从Python中删除字符串标点符号的最佳方法

Best way to strip punctuation from a string in Python

似乎有一个比以下更简单的方法:

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

有?


从效率的角度看,你不会打败

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s.translate(None, string.punctuation)

对于较高版本的python,请使用以下代码:

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

它使用一个查找表在C中执行原始字符串操作——除了编写自己的C代码,没有什么比这更好的了。

如果速度不是问题,另一个选择是:

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

这比s.replace替换为每个字符要快,但不能像下面计时中看到的那样执行非纯Python方法,如regexes或string.translate。对于这种类型的问题,在尽可能低的水平上做它会有回报。

定时代码:

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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)

结果如下:

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


正则表达式足够简单,如果你知道的话。

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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)所有非[字母数字字符(w)和空格(s)]。因此。还有?通过regex运行s变量后,变量"s"中将不存在标点符号。


为了方便使用,我总结了在python 2和python 3中从字符串中去掉标点符号的注意事项。详细描述请参考其他答案。

Python 2

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import string

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

Python 3

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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


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myString.translate(None, string.punctuation)


我通常用这样的东西:

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


string.punctuation只是ASCII码!一种更正确(但也更慢)的方法是使用unicodedata模块:

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# -*- 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


不一定简单,但如果你更熟悉这个家庭的话,那就另当别论了。

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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。

删除(一些?)标点符号,使用:

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import string

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

dict.fromkeys()类方法使得创建映射变得简单,根据键的顺序将所有值设置为None

要删除所有标点符号,而不仅仅是ASCII标点符号,您的表需要大一点;请参见J.F.Sebastian的答案(python 3版本):

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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漏掉了现实世界中常用的大量标点符号。一个适用于非ASCII标点的解决方案如何?

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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()

就我个人而言,我认为这是从python字符串中删除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,如果要删除标点符号,可以删除\{S},但保留类似$的符号。
  • 您可以对要保留的内容和要删除的内容进行具体说明,例如,\{Pd}只删除破折号。
  • 此regex还规范化空白。它将标签、回车和其他奇怪的东西映射到漂亮的单个空间。

这使用了Unicode字符属性,您可以在维基百科上了解更多信息。


下面是一个针对python 3.5的一行程序:

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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}))

这可能不是最好的解决方案,但我就是这样做的。

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

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:

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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

我还没看到这个答案。只需使用regex;它会删除除单词字符(\w和数字字符(\d外的所有字符,后跟空白字符(\s):

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


正如更新一样,我在python3中重写了@brian示例,并对其进行了更改,以将regex编译步骤移到函数内部。我在这里想的是时间的每一个步骤需要使功能工作。也许您使用的是分布式计算,并且不能在您的工作人员之间共享regex对象,并且需要在每个工作人员处执行re.compile步骤。另外,我还想知道针对python3的两种不同的maketrans实现的时间。

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

VS

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

另外,我还添加了另一个方法来使用set,在这里我利用交集函数来减少迭代次数。

这是完整的代码:

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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:

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


这里有一个没有regex的解决方案。

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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
  • 将单词之间的多个空格替换为单个空格
  • 删除尾随空格(如果有)条()

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


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


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

在不太严格的情况下,使用一行程序可能会有所帮助:

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

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#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


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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='
'
)
    print ("____________")

使用regex函数进行搜索和替换,如下所示。如果必须重复执行该操作,则可以保留一个已编译的regex模式(标点符号)副本,这将加快速度。


使用python从文本文件中删除停止字

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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='
'
)

这是如何把我们的文件改成大写的或小写字母。

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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())

我喜欢使用这样的函数:

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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