How term frequency is calculated in TfidfVectorizer?

Question

I searched a lot for understanding this but I am not able to. I understand that by default TfidfVectorizer will apply l2 normalization on term frequency. This a

Answer 1

Ok, Now lets go through the documentation I gave in comments step by step:

Documents:

`ખુબ વખાણ કરે છે
 ખુબ વધારે છે`

1. Get all unique terms (features): ['કરે', 'ખુબ', 'છે.', 'વખાણ', 'વધારે']

2. Calculate frequency of each term in documents:-

   a. Each term present in document1 [ખુબ વખાણ કરે છે] is present once, and વધારે is not present.`

   b. So the term frequency vector (sorted according to features): [1 1 1 1 0]

   c. Applying steps a and b on document2, we get [0 1 1 0 1]

   d. So our final term-frequency vector is [[1 1 1 1 0], [0 1 1 0 1]]

   Note: This is the term frequency you want

3. Now find IDF (This is based on features, not on document basis):

   idf(term) = log(number of documents/number of documents with this term) + 1

   1 is added to the idf value to prevent zero divisions. It is governed by "smooth_idf" parameter which is True by default.

   idf('કરે') = log(2/1)+1 = 0.69314.. + 1 = 1.69314..
   
   idf('ખુબ') = log(2/2)+1 = 0 + 1 = 1
   
   idf('છે.') = log(2/2)+1 = 0 + 1 = 1
   
   idf('વખાણ') = log(2/1)+1 = 0.69314.. + 1 = 1.69314..
   
   idf('વધારે') = log(2/1)+1 = 0.69314.. + 1 = 1.69314..
   

   Note: This corresponds to the data you showed in question.

4. Now calculate TF-IDF (This again is calculated document-wise, calculated according to sorting of features):

   a. For document1:

    For 'કરે', tf-idf = tf(કરે) x idf(કરે) = 1 x 1.69314 = 1.69314
   
    For 'ખુબ', tf-idf = tf(કરે) x idf(કરે) = 1 x 1 = 1
   
    For 'છે.', tf-idf = tf(કરે) x idf(કરે) = 1 x 1 = 1
   
    For 'વખાણ', tf-idf = tf(કરે) x idf(કરે) = 1 x 1.69314 = 1.69314
   
    For 'વધારે', tf-idf = tf(કરે) x idf(કરે) = 0 x 1.69314 = 0
   

   So for document1, the final tf-idf vector is [1.69314 1 1 1.69314 0]

   b. Now normalization is done (l2 Euclidean):

   dividor = sqrt(sqr(1.69314)+sqr(1)+sqr(1)+sqr(1.69314)+sqr(0))
            = sqrt(2.8667230596 + 1 + 1 + 2.8667230596 + 0)
            = sqrt(7.7334461192)
            = 2.7809074272977876...
   

   Dividing each element of the tf-idf array with dividor, we get:

   [0.6088445 0.3595948 0.3595948548 0.6088445 0]

   Note: This is the tfidf of firt document you posted in question.

   c. Now do the same steps a and b for document 2, we get:

   [ 0. 0.453294 0.453294 0. 0.767494]

Update: About sublinear_tf = True OR False

Your original term frequency vector is [[1 1 1 1 0], [0 1 1 0 1]] and you are correct in your understanding that using sublinear_tf = True will change the term frequency vector.

new_tf = 1 + log(tf)

Now the above line will only work on non zero elements in the term-frequecny. Because for 0, log(0) is undefined.

And all your non-zero entries are 1. log(1) is 0 and 1 + log(1) = 1 + 0 = 1`.

You see that the values will remain unchanged for elements with value 1. So your new_tf = [[1 1 1 1 0], [0 1 1 0 1]] = tf(original).

Your term frequency is changing due to the sublinear_tf but it still remains the same.

And hence all below calculations will be same and output is same if you use sublinear_tf=True OR sublinear_tf=False.

Now if you change your documents for which the term-frequecy vector contains elements other than 1 and 0, you will get differences using the sublinear_tf.

Hope your doubts are cleared now.