31

我有一个句子列表：

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

我需要形成二元对并将它们存储在一个变量中。问题是当我这样做时，我得到的是一对句子而不是单词。这是我所做的：

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

产生

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

火车站和城堡酒店不能合二为一。我想要的是

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

第一句的最后一个词不能和第二句的第一个词合并。我应该怎么做才能让它工作？

4

10 回答 10

52

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')]
于 2014-02-18T05:04:29.833 回答
13 
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
于 2018-02-19T18:30:32.733 回答
8

与其将文本转换为字符串列表，不如将每个句子单独作为一个字符串开始。我还删除了标点符号和停用词，如果与您无关，只需删除这些部分：

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，这更进一步，实际上对二元组进行了统计评分（这将在训练模型时派上用场）。

于 2014-02-18T04:55:32.890 回答
5

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']]]
于 2014-02-18T05:00:53.097 回答
3 
>>> 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')]

使用枚举和拆分功能。

于 2014-02-18T06:21:45.410 回答
2

读取数据集

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

收集所有可用月份

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

每月创建所有推文的代币

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

每月创建二元组

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

每月计算 bigrams

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

将结果包装在整洁的数据框中

month1 = pd.DataFrame(data = count_bigrams[0], index= bigrams[0], columns= ["Count"])
month2 = pd.DataFrame(data = count_bigrams[1], index= bigrams[1], columns= ["Count"])
于 2018-05-09T20:00:15.523 回答
1

只是修复丹的代码：

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
于 2016-10-02T20:34:01.187 回答
1

最好的方法是使用“zip”函数来生成 n-gram。其中2 in range函数是克数

test = [1,2,3,4,5,6,7,8,9]
print(test[0:])
print(test[1:])
print(list(zip(test[0:],test[1:])))
%timeit list(zip(*[test[i:] for i in range(2)]))

o/p：

[1, 2, 3, 4, 5, 6, 7, 8, 9]  
[2, 3, 4, 5, 6, 7, 8, 9]  
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)]  
1000000 loops, best of 3: 1.34 µs per loop
于 2020-08-27T15:45:32.423 回答
0

有很多方法可以解决它，但我是这样解决的：

>>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')]
于 2018-11-28T20:42:54.510 回答
0

我认为最好和最通用的方法如下：

n      = 2
ngrams = []

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

或者换句话说：

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

这应该适用于任何n和任何序列l。如果没有 ngram 长度n，则返回空列表。

于 2019-10-21T10:09:14.173 回答