>>> import re
>>> string = "Kl13@£$%[};'\""
>>> pattern = re.compile('\W')
>>> string = re.sub(pattern, '', string)
>>> print string
Kl13

如果我理解正确，最简单的方法是使用正则表达式，因为它为您提供了很大的灵活性，但另一个简单的方法是使用循环以下是示例代码，我还计算了单词的出现并存储在字典中。

s = """An... essay is, generally, a piece of writing that gives the author's own 
argument — but the definition is vague, 
overlapping with those of a paper, an article, a pamphlet, and a short story. Essays 
have traditionally been 
sub-classified as formal and informal. Formal essays are characterized by "serious 
purpose, dignity, logical 
organization, length," whereas the informal essay is characterized by "the personal 
element (self-revelation, 
individual tastes and experiences, confidential manner), humor, graceful style, 
rambling structure, unconventionality 
or novelty of theme," etc.[1]"""

d = {}      # creating empty dic      
words = s.split() # spliting string and stroing in list
for word in words:
    new_word = ''
    for c in word:
        if c.isalnum(): # checking if indiviual chr is alphanumeric or not
            new_word = new_word + c
    print(new_word, end=' ')
    # if new_word not in d:
    #     d[new_word] = 1
    # else:
    #     d[new_word] = d[new_word] +1
print(d)

如果这个答案是有用的，请评价这个!

作为这里其他一些答案的衍生，我提供了一种非常简单而灵活的方法来定义您想要限制字符串内容的一组字符。在这种情况下，我允许字母数字加破折号和下划线。只需添加或删除字符从我的PERMITTED_CHARS适合您的用例。

PERMITTED_CHARS = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ_-" 
someString = "".join(c for c in someString if c in PERMITTED_CHARS)

Python 3

使用与@John Machin的答案相同的方法，但针对Python 3进行了更新:

更大的字符集 对翻译工作方式的轻微改变。

现在假定Python代码是用UTF-8编码的 (来源:PEP 3120)

这意味着包含你想要删除的所有字符的字符串会变得更大:

    
del_chars = ''.join(c for c in map(chr, range(1114111)) if not c.isalnum())
    

翻译方法现在需要使用一个翻译表，我们可以用maketrans()创建:

    
del_map = str.maketrans('', '', del_chars)
    

现在，像以前一样，任何你想要“捏碎”的字符串:

    
scrunched = s.translate(del_map)
    

使用来自@Joe Machin的最后一个计时例子，我们可以看到它仍然比re强一个数量级:

    
> python -mtimeit -s"d=''.join(c for c in map(chr,range(1114111)) if not c.isalnum());m=str.maketrans('','',d);s='foo-'*25" "s.translate(m)"
    
1000000 loops, best of 5: 255 nsec per loop
    
> python -mtimeit -s"import re;s='foo-'*25;r=re.compile(r'[\W_]+')" "r.sub('',s)"
    
50000 loops, best of 5: 4.8 usec per loop
    

>>> import re
>>> string = "Kl13@£$%[};'\""
>>> pattern = re.compile('\W')
>>> string = re.sub(pattern, '', string)
>>> print string
Kl13

如何:

def ExtractAlphanumeric(InputString):
    from string import ascii_letters, digits
    return "".join([ch for ch in InputString if ch in (ascii_letters + digits)])

如果InputString中的字符存在于ascii_letters和digits的组合字符串中，则可以使用列表推导来生成InputString中的字符列表。然后它将列表连接在一起，形成一个字符串。