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!

对于只允许字符串中某些字符的相反要求，可以使用带有集补操作符[^ABCabc]的正则表达式。例如，要删除除ascii字母、数字和连字符以外的所有字符:

>>> import string
>>> import re
>>>
>>> phrase = '  There were "nine" (9) chick-peas in my pocket!!!      '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)

'Therewerenine9chick-peasinmypocket'

来自python正则表达式文档:

不在范围内的字符可以通过互补来匹配 一组。如果集合的第一个字符是'^'，则所有字符 不在集合中的将被匹配。例如，[^5]将匹配 除'5'以外的任何字符，[^^]将匹配除 “^”。的第一个字符没有特殊意义 集。

这里有一些可能的方法来完成这个任务:

def attempt1(string):
    return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])


def attempt2(string):
    for v in ("a", "e", "i", "o", "u"):
        string = string.replace(v, "")
    return string


def attempt3(string):
    import re
    for v in ("a", "e", "i", "o", "u"):
        string = re.sub(v, "", string)
    return string


def attempt4(string):
    return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")


for attempt in [attempt1, attempt2, attempt3, attempt4]:
    print(attempt("murcielago"))

附注:在使用" ?.!/;:"的例子中使用元音…是的，“murcielago”在西班牙语里是蝙蝠的意思…有趣的单词，因为它包含了所有的元音:)

PS2:如果你对性能感兴趣，你可以用一个简单的代码来衡量这些尝试:

import timeit


K = 1000000
for i in range(1,5):
    t = timeit.Timer(
        f"attempt{i}('murcielago')",
        setup=f"from __main__ import attempt{i}"
    ).repeat(1, K)
    print(f"attempt{i}",min(t))

在我的盒子里，你会得到:

attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465

因此，对于这个特定的输入，尝试4似乎是最快的。

>>> line = "abc#@!?efg12;:?"
>>> ''.join( c for c in line if  c not in '?:!/;' )
'abc#@efg12'

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

试试这个:

def rm_char(original_str, need2rm):
    ''' Remove charecters in "need2rm" from "original_str" '''
    return original_str.translate(str.maketrans('','',need2rm))

这个方法在python3中很有效