Question

我正在尝试删除字符串中的某些字符，但是我有一个错误，我有不同的方式循环但它仍然无法正常工作

我的问题是我如何循环字符串并删除字符？

这是我的代码：

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

for remove in MyList:   

        mystring =re.sub(remove, "", "I am trying this code")
        print (remove)

我有这个错误：

---------------------------------------------------------------------------
error                                     Traceback (most recent call last)
<ipython-input-386-df22385a9f06> in <module>()
      5 for remove in MyList:
      6 
----> 7         mystring =re.sub(remove, "", "I am trying this code ! ? /")
      8         print (remove)

~\AppData\Local\Continuum\anaconda3\lib\re.py in sub(pattern, repl, string, count, flags)
    189     a callable, it's passed the match object and must return
    190     a replacement string to be used."""
--> 191     return _compile(pattern, flags).sub(repl, string, count)
    192 
    193 def subn(pattern, repl, string, count=0, flags=0):

~\AppData\Local\Continuum\anaconda3\lib\re.py in _compile(pattern, flags)
    299     if not sre_compile.isstring(pattern):
    300         raise TypeError("first argument must be string or compiled pattern")
--> 301     p = sre_compile.compile(pattern, flags)
    302     if not (flags & DEBUG):
    303         if len(_cache) >= _MAXCACHE:

~\AppData\Local\Continuum\anaconda3\lib\sre_compile.py in compile(p, flags)
    560     if isstring(p):
    561         pattern = p
--> 562         p = sre_parse.parse(p, flags)
    563     else:
    564         pattern = None

~\AppData\Local\Continuum\anaconda3\lib\sre_parse.py in parse(str, flags, pattern)
    853 
    854     try:
--> 855         p = _parse_sub(source, pattern, flags & SRE_FLAG_VERBOSE, 0)
    856     except Verbose:
    857         # the VERBOSE flag was switched on inside the pattern.  to be

~\AppData\Local\Continuum\anaconda3\lib\sre_parse.py in _parse_sub(source, state, verbose, nested)
    414     while True:
    415         itemsappend(_parse(source, state, verbose, nested + 1,
--> 416                            not nested and not items))
    417         if not sourcematch("|"):
    418             break

~\AppData\Local\Continuum\anaconda3\lib\sre_parse.py in _parse(source, state, verbose, nested, first)
    614             if not item or (_len(item) == 1 and item[0][0] is AT):
    615                 raise source.error("nothing to repeat",
--> 616                                    source.tell() - here + len(this))
    617             if item[0][0] in _REPEATCODES:
    618                 raise source.error("multiple repeat",

error: nothing to repeat at position 0

对不起我的新手问题

Answer 1

您不需要正则表达式。只需使用str.replace：

>>> MyList= [",", ":", "\"", "=", "&", ";", "%", "$","@", "%", "^",
...          "*", "(", ")", "{", "}","[", "]", "|", "/", "\\", ">",
...          "<", "-",'!', '?', '.', "'",'--', '---', "#"]
>>> mystring = "Helo, world!?"
>>> for s in MyList:
...     mystring = mystring.replace(s, '')
...
>>> mystring
'Helo world'

Answer 2

已经发布了一些解决方法，但还没有人解释错误。 re.sub（）的文档说：

re.sub(pattern, repl, string, ...) - Return the string obtained by replacing ... pattern in string by the replacement repl

因此，在遍历列表中的一些符号后，我们得到了这个：

re.sub("*", "", "I am trying this code")

因此，您尝试将*替换为"" - 但*是正则表达式中使用的“特殊”字符，在这种情况下，"*"是一个非法/无效的正则表达。 *是正则表达式中的量词，意味着“尽可能多地返回前一个正则表达式的重复” - 但该字符串中没有先前的正则表达式。

"*"通常与"."中的".*"一起使用，这意味着尽可能多地匹配任何单个字符（"."）（"*"）。

更新：这是我如何从字符串中删除符号：

import string

s = "I# am trying th<>is code!"
print(s.translate(None, string.punctuation))

Answer 3

如果您的目标只是从字符串中删除所有标点符号，则保留所有其他字符（_和包含空格）的单个正则表达式将为：

import re

s = "Hello, world! how are you?"

print(re.sub("[^ \w]","",s))

结果：

Hello world how are you

它比循环中的x替换调用（创建尽可能多的字符串）或带有|的正则表达式更有效。

Answer 4

这里有一个列表理解和加入的答案，不是最优雅，也不是最快：

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

my_string = "Hello, world!?"

# one liner
my_string = ''.join([c for c in my_string if c not in my_list])


print(my_string)

Answer 5

您还可以将列表组合成单个表达式：

pattern = re.compile(r"\b(" + "|".join(MyList) + ")\\W")
pattern.sub("", mystring )

循环删除一些字符串

5 个答案: