使用Range和Len方法迭代分割字符串时Python函数中的索引器

发表时间：2022-07-23 00:23:25 阅读：91

*问题陈述:我正在尝试编写一个接受字符串和列表的函数.

-我将使用的一串单词.split()方法转换为列表 -使列表中的所有单词都小写，以使检查统一 -遍历列表检查，查看字符串列表中是否包含数组中的单词. -如果是，则会删除该单词 -将非填充词列表重新连接到字符串中

回溯:

Traceback (most recent call last):
  File "main.py", line 55, in <module>
    no_filler = remove_filler_words(no_punction,filler_words);
  File "main.py", line 37, in remove_filler_words
    word_list[cap_word] = word_list[cap_word].lower()
IndexError: list index out of range

代码:

def remove_filler_words(cleaned_string:string, filler_words:list):
    cleaned_sentence = ""
    word_list = cleaned_string.split()
    for cap_word in range(len(word_list)):
        word_list[cap_word] = word_list[cap_word].lower()
        for filler_word in filler_words:
            if filler_word in word_list:
                word_list.remove(filler_word)
                line_join = " ".join(word_list)
                cleaned_sentence += line_join

 return cleaned_sentence

回复:https://replit.com/join/dpxpwamhgy-terry-brooksjr

🎖️ 优质答案

尝试创建word_list的副本，例如copy_word_list并从此变量中删除单词.

代码如下:

def remove_filler_words(cleaned_string:string, filler_words:list):
    cleaned_sentence = ""
    word_list = cleaned_string.split()
    #create duplicate
    copy_word_list = word_list
    for cap_word in range(len(word_list)):
        word_list[cap_word] = word_list[cap_word].lower()
        for filler_word in filler_words:
            if filler_word in word_list:
                #remove from copy
                copy_word_list.remove(filler_word)
                line_join = " ".join(word_list)
                cleaned_sentence += line_join

 return cleaned_sentence

希望有帮助)

以下是我提出的解决方案:

def remove_filler_words(cleaned_string, filler_words):
    word_list = cleaned_string.split()
    new_word_list = []
    for cap_word in range(len(word_list)):
        lower_word = word_list[cap_word].lower()
        new_word_list.append(lower_word)
        for filler_word in filler_words:
            if filler_word in new_word_list:
                new_word_list.remove(filler_word)
                cleaned_sentence = " ".join(new_word_list)

    return cleaned_sentence

我将所有保存到lower_word的小写单词添加到一个新列表中，以解释索引错误.然后，我没有在开始时声明空列表，而是将cleaned_sentence变量分配给连接函数，该函数在返回时生成预期结果.