# Q5A · TF–IDF matrix

Subject: Natural Language Processing

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/nlp/#tee/mocks/timed-final-2025/final-2025-q5a

Canonical: https://deep-learning-study-room.vercel.app/read/nlp/tee/mocks/timed-final-2025--final-2025-q5a/

TEE · actual previous paper · 100 marks · 180 minutes

Question. For the following two sentences:  
Sentence A: “The student read a science book in the library”  
Sentence B: “The teacher read a history book in the classroom”  
1. Construct the TF-IDF matrix for these sentences after removing the stop words {The, a, in}.  
2. Identify and comment on the important terms in each sentence based on their TF-IDF scores. [10 marks]

### Answer

### Given

Two documents, N = 2. Remove the listed stop words case-insensitively, so both “The” and “the” are removed. Use raw count TF and natural-log IDF, the convention used in the course key.

### To find

The term-by-document TF–IDF matrix and the terms that distinguish each sentence.

### Formula: which one and why



$$
\mathrm{TF}(t,d)=C(t,d),\qquad \mathrm{IDF}(t)=\ln\frac{N}{\mathrm{df}(t)}
$$





$$
\mathrm{TFIDF}(t,d)=\mathrm{TF}(t,d)\,\mathrm{IDF}(t)
$$



C(t,d) is the number of occurrences of term t in document d. df(t) is the number of documents containing t, not its total token count. N is the number of documents. IDF downweights terms shared by many documents.

### Solution

### Step 1. Clean and list the tokens

A: student, read, science, book, library.  
B: teacher, read, history, book, classroom. Each retained token appears once in its sentence.

### Step 2. Fix one common vocabulary order

student, read, science, book, library, teacher, history, classroom. Each row must refer to the same term in both documents.

### Step 3. Count document frequency and calculate IDF

| Term type | df | IDF calculation |
| --- | --- | --- |
| read, book | 2 | ln(2/2) = ln(1) = 0 |
| Every other term | 1 | ln(2/1) = ln(2) ≈ 0.693147 |

A term absent from a document has TF = 0. A present term has TF = 1 here.

### Step 4. Multiply TF by IDF

| Term | TF in A | TF in B | IDF | TF–IDF A | TF–IDF B |
| --- | --- | --- | --- | --- | --- |
| student | 1 | 0 | 0.693147 | 0.693147 | 0 |
| read | 1 | 1 | 0 | 0 | 0 |
| science | 1 | 0 | 0.693147 | 0.693147 | 0 |
| book | 1 | 1 | 0 | 0 | 0 |
| library | 1 | 0 | 0.693147 | 0.693147 | 0 |
| teacher | 0 | 1 | 0.693147 | 0 | 0.693147 |
| history | 0 | 1 | 0.693147 | 0 | 0.693147 |
| classroom | 0 | 1 | 0.693147 | 0 | 0.693147 |

### Step 5. Interpret the weights

A is distinguished by student, science and library. B is distinguished by teacher, history and classroom. “read” and “book” have zero IDF in this tiny unsmoothed two-document collection because they appear in every document; that does not mean those words are meaningless in English.

### Final answer

All six document-specific terms receive **0.693147** in their own document and 0 in the other. Shared read and book receive 0 in both.

**Yaad rakho:** TF: is document mein kitni baar? IDF: kitne documents mein common? Rare-across-documents term ko zyada weight.

If normalized TF is explicitly requested, divide each present-token count by five; each nonzero weight becomes 0.693147/5 = 0.1386294. Do not silently switch conventions.