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/

10-12 19:35