从字符串中删除标点符号的最佳方法

似乎应该有一个比以下方法更简单的方法：

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

在那儿？

从效率的角度来看，您不会被击败

s.translate(None, string.punctuation)

对于更高版本的Python，请使用以下代码：

s.translate(str.maketrans('', '', string.punctuation))

它使用查找表在C语言中执行原始字符串操作-除了编写自己的C代码之外，没有什么比这更好的了。

如果不担心速度，那么另一个选择是：

exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)

这比每个char的s.replace更快，但效果不如regexes或string.translate等非纯python方法，如下面的时序所示。对于这种类型的问题，在尽可能低的水平上进行操作会有所回报。

时间码：

import re, string, timeit

s = "string. With. Punctuation"
exclude = set(string.punctuation)
table = string.maketrans("","")
regex = re.compile('[%s]' % re.escape(string.punctuation))

def test_set(s):
    return ''.join(ch for ch in s if ch not in exclude)

def test_re(s):  # From Vinko's solution, with fix.
    return regex.sub('', s)

def test_trans(s):
    return s.translate(table, string.punctuation)

def test_repl(s):  # From S.Lott's solution
    for c in string.punctuation:
        s=s.replace(c,"")
    return s

print "sets      :",timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000)
print "regex     :",timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000)
print "translate :",timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000)
print "replace   :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000)

得到以下结果：

sets      : 19.8566138744
regex     : 6.86155414581
translate : 2.12455511093
replace   : 28.4436721802

如果您知道正则表达式，就足够简单了。

import re
s = "string. With. Punctuation?"
s = re.sub(r'[^\w\s]','',s)

为了方便使用，我在Python 2和Python 3中总结了从字符串中删除标点符号的注意事项。有关详细说明，请参阅其他答案。

Python 2

import string

s = "string. With. Punctuation?"
table = string.maketrans("","")
new_s = s.translate(table, string.punctuation)      # Output: string without punctuation

Python 3

import string

s = "string. With. Punctuation?"
table = str.maketrans(dict.fromkeys(string.punctuation))  # OR {key: None for key in string.punctuation}
new_s = s.translate(table)                          # Output: string without punctuation

myString.translate(None, string.punctuation)

我通常使用这样的东西：

>>> s = "string. With. Punctuation?" # Sample string
>>> import string
>>> for c in string.punctuation:
...     s= s.replace(c,"")
...
>>> s
'string With Punctuation'

string.punctuation是ASCII 只！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

您也可以概括和去除其他类型的字符：

''.join(ch for ch in s if category(ch)[0] not in 'SP')

它还会~*+§$根据个人的视点去掉那些可能为“标点”或不为“标点”的字符。

如果您对re家族更加熟悉，则不一定会更简单，但会采用另一种方式。

import re, string
s = "string. With. Punctuation?" # Sample string 
out = re.sub('[%s]' % re.escape(string.punctuation), '', s)

对于Python 3 str或Python 2 unicode值，str.translate()只需要一个字典；在该映射中查找代码点（整数），并None删除所有映射到的代码点。

然后要删除（某些？）标点符号，请使用：

import string

remove_punct_map = dict.fromkeys(map(ord, string.punctuation))
s.translate(remove_punct_map)

使用dict.fromkeys()class方法可以轻松创建映射，并None根据键序列将所有值设置为。

要删除所有标点符号，而不仅仅是ASCII标点符号，您的表需要更大一些。参见JF Sebastian的答案（Python 3版本）：

import unicodedata
import sys

remove_punct_map = dict.fromkeys(i for i in range(sys.maxunicode)
                                 if unicodedata.category(chr(i)).startswith('P'))

string.punctuation错过了现实世界中常用的大量标点符号。一种适用于非ASCII标点的解决方案怎么样？

import regex
s = u"string. With. Some・Really Weird、Non？ASCII。 「（Punctuation）」?"
remove = regex.compile(ur'[\p{C}|\p{M}|\p{P}|\p{S}|\p{Z}]+', regex.UNICODE)
remove.sub(u" ", s).strip()

我个人认为这是从Python中的字符串中删除标点符号的最佳方法，因为：

• 删除所有Unicode标点符号
• 它很容易修改，例如，\{S}如果要删除标点符号，则可以将其删除，但要保留诸如$。
• 您可以真正确定要保留的内容和要删除的内容，例如\{Pd}仅删除破折号。
• 此正则表达式还规范了空格。它将制表符，回车符和其他奇数映射到漂亮的单个空格。

它使用Unicode字符属性，您可以在Wikipedia上了解更多信息。

我还没有看到这个答案。只需使用正则表达式即可；它会删除单词字符（\w）和数字字符（\d）之外的所有字符，然后删除空格字符（\s）：

import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(ur'[^\w\d\s]+', '', s)

