从字符串中去除标点符号的最佳方法

Question

似乎应该有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

不一定更简单，但如果你对 re 家族更熟悉的话，可以换一种方式。

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

Answer 2

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 3

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))

它使用查找表在 C 中执行原始字符串操作 - 除了编写您自己的 C 代码之外，没有什么比它更好的了。

如果不担心速度，另一种选择是：

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

这比使用每个字符的 s.replace 更快，但性能不如正则表达式或 string.translate 等非纯 python 方法，正如您从下面的时序中看到的那样。对于这种类型的问题，尽可能地降低水平是有回报的。

计时码：

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 4

myString.translate(None, string.punctuation)

Answer 5

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])

Answer 6

string.punctuation 是 ASCII 只有！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

您也可以概括和剥离其他类型的字符：

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

它还会删除像 ~*+§$ 这样的字符，这可能是也可能不是“标点符号”，这取决于一个人的观点。

Answer 7

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

Answer 8

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)

Answer 9

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），并删除任何映射到 None 的内容。

要删除（一些？）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 使创建映射变得简单，根据键的顺序将所有值设置为 None。

要删除 all 标点，而不仅仅是 ASCII 标点，您的表格需要大一点；见J.F. Sebastian's answer（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'))

Answer 10

这是我写的一个函数。它不是很有效，但很简单，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 11

在不太严格的情况下，单行可能会有所帮助：

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

Answer 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

为了使用方便，我总结了Python 2和Python 3中从字符串中剥离标点符号的注意事项，详细说明请参考其他答案。

---

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

Answer 14

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

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

Answer 15

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


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

Answer 16

string.punctuation 遗漏了大量现实世界中常用的标点符号。一个适用于非 ASCII 标点符号的解决方案怎么样？

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()

就个人而言，我认为这是在 Python 中从字符串中删除标点符号的最佳方法，因为：

• 它会删除所有 Unicode 标点符号
• 它很容易修改，例如如果要删除标点符号，可以删除 \{S}，但保留 $ 之类的符号。
• 您可以非常具体地确定要保留的内容和要删除的内容，例如 \{Pd} 只会删除破折号。
• 此正则表达式还标准化空白。它将制表符、回车和其他奇怪的东西映射到漂亮的单个空格。

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

Answer 17

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
• 用一个空格替换单词之间的多个空格
• 删除尾随空格，如果有的话 剥离（）

Answer 18

# 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 19

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 20

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 21

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

Answer 22

作为更新，我在 Python 3 中重写了 @Brian 示例并对其进行了更改以将正则表达式编译步骤移至函数内部。我的想法是为使功能正常工作所需的每一步都计时。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都有re.compile 步骤。另外，我很好奇为 Python 3 计时两个不同的 maketrans 实现

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

对比

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

另外我添加了另一种使用集合的方法，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 23

为什么你们都不用这个？

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

太慢了？

Answer 24

考虑 unicode。在 python3 中检查代码。

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

Answer 25

这是使用 RegEx 的另一种简单方法

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'

Answer 26

试试那个:)

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

Answer 27

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 28

你也可以这样做：

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

Answer 29

这个问题没有太多细节，所以我采取的方法是想出一个对问题最简单解释的解决方案：去掉标点符号。

请注意，提出的解决方案不考虑缩略词（例如，you're）或连字符（例如，anal-retentive）...关于它们是否应该被视为标点符号的争论...也不考虑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，即technically correct...但对我来说，就手头的问题而言，它是零意义的。

# 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)

Answer 30

显然我无法对所选答案进行编辑，所以这里有一个适用于 Python 3 的更新。translate 方法仍然是进行重要转换时最有效的选择。

将最初的繁重工作归功于上面的@Brian。感谢@ddejohn 对改进原始测试的出色建议。

#!/usr/bin/env python3

"""Determination of most efficient way to remove punctuation in Python 3.

Results in Python 3.8.10 on my system using the default arguments:

set       : 51.897
regex     : 17.901
translate :  2.059
replace   : 13.209
"""

import argparse
import re
import string
import timeit

parser = argparse.ArgumentParser()
parser.add_argument("--filename", "-f", default=argparse.__file__)
parser.add_argument("--iterations", "-i", type=int, default=10000)
opts = parser.parse_args()
with open(opts.filename) as fp:
    s = fp.read()
exclude = set(string.punctuation)
table = str.maketrans("", "", string.punctuation)
regex = re.compile(f"[{re.escape(string.punctuation)}]")

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

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

def test_translate(s):
    return s.translate(table)

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

opts = dict(globals=globals(), number=opts.iterations)
solutions = "set", "regex", "translate", "replace"
for solution in solutions:
    elapsed = timeit.timeit(f"test_{solution}(s)", **opts)
    print(f"{solution:<10}: {elapsed:6.3f}")