字符串方法replace不会修改原始字符串。它保留原始文件并返回修改后的副本。

你需要的是这样的:line = line.replace(char， ")

def replace_all(line, )for char in line:
    if char in " ?.!/;:":
        line = line.replace(char,'')
    return line

然而，每次删除一个字符都创建一个新的字符串是非常低效的。我推荐以下方法:

def replace_all(line, baddies, *):
    """
    The following is documentation on how to use the class,
    without reference to the implementation details:

    For implementation notes, please see comments begining with `#`
    in the source file.

    [*crickets chirp*]

    """

    is_bad = lambda ch, baddies=baddies: return ch in baddies
    filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
    mahp = replace_all.map(filter_baddies, line)
    return replace_all.join('', join(mahp))

    # -------------------------------------------------
    # WHY `baddies=baddies`?!?
    #     `is_bad=is_bad`
    # -------------------------------------------------
    # Default arguments to a lambda function are evaluated
    # at the same time as when a lambda function is
    # **defined**.
    #
    # global variables of a lambda function
    # are evaluated when the lambda function is
    # **called**
    #
    # The following prints "as yellow as snow"
    #
    #     fleece_color = "white"
    #     little_lamb = lambda end: return "as " + fleece_color + end
    #
    #     # sometime later...
    #
    #     fleece_color = "yellow"
    #     print(little_lamb(" as snow"))
    # --------------------------------------------------
replace_all.map = map
replace_all.join = str.join

令我惊讶的是，还没有人推荐使用内置的过滤功能。

    import operator
    import string # only for the example you could use a custom string

    s = "1212edjaq"

假设我们想过滤掉所有不是数字的东西。使用过滤器内置方法“…等效于生成器表达式(item for item在可迭代if函数(item)中)"[Python 3 Builtins: Filter]

    sList = list(s)
    intsList = list(string.digits)
    obj = filter(lambda x: operator.contains(intsList, x), sList)))

在Python 3中返回

    >>  <filter object @ hex>

要得到打印的字符串，

    nums = "".join(list(obj))
    print(nums)
    >> "1212"

我不确定过滤器在效率方面的排名，但在做列表理解等时，知道如何使用是一件好事。

更新

从逻辑上讲，既然过滤器可以工作，你也可以使用列表理解，从我所读到的，它应该更有效，因为lambdas是编程函数世界的华尔街对冲基金经理。另一个优点是它是一个单行程序，不需要任何导入。例如，使用上面定义的字符串's'，

      num = "".join([i for i in s if i.isdigit()])

就是这样。返回值将是原始字符串中所有数字组成的字符串。

如果你有一个特定的可接受/不可接受字符列表，你只需要调整列表理解的' If '部分。

      target_chars = "".join([i for i in s if i in some_list]) 

或者,

      target_chars = "".join([i for i in s if i not in some_list])

#!/usr/bin/python
import re

strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr

如果你想让你的字符串只允许使用ASCII码，你可以使用这段代码:

for char in s:
    if ord(char) < 96 or ord(char) > 123:
        s = s.replace(char, "")

它将删除....以外的所有字符Z是大写的。

递归分割: s =字符串;Chars =要删除的字符

def strip(s,chars):
if len(s)==1:
    return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) +  strip(s[int(len(s)/2):len(s)],chars)

例子:

print(strip("Hello!","lo"))    #He!

您还可以使用函数来替换不同类型的正则表达式或使用列表的其他模式。这样，您就可以混合正则表达式、字符类和真正基本的文本模式。当您需要替换大量元素(如HTML元素)时，它非常有用。

*注意:适用于Python 3.x

import re  # Regular expression library


def string_cleanup(x, notwanted):
    for item in notwanted:
        x = re.sub(item, '', x)
    return x

line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)

# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)

# Get rid of special characters
special_chars = ["[!@#$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)

在函数string_cleanup中，它以字符串x和未修饰的列表作为参数。对于元素或模式列表中的每一项，如果需要替代品，就会进行替换。

输出:

Uncleaned:  <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean:  My example: A text %very% $clean!!
2nd clean:  My example: A text very clean