List the words in a vocabulary according to occurrence in a text corpus, with Scikit-Learn CountVectorizer
Asked Answered
S

2

14

I have fitted a CountVectorizer to some documents in scikit-learn. I would like to see all the terms and their corresponding frequency in the text corpus, in order to select stop-words. For example

'and' 123 times, 'to' 100 times, 'for' 90 times, ... and so on

Is there any built-in function for this?

Ssw answered 18/4, 2013 at 8:27 Comment(0)
P
23

If cv is your CountVectorizer and X is the vectorized corpus, then

zip(cv.get_feature_names(),
    np.asarray(X.sum(axis=0)).ravel())

returns a list of (term, frequency) pairs for each distinct term in the corpus that the CountVectorizer extracted.

(The little asarray + ravel dance is needed to work around some quirks in scipy.sparse.)

Preferment answered 18/4, 2013 at 9:1 Comment(4)
Thanks! But they are not ordered, but I managed to do that: for tuple in sorted( occ_list ,key=lambda idx: idx[1] ): print tuple[0] +' ' + str(tuple[1]). The problem is that characters åäö are not printed out. I have set the coding to utf8.Ssw
Also are you sure that get_feature_names() will have the terms ordered according to their index in the term-frequency matrix? I have found out that cv.get_feature_names() and cv.vocabulary_.keys() does not have the same order.Ssw
@user1506145: dict.keys doesn't guarantee any order; that's exactly why get_feature_names exists.Preferment
Sorry to dredge this topic up, but how would you make a vectorized corpus, X, from a simple string like "This is the example that we will make an example of."Lymphocyte
I
5

There is no built-in. I have found a faster way to do it based on Ando Saabas's answer:

from sklearn.feature_extraction.text import CountVectorizer 
texts = ["Hello world", "Python makes a better world"]
vec = CountVectorizer().fit(texts)
bag_of_words = vec.transform(texts)
sum_words = bag_of_words.sum(axis=0)
words_freq = [(word, sum_words[0, idx]) for word, idx in vec.vocabulary_.items()]
sorted(words_freq, key = lambda x: x[1], reverse=True)

output

[('world', 2), ('python', 1), ('hello', 1), ('better', 1), ('makes', 1)]
Innermost answered 28/1, 2018 at 18:45 Comment(0)

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