用Python编写句子列表中的单词Bigrams

Question

我有一个句子列表:

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

我需要形成双字节对并将它们存储在变量中.问题是,当我这样做时,我会得到一对句子而不是单词.这是我做的:

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

产量

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

不能火车站和城堡酒店组成一个二元组.我想要的是

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

第一句的最后一个单词不应与第二句的第一个单词合并.我该怎么做才能让它发挥作用？

Answer 1

使用列表推导和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 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

Answer 3

不是将文本转换为字符串列表,而是将每个句子分别作为字符串单独开头.我还删除了标点符号和停用词,如果与您无关,请删除这些部分:

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

要使用它,请这样做:

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

请注意,这会更进一步,实际上统计得分为bigrams(这将在训练模型时派上用场).

Answer 4

如果没有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']]]