Forming Bigrams of words in list of sentences with

问题:

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?

回答1:

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')]

回答2:

from nltk import word_tokenize 
from nltk.util import ngrams


text = ['cant railway station', 'citadel hotel', 'police stn']
for line in text:
    token = nltk.word_tokenize(line)
    bigram = list(ngrams(token, 2)) 

    # the '2' represents bigram...you can change it to get ngrams with different size

回答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).

回答4:

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']]]

回答5:

>>> text = ['cant railway station','citadel hotel',' police stn']
>>> bigrams = [(ele, tex.split()[i+1]) for tex in text  for i,ele in enumerate(tex.split()) if i < len(tex.split())-1]
>>> bigrams
[('cant', 'railway'), ('railway', 'station'), ('citadel', 'hotel'), ('police', 'stn')]

Using enumerate and split function.

回答6:

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

回答7:

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"])

回答8:

There are a number of ways to solve it but I solved in this way:

>>text = ['cant railway station','citadel hotel',' police stn']
>>text2 = [[word for word in line.split()] for line in text]
>>text2
[['cant', 'railway', 'station'], ['citadel', 'hotel'], ['police', 'stn']]
>>output = []
>>for i in range(len(text2)):
    output = output+list(bigrams(text2[i]))
>>#Here you can use list comphrension also
>>output
[('cant', 'railway'), ('railway', 'station'), ('citadel', 'hotel'), ('police', 'stn')]

回答9:

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.