Question

似乎应该有一个比以下更简单的方法：

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

有吗？

Answer 1

从效率的角度来看，你不会打败

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))

它正在使用查找表在C中执行原始字符串操作 - 除了编写自己的C代码之外，没有什么能比这更好。

如果速度不是担心，另一个选择是：

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

这比使用每个char的s.replace更快，但是不能像非纯python方法那样运行，例如正则表达式或string.translate，正如您可以从下面的时间中看到的那样。对于这种类型的问题，尽可能低的水平做到这一点是值得的。

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)

这给出了以下结果：

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

Answer 2

正则表达式很简单，如果你知道的话。

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

在上面的代码中，我们用空字符串替换（re.sub）所有NON [字母数字字符（\ w）和空格（\ s）]。
因此。和？通过正则表达式运行s变量后，变量's'中不会出现标点符号。

Answer 3

为了方便使用，我总结了Python 2和Python 3中字符串条带化标点符号的注释。请参阅其他答案以获取详细说明。

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({key: None for key in string.punctuation})
new_s = s.translate(table)                          # Output: string without punctuation

Answer 4

myString.translate(None, string.punctuation)

Answer 5

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'

Answer 6

string.punctuation仅为ASCII ！更正确（但也更慢）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

Answer 7

如果你对这个家庭更熟悉的话，不一定更简单，但不一样。

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

Answer 8

对于Python 3 str或Python 2 unicode值，str.translate()只接受字典;在该映射中查找代码点（整数），并删除映射到None的任何内容。

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)

dict.fromkeys() class method使创建映射变得微不足道，根据键序列将所有值设置为None。

要删除所有标点符号，而不仅仅是ASCII标点符号，您的表格需要更大一些;见J.F. Sebastian's answer（Python 3版）：

import unicodedata
import sys

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

Answer 9

string.punctuation错过了现实世界中常用的标点符号。如何使用适用于非ASCII标点符号的解决方案？

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()

就个人而言，我认为这是从Python中删除字符串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，例如如果您要删除标点符号，可以删除\{S}，但请保留$之类的符号。
您可以真正了解要保留的内容以及要删除的内容，例如\{Pd}只会删除破折号。
这个正则表达式也规范了空白。它将标签，回车和其他奇怪的地方映射到漂亮的单一空间。

这使用Unicode字符属性，you can read more about on Wikipedia。

Answer 10

这可能不是最佳解决方案，但这就是我的方法。

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

Answer 11

这是我写的一个函数。这不是很有效，但它很简单，您可以添加或删除任何您想要的标点符号：

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

Answer 12

这是Python 3.5的单行代码：

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}))

Answer 13

我还没有看到这个答案。只需使用正则表达式;它会删除除字符（\w）和数字字符（\d）之外的所有字符，后跟空格字符（\s）：

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

Answer 14

就像更新一样，我在Python 3中重写了@Brian示例并对其进行了更改以在函数内部移动正则表达式编译步骤。我的想法是为了使功能发挥作用所需的每一步。也许您正在使用分布式计算，并且不能在您的工作人员之间共享正则表达式对象，并且需要在每个工作人员处执行re.compile步骤。另外，我很想为Python 3的两个不同的maketrans实现计时

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

vs

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

另外我添加了另一种使用set的方法，其中我利用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

Answer 15

这是一个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
删除尾随空格（如果有）条（）

Answer 16

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

Answer 17

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


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

Answer 18

在非常严格的情况下，单行可能会有所帮助：

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

Answer 19

尝试一个：）

regex.sub(r'\p{P}','', s)

Answer 20

这是使用RegEx进行操作的另一种简便方法

$joined_path2=Join-Path -LiteralPath '\\?\UNC\' -ChildPath $joined_path

Answer 21

我一直在寻找一个非常简单的解决方案。这是我得到的：

import re 

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

print(s)
'string  With  Punctuation '

Answer 22

使用正则表达式函数进行搜索和替换，如here.所示。如果你必须重复执行操作，你可以保留正则表达式模式（你的标点符号）的编译副本，这会加快速度。

Answer 23

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 ("____________")

Answer 24

#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

Answer 25

为什么你们都不使用这个？

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

太慢了吗？

Answer 26

考虑unicode。代码已在python3中检查。

df = (df.rename(columns = lambda x: x.replace('_x', ''))
        .fillna(df.filter(regex='_y$')
                  .rename(columns = lambda x: x.replace('_y', '')))
        .filter(regex=r'.*(?<!_y)$'))
print (df)
    A    B    C
0  R1  0.0  6.0
1  R2  4.0  8.0
2  R3  2.0  2.0

Answer 27

您也可以这样做：

import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())

Answer 28

这是如何将我们的文档更改为大写或小写。

print('@@@@This is lower case@@@@')

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

    str1=myFile.read()
    str1.lower()
print(str1.lower())

print('*****This is upper case****')

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

    str1=myFile.read()

    str1.upper()

print(str1.upper())

Answer 29

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')

Answer 30

def punctuationRemover(p):
'''
    Input: Takes a string. You may have to use str() to force it. 
    Removes all punctuation.
    Output: Returns a string.
'''
    punctuations = '''!()-[]{};:'"\,<>./?@#$%^&*_~'''
    no_punctuations = ''

    for words in p: # You may not have to loop this high
        for char in p:
            if char not in punctuations:
                no_punctuations = no_punctuations + char

return(no_punctuations)

Answer 31

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

从Python中删除字符串标点符号的最佳方法

31 个答案: