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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.
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']
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).
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']
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)
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()]
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.
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.
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")
You can do it in one line without nltk (python 3.x).
import string
string_text= string_text.translate(str.maketrans('','',string.punctuation))
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!')
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']
what is the code to split a sentence into a list of its constituent words AND punctuation? Most text preprocessing programs tend to remove punctuations.
For example, if I enter this:
"Punctuations to be included as its own unit."
The desired output would be:
result = ['Punctuations', 'to', 'be', 'included', 'as', 'its', 'own',
'unit', '.']
many thanks!
You might want to consider using a Natural Language Toolkit or nltk.
Try this:
import nltk
sentence = "Punctuations to be included as its own unit."
tokens = nltk.word_tokenize(sentence)
print(tokens)
Output: ['Punctuations', 'to', 'be', 'included', 'as', 'its', 'own', 'unit', '.']
The following snippet can be used using regular expression to separate the words and punctuation in a list.
import string
import re
punctuations = string.punctuation
regularExpression="[\w]+|" + "[" + punctuations + "]"
content="Punctuations to be included as its own unit."
splittedWords_Puncs = re.findall(r""+regularExpression, content)
print(splittedWords_Puncs)
Output: ['Punctuations', 'to', 'be', 'included', 'as', 'its', 'own', 'unit', '.']
There is paragraph, and I want to use regular expression to extract all the words inside.
a bdag agasg it's the cookies for dogs',don't you think so? the word 'wow' in english means.you hey b 097 dag final
I have tried several regexes with re.findall(regX,str), and found one that can match most words.
regX = "[ ,\.\?]?([a-z]+'?[a-z]?)[ ,\.\?]?"
['a', 'bdag', 'agasg', "it's", 'the', 'cookies', 'for', "dogs'", "don't", 'you', 'think', 'so', 'the', 'word', "wow'", 'in', 'english', 'means', 'you', 'hey', 'b', 'dag', 'final']
All are good except **wow'**.
I wonder if regular expression could explain the logic "it can be a comma/space/period/etc but can't be an apostrophe".
Can someone advise?
Try:
[ ,\.\?']?([a-z]*('\w)?)[\' ,\.\?]?
Added another group so you'll have to select only group 1.
I didn't fully understand what you wanted the output to be but,
try this:
[ ,\.\?]?(["-']?+[a-z]+["-']?[a-z]?)[ ,\.\?]?
using this regex lets you get the ' and " within the text.
if this still was not what you wanted please let me know so I can update my answer.
I collected some tweets through twitter api. Then I counted the words using split(' ') in python. However, some words appear like this:
correct!
correct.
,correct
blah"
...
So how can I format the tweets without punctuation? Or maybe I should try another way to split tweets? Thanks.
You can do the split on multiple characters using re.split...
from string import punctuation
import re
puncrx = re.compile(r'[{}\s]'.format(re.escape(punctuation)))
print filter(None, puncrx.split(your_tweet))
Or, just find words that contain certain contiguous characters:
print re.findall(re.findall('[\w##]+', s), your_tweet)
eg:
print re.findall(r'[\w##]+', 'talking about #python with #someone is so much fun! Is there a 140 char limit? So not cool!')
# ['talking', 'about', '#python', 'with', '#someone', 'is', 'so', 'much', 'fun', 'Is', 'there', 'a', '140', 'char', 'limit', 'So', 'not', 'cool']
I did originally have a smiley in the example, but of course these end up getting filtered out with this method, so that's something to be wary of.
Try removing the punctuation from the string before doing the split.
import string
s = "Some nice sentence. This has punctuation!"
out = s.translate(string.maketrans("",""), string.punctuation)
Then do the split on out.
I would advice to clean text from special symbols before splitting it using this code:
tweet_object["text"] = re.sub(u'[!?##$.,#:\u2026]', '', tweet_object["text"])
You would need to import re before using function sub
import re
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.
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']
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).
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']
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)
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()]
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.
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.
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")
You can do it in one line without nltk (python 3.x).
import string
string_text= string_text.translate(str.maketrans('','',string.punctuation))
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!')
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']