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.
问题:
回答1:
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2:
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3:
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:
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5:
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6:
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7:
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:
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']