Scikit学习TfidfVectorizer：如何获得tf-idf得分最高的前n个词

Question

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？

Answer 1

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')

该argsort电话确实是有用的，这里有它的文档。我们必须这样做，[::-1]因为argsort仅支持从小到大的排序。我们呼吁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，flatten仅当您一次测试一个文档时，包含to的调用才起作用。

另外，从另一个角度来说，您的意思是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”。使用\n\n代替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。