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?
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
这里有一个使用RegEx的简单方法
import re
punct = re.compile(r'(\w+)')
sentence = 'This ! is : a # sample $ sentence.' # Text with punctuation
tokenized = [m.group() for m in punct.finditer(sentence)]
sentence = ' '.join(tokenized)
print(sentence)
'This is a sample sentence'
其他回答
这里有一个使用RegEx的简单方法
import re
punct = re.compile(r'(\w+)')
sentence = 'This ! is : a # sample $ sentence.' # Text with punctuation
tokenized = [m.group() for m in punct.finditer(sentence)]
sentence = ' '.join(tokenized)
print(sentence)
'This is a sample sentence'
字符串标点符号仅为ASCII!一种更正确(但也慢得多)的方法是使用unicodedata模块:
# -*- coding: utf-8 -*-
from unicodedata import category
s = u'String — with - «punctation »...'
s = ''.join(ch for ch in s if category(ch)[0] != 'P')
print 'stripped', s
您还可以概括和剥离其他类型的字符:
''.join(ch for ch in s if category(ch)[0] not in 'SP')
它还将删除~*+§$等字符,这些字符可能是“标点符号”,也可能不是“标点符号。
# 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
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()])
这个问题没有太多细节,所以我采取的方法是用最简单的解释来解决这个问题:去掉标点符号。
请注意,所提供的解决方案不考虑收缩词(例如,you are)或连字符词(例如肛门保留)。。。这是关于它们是否应该被视为标点符号的争论。。。也不能解释非英语字符集或类似的东西。。。因为问题中没有提到这些细节。有人认为空格是标点符号,这在技术上是正确的。。。但对我来说,这在当前问题的背景下毫无意义。
# using lambda
''.join(filter(lambda c: c not in string.punctuation, s))
# using list comprehension
''.join('' if c in string.punctuation else c for c in s)