这是Python 3.5的一线式：

import string
"l*ots! o(f. p@u)n[c}t]u[a'ti\"on#$^?/".translate(str.maketrans({a:None for a in string.punctuation}))

这可能不是最佳解决方案，但是这就是我的方法。

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

这是我编写的函数。它不是很有效，但是很简单，您可以添加或删除所需的标点符号：

def stripPunc(wordList):
    """Strips punctuation from list of words"""
    puncList = [".",";",":","!","?","/","\\",",","#","@","$","&",")","(","\""]
    for punc in puncList:
        for word in wordList:
            wordList=[word.replace(punc,'') for word in wordList]
    return wordList

import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(r'[^a-zA-Z0-9\s]', '', s)

作为更新，我重写了Python 3中的@Brian示例并对其进行了更改，以将regex编译步骤移至函数内部。我的想法是计时使该功能起作用所需的每个步骤。也许您使用的是分布式计算，并且您的工作人员之间无法共享正则表达式对象，因此需要re.compile在每个工作人员中走一步。另外，我很好奇地为Python 3的maketrans的两种不同实现计时了

table = str.maketrans({key: None for key in string.punctuation})

与

table = str.maketrans('', '', string.punctuation)

另外，我添加了另一种使用set的方法，其中利用了交集函数来减少迭代次数。

这是完整的代码：

import re, string, timeit

s = "string. With. Punctuation"


def test_set(s):
    exclude = set(string.punctuation)
    return ''.join(ch for ch in s if ch not in exclude)


def test_set2(s):
    _punctuation = set(string.punctuation)
    for punct in set(s).intersection(_punctuation):
        s = s.replace(punct, ' ')
    return ' '.join(s.split())


def test_re(s):  # From Vinko's solution, with fix.
    regex = re.compile('[%s]' % re.escape(string.punctuation))
    return regex.sub('', s)


def test_trans(s):
    table = str.maketrans({key: None for key in string.punctuation})
    return s.translate(table)


def test_trans2(s):
    table = str.maketrans('', '', string.punctuation)
    return(s.translate(table))


def test_repl(s):  # From S.Lott's solution
    for c in string.punctuation:
        s=s.replace(c,"")
    return s


print("sets      :",timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000))
print("sets2      :",timeit.Timer('f(s)', 'from __main__ import s,test_set2 as f').timeit(1000000))
print("regex     :",timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000))
print("translate :",timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000))
print("translate2 :",timeit.Timer('f(s)', 'from __main__ import s,test_trans2 as f').timeit(1000000))
print("replace   :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000))

这是我的结果：

sets      : 3.1830138750374317
sets2      : 2.189873124472797
regex     : 7.142953420989215
translate : 4.243278483860195
translate2 : 2.427158243022859
replace   : 4.579746678471565

>>> s = "string. With. Punctuation?"
>>> s = re.sub(r'[^\w\s]','',s)
>>> re.split(r'\s*', s)


['string', 'With', 'Punctuation']

这是没有正则表达式的解决方案。

import string

input_text = "!where??and!!or$$then:)"
punctuation_replacer = string.maketrans(string.punctuation, ' '*len(string.punctuation))    
print ' '.join(input_text.translate(punctuation_replacer).split()).strip()

Output>> where and or then

• 用空格替换标点符号
• 用单个空格替换单词之间的多个空格
• 如果有strip（），请删除尾随空格

在不太严格的情况下，单线可能会有所帮助：

''.join([c for c in s if c.isalnum() or c.isspace()])

#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

with open('one.txt','r')as myFile:

    str1=myFile.read()

    print(str1)


    punctuation = ['(', ')', '?', ':', ';', ',', '.', '!', '/', '"', "'"] 

for i in punctuation:

        str1 = str1.replace(i," ") 
        myList=[]
        myList.extend(str1.split(" "))
print (str1) 
for i in myList:

    print(i,end='\n')
    print ("____________")

为什么你们没人使用这个？

 ''.join(filter(str.isalnum, s)) 

太慢了？

考虑unicode。代码在python3中检查。

from unicodedata import category
text = 'hi, how are you?'
text_without_punc = ''.join(ch for ch in text if not category(ch).startswith('P'))

使用Python从文本文件中删除停用词

print('====THIS IS HOW TO REMOVE STOP WORS====')

with open('one.txt','r')as myFile:

    str1=myFile.read()

    stop_words ="not", "is", "it", "By","between","This","By","A","when","And","up","Then","was","by","It","If","can","an","he","This","or","And","a","i","it","am","at","on","in","of","to","is","so","too","my","the","and","but","are","very","here","even","from","them","then","than","this","that","though","be","But","these"

    myList=[]

    myList.extend(str1.split(" "))

    for i in myList:

        if i not in stop_words:

            print ("____________")

            print(i,end='\n')

我喜欢使用这样的功能：

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