Question

我正在尝试在字符串中找到一个单字或n词短语，然后将其替换为星号。面临的挑战是，即使某个单词或n个单词的短语被某些字符所混淆，我也想这样做。

假设以下内容。 REPLACE_CHAR是我要用来替换单词或n单词短语的字符。 ILLEGAL_CHAR是我要忽略的字符。我也想忽略这种情况。

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

示例一

在这里，我想用星号代替“美元”。在字符串中，您可以看到存在“美元”，但是它被随机符号和大写字母所混淆。

string = "Lorem ipsum %@do^l&oR sit amet"
find = "dolor"

提示结果为"Lorem ipsum ***** sit amet"，其中星号数量与找到的单词的长度匹配。

示例二

在这里，我要在保留空格的同时用星号替换“ dolour sit”。在字符串中，您可以看到存在“美元坐在”，但是它被随机符号和大写字母所混淆。

string = "Lorem ipsum %@do^l&oR s%)i!T~ amet"
find = "dolor sit"

提示结果为"Lorem ipsum ***** *** amet"，其中星号数量与找到的单词的长度匹配。

更新＃1

此解决方案基于@ Ajax1234响应。

我们使用re.sub并在函数外部构建表，而不是使用ILLEGAL_CHAR来删除translate。这会稍微提高性能。

import re

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

trans = str.maketrans("", "", ILLEGAL_CHAR)
text = "Lorem ipsum %@do^l&oR sit amet"
token = "dolor sit"

def replace(data, token):
    data = data.translate(trans)
    return re.sub(token, lambda x:' '.join('*'*len(i) for i in x.group().split(' ')), data, flags=re.I)

print(replace(text, token))

Answer 1

import re

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

string = "Lorem ipsum %@do^l&oR s%)i!T~ amet"

clean_string = "".join(char for char in string if char not in ignore_chars)

bad_words = ["dolor", "sit"]

for bad_word in bad_words:
    pattern = f"\\b{bad_word}\\b"
    replace = "*" * len(bad_word)
    clean_string = re.sub(pattern, replace, clean_string, flags=re.IGNORECASE)

print(clean_string)

输出：

Lorem ipsum ***** *** amet

Answer 2

您可以使用re.sub删除非法字符，然后再用re.sub加上另一个re.I：

import re
def replace(word, target):
   w = re.sub('[\!"#\$%\&\'\(\)\*\+,\-\./:;\<\=\>\?@\[\]\^_`\{\|\}~]+', '', word)
   return re.sub(target, lambda x:' '.join('*'*len(i) for i in x.group().split(' ')), w, flags=re.I)

string = "Lorem ipsum %@do^l&oR sit amet"
find = "dolor"
r = replace(string, find)

输出：

'Lorem ipsum ***** sit amet'

string = "Lorem ipsum %@do^l&oR s%)i!T~ amet"
find = "dolor sit"
r = replace(string, find)

输出：

'Lorem ipsum ***** *** amet'

Answer 3

您可以随意使用re.sub来完成单词的混淆和重新混淆处理！这里已经有很多好的答案；该脚本的设计易于编辑，尤其是当您计划从用户或其他外部来源获得输入时。

#we'll be using regex to solve this problem
import re


#establish some constants - these can be changed later, or even read as user input
REPLACE_CHAR = "*"
ILLEGAL_CHAR = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'


#your search string - this can be read as user input
search = "Lorem ipsum %@do^l&oR sit amet"

#this regex will remove the illegal characters - specifically, it substitutes an empty 
#character ('') in place of any illegal character we find. 
#note that since the brackets are included here, the user can directly input illegal 
#symbols themselves without worrying about formatting
strip = re.sub('['+ILLEGAL_CHAR+']', '', search)


#the string to obfuscate - this can also be read as user input
find = "ipsum dolor sit"

#this splits the words on spaces, so there's spaces between tee asterisks
find_words = find.split(' ')


#now we'll check each find_word - we'll look for it in the string, and if we find it,
#we'll replace it with asterisks of the same length as the original word. 
#(we'll use a ranged for loop to go over the words)
for f_word in find_words:

  #check each f_word to see if it appears in the string. note "flags=re.I" - this 
  #tells our regex to use case-insensitive matching
  if(re.search(f_word, strip, flags=re.I)):

    #we found a word! check the length of the word, then substitute an equal number of
    #REPLACE_CHARs
    strip = re.sub(f_word, (REPLACE_CHAR * len(f_word)), strip, flags=re.I)

#ta-daa!
print(strip)

Try it here!

查找并替换字符串中混淆的单词或短语

示例一

示例二

更新＃1

3 个答案: