我试图进入编程并提出一个简单的问题：
如果我有
def break_words(stuff):
words = stuff.split(' ')
return words
def sort_words(words):
##Sorts the words."""
words = break_words(words)
return sorted(words)
def print_first_word(sentence):
words = break_words(sentence)
words = sorted(words)
return words.pop(0)
sentence = "Tequila Mariachi Sangria"
print break_words(sentence)
print sort_words(sentence)
print print_first_word(sentence)
当我运行它时，我的代码很好，而如果我写
####################Test################
def break_words(stuff):
words = stuff.split(' ')
return words
def sort_words(words):
##Sorts the words."""
words = break_words(words)
return sorted(words)
def print_first_word(sentence):
words = break_words(sentence)
words = sort_words(words)
return words.pop(0)
sentence = "Tequila Mariachi Sangria"
print break_words(sentence)
print sort_words(sentence)
print print_first_word(sentence)
我会得到
AttributeError: "list" object has no attribute "split"
两个函数break_words和sort_words都创建了列表对象，为什么我在第二种情况下得到错误？
答案 0 :(得分：2)
print_first_word(sentence)来电
words = break_words(sentence)
此处，words现在是一个列表，然后传递给sort_words(words)，然后传递给break_words(words)，其中调用
words = stuff.split(' ')
其中stuff是一个列表，导致错误。
答案 1 :(得分：0)
我认为第二个源代码应该是：
####################Test################
def break_words(stuff):
words = stuff.split(' ')
return words
def sort_words(words):
##Sorts the words."""
words = break_words(words)
return sorted(words)
def print_first_word(sentence):
words = sort_words(sentence)
return words.pop(0)
sentence = "Tequila Mariachi Sangria"
print break_words(sentence)
print sort_words(sentence)
print print_first_word(sentence)
我将words = sort_words(words)更改为words = sort_words(sentence)