IndexError：使用弹出功能时列表索引超出范围[重复]

Question

我正在制作一个制作词云的程序。我想要一个没有标点符号和常用词的单词列表。我删除了标点符号，使用函数removepunc；它工作正常。现在我正在创建第二个函数来删除常用词（我没有使用以前的逻辑，因为它从程序中删除了字母 I 以及代词 I），我收到错误 IndexError: list index out of range，我将文件转换为列表。

代码：

def removepunc(z):
test_str=z
punc = '''!()-[]{};:'""\,<>./?@#$%^&*_~'''
for ele in test_str:
    if ele in punc:
        test_str = test_str.replace(ele, "")
return test_str

def removebad(f):
    print(type(f))
    z=[]
    badword2 = ["the", "a", "to", "if", "is", "it", "of", "and", "or", "an", "as", "i", "me", 
    "my","we", "our", "ours", "you", "your", "yours", "he", "she", "him", "his", "her", 
    "hers", "its", "they","them","their", "what", "which", "who", "whom", "this", "that", 
    "am", "are", "was", "were", "be", "been","being","have", "has", "had", "do", "does", 
    "did", "but", "at", "by", "with", "from", "here", "when", "where","how","all", "any", 
    "both", "each", "few", "more", "some", "such", "no", "nor", "too", "very", "can", 
    "will","just"]
    for i in range (len(f)-1):
        if f[i] in badword2:
            x=f.pop(i)
            z.append(x)
        else:
            continue
    return f



file=open("openfile.txt")
a=file.read()
a=a.lower()
unqword=removepunc(a)
ab=unqword.split()
print(type(ab))
unqword1=removebad(ab)
print(unqword1)

`

输出：

C:\Users\Nitin\PycharmProjects\pythonProject1\venv\Scripts\python.exe C:/Users/Nitin/PycharmProjects/pythonProject1/prjt.py
<class 'list'>
<class 'list'>
Traceback (most recent call last):
  File "C:/Users/Nitin/PycharmProjects/pythonProject1/prjt.py", line 29, in <module>
    unqword1=removebad(ab)
  File "C:/Users/Nitin/PycharmProjects/pythonProject1/prjt.py", line 14, in removebad
    if f[i] in badword2:
IndexError: list index out of range

Process finished with exit code 1

我还没有为 wordcloud 编写逻辑，稍后我会在摆脱它时执行此操作

Answer 1

from typing import List


def removepunc(z):
    test_str = z
    punc = r'''!()-[]{};:'""\,<>./?@#$%^&*_~'''
    for ele in test_str:
        if ele in punc:
            test_str = test_str.replace(ele, "")
    return test_str


def removebad(f: List):
    print(type(f))
    z = []
    badword2 = {"the", "a", "to", "if", "is", "it", "of", "and", "or", "an", "as", "i", "me",
                "my", "we", "our", "ours", "you", "your", "yours", "he", "she", "him", "his", "her",
                "hers", "its", "they", "them", "their", "what", "which", "who", "whom", "this", "that",
                "am", "are", "was", "were", "be", "been", "being", "have", "has", "had", "do", "does",
                "did", "but", "at", "by", "with", "from", "here", "when", "where", "how", "all", "any",
                "both", "each", "few", "more", "some", "such", "no", "nor", "too", "very", "can",
                "will", "just"}
    returned_f = []
    for item in f:
        if item in badword2:
            z.append(item)
        else:
            returned_f.append(item)
    return returned_f


file = open("openfile.txt")
a = file.read()
a = a.lower()
unqword = removepunc(a)
ab = unqword.split()
print(type(ab))
unqword1 = removebad(ab)
print(unqword1)

Answer 2

创建一个包含要保留的单词的新列表：

def removebad(f):
    print(type(f))
    badword2 = ["the", "a", "to", "if", "is", "it", "of", "and", "or", "an", "as", "i", "me", 
    "my","we", "our", "ours", "you", "your", "yours", "he", "she", "him", "his", "her", 
    "hers", "its", "they","them","their", "what", "which", "who", "whom", "this", "that", 
    "am", "are", "was", "were", "be", "been","being","have", "has", "had", "do", "does", 
    "did", "but", "at", "by", "with", "from", "here", "when", "where","how","all", "any", 
    "both", "each", "few", "more", "some", "such", "no", "nor", "too", "very", "can", 
    "will","just"]
    keep = []
    toss = []
    for w in f:
        if w in badword2:
            toss.append( w )
        else:
            keep.append( w )
    return keep

当然，我不明白您为什么要保留“折腾”字样（您的z 列表），因为您不使用它做任何事情。没有它，它可能只是：

    return [w for w in f if w not in badword2]