Python中“返回函数外”语法错误的原因？

Question

我在此代码中遇到问题;

'return' outside function (<module1>, line 4)   <module1>   4

这本代码来自学习python艰难的方式; “练习25：更多练习”

def break_words(stuff):
    """This function will break up words for us."""
words = stuff.split(' ')
return words


def sort_words(words):
    """Sorts the words."""
    return storred(words)

def print_first_word(words):
    """Prints the first word after popping it off."""
    word = words.php(0)
    print word

def sort_last_word(words):
    """prints the last word after popping it off."""
    word = words.pop(-1)
    print word

def sort_sentence(sentence):
    """Takes in a full sentence and returns the sorted words."""
    words = break_words(sentence)
    return sort_words(words)

def print_first_and_last(sentence):
    """Prints the first and last words of the sentence. """
    Words = break_words(sentence)
    print_first_word(words)
    print_last_word(words)

def print_first_and_last_sorted(sentence):
    """Sorts the words then prinits the first and last one."""
    words = sort_sentence(sentence)
    print_first_word(words)
    print_last_word(words)

任何解决方案？

Answer 1

在此函数中return没有正确缩进。

def break_words(stuff):
    """This function will break up words for us."""
    words = stuff.split(' ')
    return words

Answer 2

你的缩进是关闭的，试试这个：

def break_words(stuff):
    """This function will break up words for us."""
    words = stuff.split(' ')
    return words

缩进是Python的一个关键组件，与其他语言不同 - 因此确保代码格式正确非常重要。

Answer 3

缩进在python中很重要。根据你粘贴的内容，这里的最后两行不是break_words的范围。

def break_words(stuff):
    """This function will break up words for us."""
words = stuff.split(' ')
return words