Python sklearn CountVectorZor的词汇表选项和bigrams返回零数组
我想从一个数组中提取bigram,获取频率大于100的所有bigram,然后使用减少的词汇表为第二个数组打分 看起来词汇选项应该能满足我的需要,但它似乎不起作用。即使将其中一个的输出直接馈送给另一个,也只会产生一个(形状正确的)零数组Python sklearn CountVectorZor的词汇表选项和bigrams返回零数组,python,scikit-learn,n-gram,Python,Scikit Learn,N Gram,我想从一个数组中提取bigram,获取频率大于100的所有bigram,然后使用减少的词汇表为第二个数组打分 看起来词汇选项应该能满足我的需要,但它似乎不起作用。即使将其中一个的输出直接馈送给另一个,也只会产生一个(形状正确的)零数组 from sklearn.feature_extraction.text import CountVectorizer docs = ['run fast into a bush','run fast into a tree','run slow','run f
from sklearn.feature_extraction.text import CountVectorizer
docs = ['run fast into a bush','run fast into a tree','run slow','run fast']
# Collect bigrams
vectorizer = CountVectorizer(ngram_range = (2,2))
vectorizer.fit(docs)
vocab = vectorizer.vocabulary_
# Score the exact same data
vectorizer = CountVectorizer(vocabulary=vocab)
output = vectorizer.transform(docs)
# Demonstrate that the array is all zeros
print "Length of vocab", len(vocab)
print output.A
Length of vocab 5
[[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]]
刚抓住它。您需要在第二个实例中指定ngram_范围(它不会自动解释unigrams和bigrams。)
创建标记化单词的语料库。如果有奇数个单词,将ngram设置为3,否则设置为2表示偶数。创建一个单词包矩阵。使用bag of word矩阵创建一个数据帧,并将矢量器单词作为特征
from sklearn.feature_extraction.text import CountVectorizer
import nltk
docs = ['run fast into a bush','run fast into a tree','run slow','run fast']
str_buffer=" ".join(docs)
#print(str_buffer)
corpus=nltk.word_tokenize(str_buffer)
vectorizer_ng2=CountVectorizer(ngram_range=range(1,3),stop_words='english')
bow_matrix=vectorizer_ng2.fit_transform(corpus)
print(bow_matrix.toarray())
bow_df = pd.DataFrame(bow_matrix.toarray())
bow_df.columns = vectorizer_ng2.get_feature_names()
print(bow_df)
输出:
bush fast run slow tree
0 0 0 1 0 0
1 0 1 0 0 0
2 0 0 0 0 0
3 0 0 0 0 0
4 1 0 0 0 0
5 0 0 1 0 0
6 0 1 0 0 0
7 0 0 0 0 0
8 0 0 0 0 0
9 0 0 0 0 1
10 0 0 1 0 0
11 0 0 0 1 0
12 0 0 1 0 0
13 0 1 0 0 0
bush fast run slow tree
0 0 0 1 0 0
1 0 1 0 0 0
2 0 0 0 0 0
3 0 0 0 0 0
4 1 0 0 0 0
5 0 0 1 0 0
6 0 1 0 0 0
7 0 0 0 0 0
8 0 0 0 0 0
9 0 0 0 0 1
10 0 0 1 0 0
11 0 0 0 1 0
12 0 0 1 0 0
13 0 1 0 0 0