从Python中删除字符串标点符号的最佳方法

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)

在那儿？

Answer 1

从效率的角度来看,你不会打败

s.translate(None, string.punctuation)

它使用查找表在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:

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

这比使用每个char的s.replace更快,但是不能像非纯python方法那样执行,例如regexes或string.translate,正如您可以从下面的时间看到的那样.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:

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

这给出了以下结果:

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)

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(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

我经常使用这样的东西:

>>> 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是ASCII 只!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

如果你对家庭更熟悉,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(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)

所述dict.fromkeys()类方法使得它琐碎创建映射,所有的值设置为None基于密钥的序列.

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; 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:

• 它删除所有Unicode标点符号
• 它很容易修改,例如你可以删除\{S}如果你想删除标点,但保持符号$.
• 您可以非常具体地了解要保留的内容以及要删除的内容,例如,\{Pd}只删除短划线.
• 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个空间.

这使用Unicode字符属性,您可以在维基百科上阅读更多信息.

Answer 10

这是Python 3.5的单行程序:

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 11

我还没有看到这个答案.只需使用正则表达式; 它除了单词字符(\w)和数字字符(\d)之外的所有字符,后跟一个空白字符(\s):

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

Answer 12

这可能不是最好的解决方案,但这就是我做到的.

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 14

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 15

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


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

Answer 16

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
• 使用 strip() 删除尾随空格（如果有）

Answer 17

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

Answer 18

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

Answer 19

为什么你们没人用这个？

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

太慢了？

Answer 20

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 '