How to remove ' in strings with RegexpTokenizer - python

from nltk.tokenize import RegexpTokenizer
text="That's some text, you know!"
tokens=[]
tokenizer = RegexpTokenizer(r'\w+')
tokens+=tokenizer.tokenize(text.lower())
Currently returns: text = ['that', 's', 'some', 'text', 'you', 'know']
I need it to return: Currently returns: text = ['thats', 'some', 'text', 'you', 'know'] (The "thats" is one word)

There are 2 solutions. Either you want to preprocess your text variable with:
text = text.replace("'", "")
or you want to match "that's" as a single word with this modification:
tokenizer = RegexpTokenizer(r'[\w\']+')

Related

How to remove punctuations using NLTK? [duplicate]

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']

How do you apply tokenization and stopwords on multiple lines of text from a csv file using python

I have a csv file that contains a column of multiple lines of sentences. I need to tokenize, identify and remove the stopwords from every row of text using python. I can do one line of text at a time but how do you use python to make it read each row and apply tokenization and stopwords?
This is what my code is for one sentence:
from nltk.tokenize import word_tokenize
covid_text = "i think i need to self isolate because i don't feel good :("
stop_words = set(stopwords.words('english'))
word_tokens = word_tokenize(covid_text)
filtered_sentence = [w for w in word_tokens if not w in stop_words]
filtered_sentence = []
for w in word_tokens:
if w not in stop_words:
filtered_sentence.append(w)
print(word_tokens)
print(filtered_sentence)```
Output:
['i', 'think', 'i', 'need', 'to', 'self', 'isolate', 'because', 'i', 'do', "n't", 'feel', 'good', ':', '(']
['think', 'need', 'self', 'isolate', "n't", 'feel', 'good', ':', '(']

How to remove punctuation and numbers during TweetTokenizer step in NLP?

I am relatively new to NLP so please be gentle. I
have a complete list of the text from Trump's tweets since taking office and I am tokenizing the text to analyze the content.
I am using the TweetTokenizer from the nltk library in python and I'm trying to get everything tokenized except for numbers and punctuation. Problem is my code removes all the tokens except one.
I have tried using the .isalpha() method but this did not work, which I thought would as should only be True for strings composed from the alphabet.
#Create a content from the tweets
text= non_re['text']
#Make all text in lowercase
low_txt= [l.lower() for l in text]
#Iteratively tokenize the tweets
TokTweet= TweetTokenizer()
tokens= [TokTweet.tokenize(t) for t in low_txt
if t.isalpha()]
My output from this is just one token.
If I remove the if t.isalpha() statement then I get all of the tokens including numbers and punctuation, suggesting the isalpha() is to blame from the over-trimming.
What I would like, is a way to get the tokens from the tweet text without punctuation and numbers.
Thanks for your help!
Try something like below:
import string
import re
import nltk
from nltk.tokenize import TweetTokenizer
tweet = "first think another Disney movie, might good, it's kids movie. watch it, can't help enjoy it. ages love movie. first saw movie 10 8 years later still love it! Danny Glover superb could play"
def clean_text(text):
# remove numbers
text_nonum = re.sub(r'\d+', '', text)
# remove punctuations and convert characters to lower case
text_nopunct = "".join([char.lower() for char in text_nonum if char not in string.punctuation])
# substitute multiple whitespace with single whitespace
# Also, removes leading and trailing whitespaces
text_no_doublespace = re.sub('\s+', ' ', text_nopunct).strip()
return text_no_doublespace
cleaned_tweet = clean_text(tweet)
tt = TweetTokenizer()
print(tt.tokenize(cleaned_tweet))
output:
['first', 'think', 'another', 'disney', 'movie', 'might', 'good', 'its', 'kids', 'movie', 'watch', 'it', 'cant', 'help', 'enjoy', 'it', 'ages', 'love', 'movie', 'first', 'saw', 'movie', 'years', 'later', 'still', 'love', 'it', 'danny', 'glover', 'superb', 'could', 'play']
# Function for removing Punctuation from Text and It gives total no.of punctuation removed also
# Input: Function takes Existing fie name and New file name as string i.e 'existingFileName.txt' and 'newFileName.txt'
# Return: It returns two things Punctuation Free File opened in read mode and a punctuation count variable.
def removePunctuation(tokenizeSampleText, newFileName):
from nltk.tokenize import word_tokenize
existingFile = open(tokenizeSampleText, 'r')
read_existingFile = existingFile.read()
tokenize_existingFile = word_tokenize(read_existingFile)
puncRemovedFile = open(newFileName, 'w+')
import string
stringPun = list(string.punctuation)
count_pun = 0
for word in tokenize_existingFile:
if word in stringPun:
count_pun += 1
else:
word = word + ' '
puncRemovedFile.write(''.join(word))
existingFile.close()
puncRemovedFile.close()
return open(newFileName, 'r'), count_pun
punRemoved, punCount = removePunctuation('Macbeth.txt', 'Macbeth-punctuationRemoved.txt')
print(f'Total Punctuation : {punCount}')
punRemoved.read()

TfidfVectorizer to respect hyphenated compounds (words that are joined with a hyphen)

I have a list of strings that look like this:
df_train = ['Hello John-Smith it is nine o'clock','This is a completely random-sequence']
I would like sklearn TfidfVectorizer to treat words joined with a hyphen as a whole word. When I apply the following code, the words separated by hyphen (or other punctuation) are treated as separate words:
vectorizer_train = TfidfVectorizer(analyzer = 'word',
min_df=0.0,
max_df = 1.0,
strip_accents = None,
encoding = 'utf-8',
preprocessor=None,
token_pattern=r"(?u)\b\w\w+\b")
vectorizer_train.fit_transform(df_train)
vectorizer_train.get_feature_names()
I have changed the parameter token_pattern but with no success. Any idea of how I could solve this issue? In addition, is it possible to treat as a single entity words that are separated by any punctuation? (e.g. 'Hi.there How_are you:doing')
It seems you need to split on white space only, try switch the pattern to (?u)\S\S+, which captures consecutive non white space characters as a single word:
df_train = ["Hello John-Smith it is nine o'clock",
"This is a completely random-sequence",
"Hi.there How_are you:doing"]
vectorizer_train = TfidfVectorizer(analyzer = 'word',
min_df=0.0,
max_df = 1.0,
strip_accents = None,
encoding = 'utf-8',
preprocessor=None,
token_pattern=r"(?u)\S\S+")
​
vectorizer_train.fit_transform(df_train)
vectorizer_train.get_feature_names()
gives:
['completely',
'hello',
'hi.there',
'how_are',
'is',
'it',
'john-smith',
'nine',
"o'clock",
'random-sequence',
'this',
'you:doing']
To respect hyphenated compounds only, you can use (?u)\b\w[\w-]*\w\b:
['clock',
'completely',
'doing',
'hello',
'hi',
'how_are',
'is',
'it',
'john-smith',
'nine',
'random-sequence',
'there',
'this',
'you']

How to get rid of punctuation using NLTK tokenizer?

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']

Categories