guessed_letter = prompt_for_letter_somehow()
    if guessed_letter not in the_word:
        add_hangman_piece() # put another arm/leg/whatever on your dude
    guessed_letters.add(guessed_letter)
        # add that letter to the guessed_letters set anyway
    # validation would be nice here so you can't just keep guessing the same letters

Answer 2

这是我很久以前为计算机科学的mitx介绍所做的一个刽子手游戏，你可能会发现它很有用，它展示了如何在不需要任何全局声明的情况下制作游戏：

from string import ascii_lowercase


def is_word_guessed(secret_word, letters_guessed):
    return all(x in letters_guessed for x in secret_word)


def get_available_letters(letters_guessed):
    return "".join([x for x in ascii_lowercase if x not in letters_guessed])


def get_guessed_word(secret_word, letters_guessed):
    return "".join([letter if letter in letters_guessed else "_" for letter in secret_word])


def hangman(secret_word):
    print "Welcome to the game Hangman!"
    print "I am thinking of a word that is {} letters long\n-----------".format(len(secret_word))
    guesses = 8
    letters_guessed = []
    missed_l = ''
    correct_l = ''
    while True:
        if is_word_guessed(secret_word, letters_guessed):
            print "Congratulations, you won!"
            break
        elif len(missed_l) == 8:
            print "Sorry, you ran out of guesses. The word was {}.".format(secret_word)
            break
        print "You have {} guesses left".format(guesses)
        print "Available Letters: {}".format(get_available_letters(letters_guessed))
        l = raw_input("Please guess a letter: ").lower()
        if l in letters_guessed:
            print "Oops! You've already guessed that letter: {}\n-----------".format(
                get_guessed_word(secret_word, letters_guessed))
        elif l in secret_word:
            letters_guessed.append(l)
            get_guessed_word(secret_word, letters_guessed)
            correct_l += l
            print( "Good guess: {}\n-----------".format(
                "".join([x if x in letters_guessed else "_" for x in secret_word])))
        elif l not in secret_word:
            letters_guessed.append(l)
            guesses -= 1
            print "Oops! That letter is not in my word: {}\n-----------".format(
                get_guessed_word(secret_word, letters_guessed))
            missed_l += l

hangman("foobar")

如何在python中引用列表中的字符串

2 个答案: