如何使用NLTK标记器消除标点符号?

问题:如何使用NLTK标记器消除标点符号?

我刚刚开始使用NLTK,但我不太了解如何从文本中获取单词列表。如果使用nltk.word_tokenize(),则会得到单词和标点的列表。我只需要这些词。我如何摆脱标点符号?同样word_tokenize不适用于多个句子:点号会添加到最后一个单词中。

I’m just starting to use NLTK and I don’t quite understand how to get a list of words from text. If I use nltk.word_tokenize(), I get a list of words and punctuation. I need only the words instead. How can I get rid of punctuation? Also word_tokenize doesn’t work with multiple sentences: dots are added to the last word.


回答 0

看看nltk 在此处提供的其他标记化选项。例如,您可以定义一个令牌生成器,该令牌生成器将字母数字字符序列选作令牌,并丢弃其他所有内容:

from nltk.tokenize import RegexpTokenizer

tokenizer = RegexpTokenizer(r'\w+')
tokenizer.tokenize('Eighty-seven miles to go, yet.  Onward!')

输出:

['Eighty', 'seven', 'miles', 'to', 'go', 'yet', 'Onward']

Take a look at the other tokenizing options that nltk provides here. For example, you can define a tokenizer that picks out sequences of alphanumeric characters as tokens and drops everything else:

from nltk.tokenize import RegexpTokenizer

tokenizer = RegexpTokenizer(r'\w+')
tokenizer.tokenize('Eighty-seven miles to go, yet.  Onward!')

Output:

['Eighty', 'seven', 'miles', 'to', 'go', 'yet', 'Onward']

回答 1

您实际上并不需要NLTK来删除标点符号。您可以使用简单的python将其删除。对于字符串:

import string
s = '... some string with punctuation ...'
s = s.translate(None, string.punctuation)

或对于unicode:

import string
translate_table = dict((ord(char), None) for char in string.punctuation)   
s.translate(translate_table)

然后在令牌生成器中使用此字符串。

PS字符串模块还有一些其他可以删除的元素集(例如数字)。

You do not really need NLTK to remove punctuation. You can remove it with simple python. For strings:

import string
s = '... some string with punctuation ...'
s = s.translate(None, string.punctuation)

Or for unicode:

import string
translate_table = dict((ord(char), None) for char in string.punctuation)   
s.translate(translate_table)

and then use this string in your tokenizer.

P.S. string module have some other sets of elements that can be removed (like digits).


回答 2

下面的代码将删除所有标点符号以及非字母字符。从他们的书中复制。

http://www.nltk.org/book/ch01.html

import nltk

s = "I can't do this now, because I'm so tired.  Please give me some time. @ sd  4 232"

words = nltk.word_tokenize(s)

words=[word.lower() for word in words if word.isalpha()]

print(words)

输出

['i', 'ca', 'do', 'this', 'now', 'because', 'i', 'so', 'tired', 'please', 'give', 'me', 'some', 'time', 'sd']

Below code will remove all punctuation marks as well as non alphabetic characters. Copied from their book.

http://www.nltk.org/book/ch01.html

import nltk

s = "I can't do this now, because I'm so tired.  Please give me some time. @ sd  4 232"

words = nltk.word_tokenize(s)

words=[word.lower() for word in words if word.isalpha()]

print(words)

output

['i', 'ca', 'do', 'this', 'now', 'because', 'i', 'so', 'tired', 'please', 'give', 'me', 'some', 'time', 'sd']

回答 3

正如注释中所注意到的那样,因为word_tokenize()仅适用于单个句子,所以它以send_tokenize()开头。您可以使用filter()过滤出标点符号。如果您有一个unicode字符串,请确保它是一个unicode对象(而不是使用“ utf-8”之类的编码编码的“ str”)。

from nltk.tokenize import word_tokenize, sent_tokenize

text = '''It is a blue, small, and extraordinary ball. Like no other'''
tokens = [word for sent in sent_tokenize(text) for word in word_tokenize(sent)]
print filter(lambda word: word not in ',-', tokens)

As noticed in comments start with sent_tokenize(), because word_tokenize() works only on a single sentence. You can filter out punctuation with filter(). And if you have an unicode strings make sure that is a unicode object (not a ‘str’ encoded with some encoding like ‘utf-8’).

