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?
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')]
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).
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
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']]]
>>> 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.
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
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"])
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')]
来源:https://stackoverflow.com/questions/21844546/forming-bigrams-of-words-in-list-of-sentences-with-python