似乎应该有一种比以下更简单的方法:

import string
s = "string. With. Punctuation?" # Sample string 
out = s.translate(string.maketrans("",""), string.punctuation)

有?


当前回答

这可能不是最好的解决方案,但我就是这样做的。

import string
f = lambda x: ''.join([i for i in x if i not in string.punctuation])

其他回答

这可能不是最好的解决方案,但我就是这样做的。

import string
f = lambda x: ''.join([i for i in x if i not in string.punctuation])

在不太严格的情况下,单行线可能会有所帮助:

''.join([c for c in s if c.isalnum() or c.isspace()])
import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(r'[^a-zA-Z0-9\s]', '', s)

我喜欢使用这样的函数:

def scrub(abc):
    while abc[-1] is in list(string.punctuation):
        abc=abc[:-1]
    while abc[0] is in list(string.punctuation):
        abc=abc[1:]
    return abc
# FIRST METHOD
# Storing all punctuations in a variable    
punctuation='!?,.:;"\')(_-'
newstring ='' # Creating empty string
word = raw_input("Enter string: ")
for i in word:
     if(i not in punctuation):
                  newstring += i
print ("The string without punctuation is", newstring)

# SECOND METHOD
word = raw_input("Enter string: ")
punctuation = '!?,.:;"\')(_-'
newstring = word.translate(None, punctuation)
print ("The string without punctuation is",newstring)


# Output for both methods
Enter string: hello! welcome -to_python(programming.language)??,
The string without punctuation is: hello welcome topythonprogramminglanguage