我正在研究关键字提取问题。考虑非常普遍的情况
tfidf = TfidfVectorizer(tokenizer=tokenize, stop_words='english')
t = """Two Travellers, walking in the noonday sun, sought the shade of a widespreading tree to rest. As they lay looking up among the pleasant leaves, they saw that it was a Plane Tree.
"How useless is the Plane!" said one of them. "It bears no fruit whatever, and only serves to litter the ground with leaves."
"Ungrateful creatures!" said a voice from the Plane Tree. "You lie here in my cooling shade, and yet you say I am useless! Thus ungratefully, O Jupiter, do men receive their blessings!"
Our best blessings are often the least appreciated."""
tfs = tfidf.fit_transform(t.split(" "))
str = 'tree cat travellers fruit jupiter'
response = tfidf.transform([str])
feature_names = tfidf.get_feature_names()
for col in response.nonzero()[1]:
print(feature_names[col], ' - ', response[0, col])
这给了我
(0, 28) 0.443509712811
(0, 27) 0.517461475101
(0, 8) 0.517461475101
(0, 6) 0.517461475101
tree - 0.443509712811
travellers - 0.517461475101
jupiter - 0.517461475101
fruit - 0.517461475101
很好对于其中出现的任何新文档,是否有办法获得tfidf得分最高的前n个术语?
最佳答案
您必须做一点点的歌舞才能将矩阵转换为numpy数组,但这应该可以满足您的需求:
feature_array = np.array(tfidf.get_feature_names())
tfidf_sorting = np.argsort(response.toarray()).flatten()[::-1]
n = 3
top_n = feature_array[tfidf_sorting][:n]
这给了我:
array([u'fruit', u'travellers', u'jupiter'],
dtype='<U13')
argsort
调用确实是有用的here are the docs for it。我们必须执行[::-1]
,因为argsort
仅支持从小到大的排序。我们调用flatten
将维数减少到1d,以便可以使用排序后的索引来索引1d特征数组。请注意,仅当您一次测试一个文档时,才包括对flatten
的调用。另外,在另一个注释上,您是说类似
tfs = tfidf.fit_transform(t.split("\n\n"))
吗?否则,多行字符串中的每个术语都将被视为“文档”。相反,使用\n\n
意味着我们实际上正在查看4个文档(每行一个),这在您考虑tfidf时更有意义。关于python - Scikit学习TfidfVectorizer : How to get top n terms with highest tf-idf score,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/34232190/