One-hot encoding for text in NLP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of one-hot encoding a small text corpus

What is the output of the following code that one-hot encodes a list of words?

NLP

from sklearn.preprocessing import OneHotEncoder
import numpy as np

words = np.array([['cat'], ['dog'], ['cat'], ['bird']])
encoder = OneHotEncoder(sparse=False)
encoded = encoder.fit_transform(words)
print(encoded)

A

[[0. 1. 0.]
 [0. 0. 1.]
 [0. 1. 0.]
 [1. 0. 0.]]

B

[[1. 0. 0.]
 [0. 1. 0.]
 [1. 0. 0.]
 [0. 0. 1.]]

C

[[0. 1. 0.]
 [1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]

D

[[1. 0. 0.]
 [0. 0. 1.]
 [1. 0. 0.]
 [0. 1. 0.]]

💡 Hint

Remember that OneHotEncoder assigns columns in alphabetical order of unique words.

The unique words are ['bird', 'cat', 'dog'] alphabetically. So 'bird' is [1,0,0], 'cat' is [0,1,0], 'dog' is [0,0,1]. The input words are ['cat', 'dog', 'cat', 'bird'], so the output matches option A.

🧠 Conceptual

intermediate

Understanding one-hot encoding vocabulary size

If you one-hot encode a text corpus with 10,000 unique words, what will be the size of each one-hot vector?

A A vector of length 1 with the index of the word

B 10,000 elements with multiple elements set to 1 depending on word frequency

C A vector of length equal to the number of words in the sentence

D 10,000 elements with exactly one element set to 1 and the rest 0

💡 Hint

One-hot encoding creates a vector with one position for each unique word.

One-hot encoding creates a vector with length equal to the vocabulary size. Only one position is 1, representing the word, and all others are 0.

❓ Hyperparameter

advanced

Choosing one-hot encoding parameters for text data

Which parameter of sklearn's OneHotEncoder controls whether the output is a sparse matrix or a dense array?

A sparse

B handle_unknown

C categories

D drop

💡 Hint

This parameter decides the output format to save memory or not.

The 'sparse' parameter when set to True returns a sparse matrix, which saves memory for large vocabularies. False returns a dense numpy array.

❓ Metrics

advanced

Evaluating one-hot encoded text input for a classification model

You trained a classifier on one-hot encoded text data. Which metric best measures how well the model predicts the correct class labels?

A Perplexity

B Accuracy

C Mean Squared Error

D Silhouette Score

💡 Hint

Think about classification performance metrics.

Accuracy measures the fraction of correct predictions in classification tasks. Mean Squared Error is for regression, Silhouette Score for clustering, and Perplexity for language models.

🔧 Debug

expert

Debugging one-hot encoding with unseen words during inference

You trained a OneHotEncoder on a training set and saved it. At inference, you try to transform new text containing words not seen during training. What error will sklearn's OneHotEncoder raise by default?

A KeyError: word not found in vocabulary

B TypeError: unsupported operand type(s)

C ValueError: Found unknown categories during transform

D IndexError: index out of range

💡 Hint

Check how OneHotEncoder handles unknown categories by default.

By default, OneHotEncoder raises a ValueError if transform sees categories not present during fit. To avoid this, handle_unknown='ignore' can be set.

Practice

1. What does one-hot encoding do to words in text processing?

easy

A. Converts each word into a vector with one 1 and rest 0s

B. Replaces words with their synonyms

C. Counts the number of letters in each word

D. Sorts words alphabetically

Solution

1. Step 1: Understand one-hot encoding concept

   One-hot encoding creates a vector for each word where only one position is 1 and all others are 0.

2. Step 2: Compare options with definition

   Only Converts each word into a vector with one 1 and rest 0s matches this definition exactly.

