Question

错误跟踪：

C：\ Users \ Abhi.Abhi-PC \ Desktop \ PYE＆gt; ex25ex.py Traceback（最新版）最后调用）：文件“C：\ Users \ Abhi.Abhi-PC \ Desktop \ PYE \ ex25ex.py”，行 41，在print_last_word（句子）文件中 “C：\ Users \ Abhi.Abhi-PC \ Desktop \ PYE \ ex25ex.py”，第17行，in print_last_word word = words.pop（1）AttributeError：'str'对象没有属性'pop'

这是代码

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 sorted(words)

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

def print_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 prints the first and last one."""
    words = sort_sentence(sentence)
    print_first_word(words)
    print_last_word(words)

sentence="island has lush vegetation, area. However, summers are cooler than those  abundant."

break_words(sentence)
sort_words(sentence)
print_last_word(sentence)
print_last_word(sentence)
print_first_and_last_sorted(sentence)
print_first_and_last(sentence)

我无法解决问题的问题。我是python的新手。我正在从“如何学习python艰难的方式”这本书开始工作。

Answer 1

您的代码存在的问题是您忽略了函数的返回值，尤其是break_words，并始终对原始输入进行操作。这样可以解决问题：

words = break_words(sentence)
sort_words(words)
print_first_word(words)
print_last_word(words)
print_first_and_last_sorted(sentence)
print_first_and_last(sentence)

还有一个小问题，您应该使用words.pop(-1)来访问单词列表中的最后一项。

您如何从错误消息中识别出问题。例外是一个表示'str' object has no attribute 'pop'的AttributeError。显然，在行words.pop(0)中，单词是一个字符串，而不是字符串列表，正如变量名所示。然后只需要快速浏览一下，看看为什么这个变量是一个字符串：当你用print_last_word(sentence)调用函数时，你传递原始数据（类型为str）而不是你可能的预处理数据本打算通过。

Answer 2

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 sorted(words)

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

def print_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 prints the first and last one."""
    words = sort_sentence(sentence)
    print_first_word(words)
    print_last_word(words)

sentence="island has lush vegetation, area. However, summers are cooler than those  abundant."

wordList=break_words(sentence)
print('')
print('wordList:',wordList)
print('')
print('Sorted :',sort_words(wordList))
print('')
print_first_word(wordList)
print('')
print_last_word(wordList)
print('')
print_first_and_last_sorted(sentence)
print('')
print_first_and_last(sentence)

<强> RESULT

wordList: ['island', 'has', 'lush', 'vegetation,', 'area.', 'However,', 'summers', 'are', 'cooler', 'than', 'those', 'abundant.']

Sorted : ['However,', 'abundant.', 'are', 'area.', 'cooler', 'has', 'island', 'lush', 'summers', 'than', 'those', 'vegetation,']

island

abundant.

However,
vegetation,

island
abundant.

Answer 3

您要在此处执行的操作是从字符串中删除单词。当您尝试sentence.pop()时，您试图从不可变序列类型中删除元素。这意味着您正在尝试从字符串中删除元素，这没有多大意义。您应该做的是，将字符串分解为words.split(' ')的单词，然后删除要从中删除的任何元素。因此，替换第17行的解决方案是：

word = words.split(' ').pop(len(words.split(' '))-1)

这将删除句子中的最后一个单词并将其分配给word。

注意：这显然不是最有效的解决方案。我将其分解以使其可读。

希望这有帮助

Answer 4

请参见pop()不是字符串函数，因此可以创建一个函数，将n作为需要从字符串中弹出的字符的索引传递为str。将以下代码写入弹出窗口：

def popout(str,n):
  front = str[:n]   # up to but not including n
  back = str[n+1:]  # n+1 through end of string
  return front + back

根据我的理解，这可能会回答您的问题。如果没有，请在评论部分纠正我。

AttributeError：'str'对象没有属性'pop'

4 个答案: