使用Python从文本文件中删除停止词

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')

作为更新，我重写了Python 3中的@Brian示例，并对其进行了更改，以将正则表达式编译步骤移到函数内部。我在这里的想法是对使功能工作所需的每一步进行计时。也许您使用的是分布式计算，无法在工作人员之间共享regex对象，需要在每个工作人员处执行re.compile步骤。此外，我还很好奇地对Python 3的maketrans的两种不同实现进行计时

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集合的方法，在这里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

从效率的角度来看，你不会击败

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语言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的了。

如果速度不令人担忧，另一个选择是：

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

这比用每个字符替换s.replace更快，但不会像正则表达式或字符串转换等非纯python方法那样执行得好，正如您从下面的计时中看到的那样。对于这种类型的问题，在尽可能低的水平上解决是有回报的。

计时代码：

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)

结果如下：

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

字符串标点符号仅为ASCII！一种更正确（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

您还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')

它还将删除~*+§$等字符，这些字符可能是“标点符号”，也可能不是“标点符号。

对于严肃的自然语言处理（NLP），您应该让像SpaCy这样的库通过标记化处理标点符号，然后您可以根据需要手动调整。

例如，您希望如何处理单词中的连字符？例外情况，如缩写？开始和结束引号？URL？在NLP中，将“let’s”这样的收缩分隔为“let”和“s”以进行进一步处理通常很有用。

# 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