from nltk.tokenize import word_tokenize, sent_tokenize

text = '''It is a blue, small, and extraordinary ball. Like no other'''
tokens = [word for sent in sent_tokenize(text) for word in word_tokenize(sent)]
print filter(lambda word: word not in ',-', tokens)

回答 4

我只使用了以下代码,删除了所有标点符号:

tokens = nltk.wordpunct_tokenize(raw)

type(tokens)

text = nltk.Text(tokens)

type(text)  

words = [w.lower() for w in text if w.isalpha()]

I just used the following code, which removed all the punctuation:

tokens = nltk.wordpunct_tokenize(raw)

type(tokens)

text = nltk.Text(tokens)

type(text)  

words = [w.lower() for w in text if w.isalpha()]

回答 5

我认为您需要某种正则表达式匹配(以下代码在Python 3中):

import string
import re
import nltk

s = "I can't do this now, because I'm so tired.  Please give me some time."
l = nltk.word_tokenize(s)
ll = [x for x in l if not re.fullmatch('[' + string.punctuation + ']+', x)]
print(l)
print(ll)

输出:

['I', 'ca', "n't", 'do', 'this', 'now', ',', 'because', 'I', "'m", 'so', 'tired', '.', 'Please', 'give', 'me', 'some', 'time', '.']
['I', 'ca', "n't", 'do', 'this', 'now', 'because', 'I', "'m", 'so', 'tired', 'Please', 'give', 'me', 'some', 'time']

在大多数情况下应该可以正常使用,因为它可以删除标点符号,同时保留“ n’t”之类的令牌,而这些令牌不能从regex令牌生成器(如)获得wordpunct_tokenize

I think you need some sort of regular expression matching (the following code is in Python 3):

import string
import re
import nltk

s = "I can't do this now, because I'm so tired.  Please give me some time."
l = nltk.word_tokenize(s)
ll = [x for x in l if not re.fullmatch('[' + string.punctuation + ']+', x)]
print(l)
print(ll)

Output:

['I', 'ca', "n't", 'do', 'this', 'now', ',', 'because', 'I', "'m", 'so', 'tired', '.', 'Please', 'give', 'me', 'some', 'time', '.']
['I', 'ca', "n't", 'do', 'this', 'now', 'because', 'I', "'m", 'so', 'tired', 'Please', 'give', 'me', 'some', 'time']

Should work well in most cases since it removes punctuation while preserving tokens like “n’t”, which can’t be obtained from regex tokenizers such as wordpunct_tokenize.


回答 6

真诚的问,这是什么字?如果您假设一个单词仅由字母字符组成,那您就错了,因为如果在标记化之前删除标点符号,can't诸如的单词将被破坏成碎片(例如cant),这很可能会对程序产生负面影响。

因此,解决方案是先标记化然后删除标点标记

import string

from nltk.tokenize import word_tokenize

tokens = word_tokenize("I'm a southern salesman.")
# ['I', "'m", 'a', 'southern', 'salesman', '.']

tokens = list(filter(lambda token: token not in string.punctuation, tokens))
# ['I', "'m", 'a', 'southern', 'salesman']

……然后,如果你愿意,你可以替换某些标记,如'mam

Sincerely asking, what is a word? If your assumption is that a word consists of alphabetic characters only, you are wrong since words such as can't will be destroyed into pieces (such as can and t) if you remove punctuation before tokenisation, which is very likely to affect your program negatively.

Hence the solution is to tokenise and then remove punctuation tokens.

import string

from nltk.tokenize import word_tokenize

tokens = word_tokenize("I'm a southern salesman.")
# ['I', "'m", 'a', 'southern', 'salesman', '.']

tokens = list(filter(lambda token: token not in string.punctuation, tokens))
# ['I', "'m", 'a', 'southern', 'salesman']

…and then if you wish, you can replace certain tokens such as 'm with am.


