Forming Bigrams of words in list of sentences with

Question

I have a list of sentences:

text = ['cant railway station','citadel hotel',' police stn']. 

I need to form bigram pairs and store them in a variable. The problem is that when I do that, I get a pair of sentences instead of words. Here is what I did:

text2 = [[word for word in line.split()] for line in text]
bigrams = nltk.bigrams(text2)
print(bigrams)

which yields

[(['cant', 'railway', 'station'], ['citadel', 'hotel']), (['citadel', 'hotel'], ['police', 'stn'])

Can't railway station and citadel hotel form one bigram. What I want is

[([cant],[railway]),([railway],[station]),([citadel,hotel]), and so on...

The last word of the first sentence should not merge with the first word of second sentence. What should I do to make it work?

Answer 1

Just fixing Dan's code:

def get_bigrams(myString):
    tokenizer = WordPunctTokenizer()
    tokens = tokenizer.tokenize(myString)
    stemmer = PorterStemmer()
    bigram_finder = BigramCollocationFinder.from_words(tokens)
    bigrams = bigram_finder.nbest(BigramAssocMeasures.chi_sq, 500)

    for bigram_tuple in bigrams:
        x = "%s %s" % bigram_tuple
        tokens.append(x)

    result = [' '.join([stemmer.stem(w).lower() for w in x.split()]) for x in tokens if x.lower() not in stopwords.words('english') and len(x) > 8]
    return result

Answer 2

Using list comprehensions and zip:

>>> text = ["this is a sentence", "so is this one"]
>>> bigrams = [b for l in text for b in zip(l.split(" ")[:-1], l.split(" ")[1:])]
>>> print(bigrams)
[('this', 'is'), ('is', 'a'), ('a', 'sentence'), ('so', 'is'), ('is', 'this'), ('this',     
'one')]

Answer 3

Rather than turning your text into lists of strings, start with each sentence separately as a string. I've also removed punctuation and stopwords, just remove these portions if irrelevant to you:

import nltk
from nltk.corpus import stopwords
from nltk.stem import PorterStemmer
from nltk.tokenize import WordPunctTokenizer
from nltk.collocations import BigramCollocationFinder
from nltk.metrics import BigramAssocMeasures

def get_bigrams(myString):
    tokenizer = WordPunctTokenizer()
    tokens = tokenizer.tokenize(myString)
    stemmer = PorterStemmer()
    bigram_finder = BigramCollocationFinder.from_words(tokens)
    bigrams = bigram_finder.nbest(BigramAssocMeasures.chi_sq, 500)

    for bigram_tuple in bigrams:
        x = "%s %s" % bigram_tuple
        tokens.append(x)

    result = [' '.join([stemmer.stem(w).lower() for w in x.split()]) for x in tokens if x.lower() not in stopwords.words('english') and len(x) > 8]
    return result

To use it, do like so:

for line in sentence:
    features = get_bigrams(line)
    # train set here

Note that this goes a little further and actually statistically scores the bigrams (which will come in handy in training the model).

Answer 4

Read the dataset

df = pd.read_csv('dataset.csv', skiprows = 6, index_col = "No")

Collect all available months

df["Month"] = df["Date(ET)"].apply(lambda x : x.split('/')[0])

Create tokens of all tweets per month

tokens = df.groupby("Month")["Contents"].sum().apply(lambda x : x.split(' '))

Create bigrams per month

bigrams = tokens.apply(lambda x : list(nk.ngrams(x, 2)))

Count bigrams per month

count_bigrams = bigrams.apply(lambda x : list(x.count(item) for item in x))

Wrap up the result in neat dataframes

month1 = pd.DataFrame(data = count_bigrams[0], index= bigrams[0], columns= ["Count"])
month2 = pd.DataFrame(data = count_bigrams[1], index= bigrams[1], columns= ["Count"])

Answer 5

Without nltk:

ans = []
text = ['cant railway station','citadel hotel',' police stn']
for line in text:
    arr = line.split()
    for i in range(len(arr)-1):
        ans.append([[arr[i]], [arr[i+1]]])


print(ans) #prints: [[['cant'], ['railway']], [['railway'], ['station']], [['citadel'], ['hotel']], [['police'], ['stn']]]

Answer 6

I think the best and most general way to do it is the following:

n      = 2
ngrams = []

for l in L:
    for i in range(n,len(l)+1):
        ngrams.append(l[i-n:i])

or in other words:

ngrams = [ l[i-n:i] for l in L for i in range(n,len(l)+1) ]

This should work for any n and any sequence l. If there are no ngrams of length n it returns the empty list.