是什么导致分配错误之前引用的局部变量？

Question

我目前正在处理 MIT 问题集 2，我遇到了一个已修复的错误，但不明白为什么会发生。此代码的相关部分是变量字母表。我在is_word_guessed 和get_guessed_word 两个函数中都引用了它。但是，当我运行代码时，我收到了在 get_guessed_word 函数中分配错误之前引用的局部变量。但是，我在这两个函数之外分配了变量。最重要的是，函数is_word_guessed 不会产生任何错误。

我可以通过在函数本身中分配字母来修复错误，但我想了解错误的原因以便学习。

1. 两个函数都引用了变量Alphabet（在函数外部分配），但唯一的get_guessed_word 尝试修改值。
2. 我可以假设错误源于变量的修改。但是，当我在同一代码中修改 secret_word 时，为什么不会出现此错误？ （是不是因为 secret_word 是一个赋值参数？）

secret_word = 'apple'
letters_guessed = ['a','p','q']#['e', 'i', 'k', 'p', 'r', 's']
alphabet='abcdefghijklmnopqrstuvwxyz'


def is_word_guessed(secret_word, letters_guessed):
    '''
    secret_word: string, the word the user is guessing; assumes all letters are
      lowercase
    letters_guessed: list (of letters), which letters have been guessed so far;
      assumes that all letters are lowercase
    returns: boolean, True if all the letters of secret_word are in letters_guessed;
      False otherwise
    '''
    secret_letters=set()
    for l in secret_word:
        if l in alphabet:
            secret_letters.add(l)
            print('Secret letters are ',secret_letters)
        else:
            pass
    for l in letters_guessed:
        if l in secret_letters:
            secret_letters.remove(l)
            print('Secret letters minus guessed letters are', secret_letters)
        else:
            pass
    if len(secret_letters)==0:
        return True
    else:
        return False
        
print(is_word_guessed(secret_word, letters_guessed))

def get_guessed_word(secret_word, letters_guessed):
    '''
    secret_word: string, the word the user is guessing
    letters_guessed: list (of letters), which letters have been guessed so far
    returns: string, comprised of letters, underscores (_), and spaces that represents
      which letters in secret_word have been guessed so far.
    '''
    ###alphabet='abcdefghijklmnopqrstuvwxyz'
    for l in letters_guessed:
        if l in alphabet:
            alphabet=alphabet.replace(l,'')
            print(alphabet)
    for l in alphabet:
        if l in secret_word:
            secret_word=secret_word.replace(l,'_ ')
            print(secret_word)
    return secret_word

print(get_guessed_word(secret_word, letters_guessed))

提前感谢您的帮助。我知道这是一个常见错误，但我阅读的文章都没有解决我的具体问题。

Answer 1

您问题的答案取决于您在 get_guessed_work 函数中分配字母的事实。

当您在函数中分配不可变变量（例如字符串）时，Python 会自动将此变量视为局部变量，而不是全局变量。 因此，字母表现在是一个全局变量，您之前没有定义它。 您可以通过在函数声明中添加默认函数变量来解决此问题：

def get_guessed_word(secret_word, letters_guessed,alphabet=alphabet):

希望对您有所帮助！ 马克西姆。

Answer 2

你可以添加

global alphabet

喜欢

secret_word = 'apple'
letters_guessed = ['a','p','q']#['e', 'i', 'k', 'p', 'r', 's']
alphabet='abcdefghijklmnopqrstuvwxyz'


def is_word_guessed(secret_word, letters_guessed):
    '''
    secret_word: string, the word the user is guessing; assumes all letters are
      lowercase
    letters_guessed: list (of letters), which letters have been guessed so far;
      assumes that all letters are lowercase
    returns: boolean, True if all the letters of secret_word are in letters_guessed;
      False otherwise
    '''
    secret_letters=set()
    for l in secret_word:
        if l in alphabet:
            secret_letters.add(l)
            print('Secret letters are ',secret_letters)
        else:
            pass
    for l in letters_guessed:
        if l in secret_letters:
            secret_letters.remove(l)
            print('Secret letters minus guessed letters are', secret_letters)
        else:
            pass
    if len(secret_letters)==0:
        return True
    else:
        return False
        
print(is_word_guessed(secret_word, letters_guessed))

def get_guessed_word(secret_word, letters_guessed):
    '''
    secret_word: string, the word the user is guessing
    letters_guessed: list (of letters), which letters have been guessed so far
    returns: string, comprised of letters, underscores (_), and spaces that represent
      which letters in secret_word have been guessed so far.
    '''
    global alphabet
    for l in letters_guessed:
        if l in alphabet:
            alphabet=alphabet.replace(l,'')
            print(alphabet)
    for l in alphabet:
        if l in secret_word:
            secret_word=secret_word.replace(l,'_ ')
            print(secret_word)
    return secret_word

print(get_guessed_word(secret_word, letters_guessed))

提示：避免将函数参数命名为与变量相同的名称，因为这样会造成混淆 而且你应该总是使用

x = y

而不是

x=y

只是为了美观。

secret_word = 'apple'
letters_guessed = ['a','p','q']
alphabet = 'abcdefghijklmnopqrstuvwxyz'


def is_word_guessed(word, guess):
    '''
    secret_word: string, the word the user is guessing; assumes all letters are
      lowercase
    letters_guessed: list (of letters), which letters have been guessed so far;
      assumes that all letters are lowercase
    returns: boolean, True if all the letters of secret_word are in letters_guessed;
      False otherwise
    '''
    secret_letters = set()
    for l in word:
        if l in alphabet:
            secret_letters.add(l)
            print('Secret letters are ',secret_letters)
        else:
            pass
    for l in guess:
        if l in secret_letters:
            secret_letters.remove(l)
            print('Secret letters minus guessed letters are', secret_letters)
        else:
            pass
    if len(secret_letters) == 0:
        return True
    else:
        return False
        
print(is_word_guessed(secret_word, letters_guessed))

def get_guessed_word(word, guess):
    '''
    secret_word: string, the word the user is guessing
    letters_guessed: list (of letters), which letters have been guessed so far
    returns: string, comprised of letters, underscores (_), and spaces that represent
      which letters in secret_word have been guessed so far.
    '''
    global alphabet
    for l in guess:
        if l in alphabet:
            alphabet = alphabet.replace(l,'')
            print(alphabet)
    for l in alphabet:
        if l in word:
            word = word.replace(l,'_ ')
            print(word)
    return word

print(get_guessed_word(secret_word, letters_guessed))

希望我能帮上忙。