Question

我目前正在研究python hangman程序。我有一些基础知识，例如显示空白和检测正确的字母，但是我无法弄清楚何时正确识别了单词或句子。

EDIT

阅读评论后，我更新了代码，使用snake_case而不是camelCase命名了变量，并将输出添加到了show_word函数中。

def show_word(word, known_letters):
    word = word.lower()
    characters = ""

    for i in word:
        if i in known_letters:
            characters = characters + i + " "
        elif i == " ":
            characters = characters + "/ "
        else:
            characters = characters + "_ "

    print characters
    return characters.replace(" ", "").replace("/", " ")

def guess_word(word, known_letters):
    all_letters = "abcdefghijklmnopqrstuvwxyz"
    word = word.lower()
    guesses = 0

    while show_word(word, known_letters) != word:
        print("Guess a letter")
        guess = raw_input()

        if guess in known_letters:
            print("You have already guessed %s" % (guess))
        elif len(guess) != 1:
            print("Only type one letter")
        else:
            known_letters += guess

        guesses += 1

    print("Well done you have guessed the word(s) (%s) in %s guesses" % (word, guesses))

guess_word("The quick brown fox jumps over the lazy dog", "")

该程序现在可以正确识别何时我猜对了所有正确的字母。感谢所有提供帮助的人。如果其他人在此问题上需要帮助，则下面是我的原始代码。

def showWord(word, knownLetters):
    word = word.lower()
    characters = ""

    for i in word:
        if i in knownLetters:
            characters = characters + i + " "
        elif i == " ":
            characters = characters + "/ "
        else:
            characters = characters + "_ "

    print characters

def guessWord(word, knownLetters):
    word = word.lower()
    guesses = 0

    while showWord(word, knownLetters) != word:
        print("Guess a letter")
        guess = raw_input()

        if guess in knownLetters:
            print("You have already guessed %s" % (guess))
        elif len(guess) != 1:
            print("Only guess one letter")
        else:
            knownLetters += guess

        guesses += 1

    print("Well done, you guessed the word (%s) in %s guesses" % (word, guesses))

    showWord(word, knownLetters)

guessWord("The quick brown fox jumps over the lazy dog", "")

Answer 1

您的showWord()函数默认不会返回None，因为它不会显式返回任何内容。

您可能应该在return characters.replace(' ', '').replace('/', ' ')之后添加print characters，并且应该满足您的最终条件。

由于字母不会与showWord()完全对齐，因此您需要调整word的输出。

当showWord()实际上返回您的字符串时，以下陈述将正确地等同于

：

while showWord(word, knownLetters) != word:

然后将工作，因为可以将characters与word进行比较。

此外，Python的约定是在snake_case中而不是camelCase中命名变量。

最后，要重定向您的奖金问题，很多人都使用Python制作了ASCII Hangman，也许您可以参考，学习和采用这些方法。如果您对此有任何疑问，可以在Stack Overflow上询问另一个问题，以明确显示您尝试过的内容。

Answer 2

这是一些工作代码的模板，您可以希望使用它们来找出修改代码的方法。

def guessWord(word, knownLetters):
    word = word.lower()
    cur_word = ['_'] * len(word)
    result_to_compare = ""

    num_guesses = 0

    while result_to_compare != word:
        print("Guess a letter!")
        guess = raw_input()

        if guess in knownLetters:
            print("You have already guessed this letter!")
        else:
            knownLetters += guess
            for i in range(len(word)):
                if guess is word[i]:
                    cur_word[i] = guess

            print(cur_word)
            result_to_compare = ''.join(cur_word)            

    print("Congrats! You guessed it!")          

guessWord("severe", "")

我使用cur_word作为临时变量来显示用户的进度。经过有效猜测后，我从cur_word中的值创建字符串result_to_compare并将其与实际解决方案进行比较。希望这可以帮助。只要用户输入raw_input作为空格，这将在单词之间使用空格。您应该能够从这里想出如何对其进行修改以按自己的喜好工作。

Answer 3

这是我的版本，我使代码尽可能冗长，因此应该易于说明。

def showWord(word,knownLetters):
    word=word.lower()
    characters=""
    wordsofar=""
    for char in word:
        if char in knownLetters:
            characters = characters + char + " "
            wordsofar += char
        elif char == " ":
            characters = characters + "/ "
            wordsofar += " "
        else:
            characters += "_ "

    print characters

    return validate(word, wordsofar)

def validate(word, knownLetters):
    word_dict = {}

    for c in word:
        if c not in word_dict:
            word_dict[c] = 1
        else:
            word_dict[c] += 1

    for c in knownLetters:
        if c not in word_dict:
            return False
        else:
            word_dict[c] -=1

    for k,v in word_dict.iteritems():
        if v != 0:
            return False
    return True

然后在while循环中，您将检查showWord（word，knownLetters）是否返回false。

Hangman项目无法正确识别胜利。 Python 2.7

3 个答案: