python - 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)

有没有？

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))

它使用查找表在 C 中执行原始字符串操作 - 没有什么比这更好的了，但是编写您自己的 C 代码。

如果速度不是问题，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，从下面的时间可以看出。对于这种类型的问题，在尽可能低的水平上做是有回报的。

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 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)

Answer 3

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。

---

蟒蛇2

import string

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

---

蟒蛇 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 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

不一定更简单，而是另一种方式，如果你更熟悉 re 家族的话。

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

Answer 7

string.punctuation只是ASCII ！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~*+§$根据一个人的观点去除可能是也可能不是“标点符号”的字符。

Answer 8

对于 Python 3str或 Python 2的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删除映射到的任何内容。

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)

类dict.fromkeys()方法使创建映射变得很简单，None根据键的顺序设置所有值。

要删除所有标点符号，而不仅仅是 ASCII 标点符号，您的表格需要大一点；请参阅JF Sebastian 的回答（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 9

string.punctuation错过了现实世界中常用的大量标点符号。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，我相信这是从 Python 中的字符串中删除标点符号的最佳方法，因为：

• 它删除了所有 Unicode 标点符号
• 它很容易修改，例如，\{S}如果要删除标点符号，可以删除 ，但保留$.
• 您可以非常具体地了解要保留的内容和要删除的内容，例如\{Pd}只会删除破折号。
• 此正则表达式还规范化空格。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，您可以在 Wikipedia 上阅读更多信息。

Answer 10

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 11

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 12

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 13

这是我写的一个函数。它不是很有效，但很简单，您可以添加或删除您想要的任何标点符号：

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 14

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

Answer 15

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，并且无法在您的工作人员之间共享正则表达式对象，并且需要对re.compile每个工作人员采取措施。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 set 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 16

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()])

Answer 17

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


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

Answer 18

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
• 使用 strip() 删除尾随空格，如果有的话

Answer 19

# 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 20

为什么你们都不用这个？

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

太慢了？

Answer 21

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 22

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 23

试试那个:)

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

Answer 24

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 25

这个问题没有太多的细节，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......也没有考虑到非英文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，在手头的问题的背景下它是零意义的。

# 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 26

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在进行非平凡的转换时，该方法仍然是最有效的选择。

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}")

Answer 27

考虑unicode。在 python3 中检查的代码。

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 28

你也可以这样做：

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

Answer 29

当您处理 Unicode 字符串时，我建议使用PyPiregex模块，因为它支持 Unicode 属性类（如\p{X}/ \P{X}）和 POSIX 字符类（如[:name:]）。

只需在终端中键入pip install regex（或）并按 ENTER即可安装软件包。pip3 install regex

如果您需要删除任何类型的标点符号和符号（即除字母、数字和空格之外的任何内容），您可以使用

regex.sub(r'[\p{P}\p{S}]', '', text)  # to remove one by one
regex.sub(r'[\p{P}\p{S}]+', '', text) # to remove all consecutive punctuation/symbols with one go
regex.sub(r'[[:punct:]]+', '', text)  # Same with a POSIX character class

在线查看Python 演示：

import regex

text = 'भारत India <><>^$.,,! 002'
new_text = regex.sub(r'[\p{P}\p{S}\s]+', ' ', text).lower().strip()
# OR
# new_text = regex.sub(r'[[:punct:]\s]+', ' ', text).lower().strip()

print(new_text)
# => भारत india 002

在这里，我在字符类中添加了一个空格\s模式

Answer 30

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 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