回答 7

我使用以下代码删除标点符号:

import nltk
def getTerms(sentences):
    tokens = nltk.word_tokenize(sentences)
    words = [w.lower() for w in tokens if w.isalnum()]
    print tokens
    print words

getTerms("hh, hh3h. wo shi 2 4 A . fdffdf. A&&B ")

而且,如果您想检查令牌是否为有效的英语单词,则可能需要PyEnchant

教程:

 import enchant
 d = enchant.Dict("en_US")
 d.check("Hello")
 d.check("Helo")
 d.suggest("Helo")

I use this code to remove punctuation:

import nltk
def getTerms(sentences):
    tokens = nltk.word_tokenize(sentences)
    words = [w.lower() for w in tokens if w.isalnum()]
    print tokens
    print words

getTerms("hh, hh3h. wo shi 2 4 A . fdffdf. A&&B ")

And If you want to check whether a token is a valid English word or not, you may need PyEnchant

Tutorial:

 import enchant
 d = enchant.Dict("en_US")
 d.check("Hello")
 d.check("Helo")
 d.suggest("Helo")

回答 8

删除标点符号(它将删除和标点符号处理的一部分,使用下面的代码)

        tbl = dict.fromkeys(i for i in range(sys.maxunicode) if unicodedata.category(chr(i)).startswith('P'))
        text_string = text_string.translate(tbl) #text_string don't have punctuation
        w = word_tokenize(text_string)  #now tokenize the string 

样本输入/输出:

direct flat in oberoi esquire. 3 bhk 2195 saleable 1330 carpet. rate of 14500 final plus 1% floor rise. tax approx 9% only. flat cost with parking 3.89 cr plus taxes plus possession charger. middle floor. north door. arey and oberoi woods facing. 53% paymemt due. 1% transfer charge with buyer. total cost around 4.20 cr approx plus possession charges. rahul soni

['direct', 'flat', 'oberoi', 'esquire', '3', 'bhk', '2195', 'saleable', '1330', 'carpet', 'rate', '14500', 'final', 'plus', '1', 'floor', 'rise', 'tax', 'approx', '9', 'flat', 'cost', 'parking', '389', 'cr', 'plus', 'taxes', 'plus', 'possession', 'charger', 'middle', 'floor', 'north', 'door', 'arey', 'oberoi', 'woods', 'facing', '53', 'paymemt', 'due', '1', 'transfer', 'charge', 'buyer', 'total', 'cost', 'around', '420', 'cr', 'approx', 'plus', 'possession', 'charges', 'rahul', 'soni']

Remove punctuaion(It will remove . as well as part of punctuation handling using below code)

        tbl = dict.fromkeys(i for i in range(sys.maxunicode) if unicodedata.category(chr(i)).startswith('P'))
        text_string = text_string.translate(tbl) #text_string don't have punctuation
        w = word_tokenize(text_string)  #now tokenize the string 

Sample Input/Output:

direct flat in oberoi esquire. 3 bhk 2195 saleable 1330 carpet. rate of 14500 final plus 1% floor rise. tax approx 9% only. flat cost with parking 3.89 cr plus taxes plus possession charger. middle floor. north door. arey and oberoi woods facing. 53% paymemt due. 1% transfer charge with buyer. total cost around 4.20 cr approx plus possession charges. rahul soni

['direct', 'flat', 'oberoi', 'esquire', '3', 'bhk', '2195', 'saleable', '1330', 'carpet', 'rate', '14500', 'final', 'plus', '1', 'floor', 'rise', 'tax', 'approx', '9', 'flat', 'cost', 'parking', '389', 'cr', 'plus', 'taxes', 'plus', 'possession', 'charger', 'middle', 'floor', 'north', 'door', 'arey', 'oberoi', 'woods', 'facing', '53', 'paymemt', 'due', '1', 'transfer', 'charge', 'buyer', 'total', 'cost', 'around', '420', 'cr', 'approx', 'plus', 'possession', 'charges', 'rahul', 'soni']


回答 9

只需添加@rmalouf的解决方案,就不会包含任何数字,因为\ w +等效于[a-zA-Z0-9_]

from nltk.tokenize import RegexpTokenizer
tokenizer = RegexpTokenizer(r'[a-zA-Z]')
tokenizer.tokenize('Eighty-seven miles to go, yet.  Onward!')

Just adding to the solution by @rmalouf, this will not include any numbers because \w+ is equivalent to [a-zA-Z0-9_]

from nltk.tokenize import RegexpTokenizer
tokenizer = RegexpTokenizer(r'[a-zA-Z]')
tokenizer.tokenize('Eighty-seven miles to go, yet.  Onward!')

回答 10

您可以在没有nltk(python 3.x)的情况下一行完成此操作。

import string
string_text= string_text.translate(str.maketrans('','',string.punctuation))

You can do it in one line without nltk (python 3.x).

import string
string_text= string_text.translate(str.maketrans('','',string.punctuation))