您也可以使用regex：

df_reviews['name'] = df_reviews['name'].str.replace('Review|[:\'"#{}_]', "", regex=True)

正则表达式模式：

'Review|[:\'"#{}_]'

• Review：替换"审查"一词
• |：或
• [:\'"#{}_]-方括号[]中的任意字符

注：

如果您希望删除所有标点符号：你可以用这个代替

import string

df_reviews['name'] = df_reviews['name'].str.replace(f'Review|[{string.punctuation}]', "", regex=True)

这将删除以下字符：

!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~

试试这个：

df['name'] = df['name'].str.replace('Review| \:| \'|\"|\#| \_', "").str.strip()

import pandas as pd

REMOVE_CHARS = ["Review", ":", "#", "{", "}", "_", "'", '"']
def process_name(name: str) -> str:
    for removal_char in REMOVE_CHARS:
        try:
            print(f"removal char {removal_char}", name.index(removal_char))
            name = name.replace(removal_char,"")
        except ValueError:
            continue
    return name

def clean_code(df_reviews: pd.DataFrame):
    # Renaming `Unnamed: 0` as `index` ; `0` as `Name`
    df_reviews = df_reviews.rename(columns={"Unnamed: 0": "index", "0": "Name"})
    # todo: clarification needed
    # Here Name col contains a words separated by : so `expand=True` separate it into different columns
    # then we just read the zeroth column
    df_reviews['name'] = df_reviews["Name"].str.split(':', expand=True)[0]
    # Preprocessing of name column
    # >> if `name` contains ["Review",":","#","{","}","_","'",'"'] remove/replace it
    df_reviews['name'] = df_reviews['name'].apply(lambda x: process_name(x))
    df_reviews['name'] = df_reviews['name'].str.strip()

if __name__ == "__main__":
    df_reviews = pd.read_csv("reviews.csv")