3. Final Answer:

   Converts each word into a vector with one 1 and rest 0s -> Option A

4. Quick Check:

   One-hot encoding = vector with single 1 [OK]

Hint: One-hot means one 1 in vector, rest zeros [OK]

Common Mistakes:

• Thinking it replaces words with synonyms
• Confusing with counting letters
• Assuming it sorts words

2. Which of the following is the correct Python syntax to create a one-hot vector for the word 'cat' from vocabulary ['cat', 'dog', 'bird']?

easy

A. one_hot = [0, 0, 1]

B. one_hot = [0, 1, 0]

C. one_hot = [1, 1, 0]

D. one_hot = [1, 0, 0]

Solution

1. Step 1: Identify the index of 'cat' in vocabulary

   'cat' is at index 0 in ['cat', 'dog', 'bird'].

2. Step 2: Create one-hot vector with 1 at index 0

   The vector should have 1 at position 0 and 0 elsewhere: [1, 0, 0].

3. Final Answer:

   [1, 0, 0] -> Option D

4. Quick Check:

   Index 0 gets 1 in one-hot vector [OK]

Hint: Index of word = position of 1 in vector [OK]

Common Mistakes:

• Putting 1 in wrong index
• Using multiple 1s in vector
• Confusing word order in vocabulary

3. What will be the output of this Python code?

vocab = ['apple', 'banana', 'cherry']
word = 'banana'
one_hot = [1 if w == word else 0 for w in vocab]
print(one_hot)

medium

A. [1, 0, 0]

B. [0, 1, 0]

C. [0, 0, 1]

D. [1, 1, 0]

Solution

1. Step 1: Understand list comprehension logic

   For each word in vocab, put 1 if it matches 'banana', else 0.

2. Step 2: Apply to vocab list

   'apple' != 'banana' -> 0, 'banana' == 'banana' -> 1, 'cherry' != 'banana' -> 0, so [0, 1, 0].

3. Final Answer:

   [0, 1, 0] -> Option B

4. Quick Check:

   Only 'banana' gets 1 in vector [OK]

Hint: Check which vocab word equals target word [OK]

Common Mistakes:

• Mixing up word positions
• Using 1 for all words
• Misreading list comprehension

4. Identify the error in this one-hot encoding code snippet:

vocab = ['red', 'green', 'blue']
word = 'green'
one_hot = [0 if w == word else 1 for w in vocab]
print(one_hot)

medium

A. The list comprehension syntax is invalid

B. The vocabulary list is missing a word

C. The condition is reversed; it should assign 1 when words match

D. The print statement syntax is incorrect

Solution

1. Step 1: Analyze the list comprehension condition

   It assigns 0 if word matches, else 1, which is opposite of one-hot logic.

2. Step 2: Correct logic for one-hot encoding

   One-hot should assign 1 when words match and 0 otherwise.

3. Final Answer:

   The condition is reversed; it should assign 1 when words match -> Option C

4. Quick Check:

   Match word -> 1, else 0 [OK]

Hint: One-hot sets 1 for match, not 0 [OK]

Common Mistakes:

• Reversing 0 and 1 in condition
• Assuming syntax error instead of logic error
• Ignoring correct vocabulary

5. Given a vocabulary ['sun', 'moon', 'star'] and a sentence 'moon star sun star', which one-hot encoded matrix correctly represents the sentence?

hard

A. [[0,1,0],[0,0,1],[1,0,0],[0,0,1]]

B. [[1,0,0],[0,1,0],[0,0,1],[0,1,0]]

C. [[0,0,1],[1,0,0],[0,1,0],[1,0,0]]

D. [[1,1,0],[0,0,1],[1,0,0],[0,0,1]]

Solution

1. Step 1: Map each word to its one-hot vector

   Vocabulary indices: 'sun'->0, 'moon'->1, 'star'->2. So 'moon'=[0,1,0], 'star'=[0,0,1], 'sun'=[1,0,0].

2. Step 2: Encode sentence words in order

   Sentence words: 'moon' -> [0,1,0], 'star' -> [0,0,1], 'sun' -> [1,0,0], 'star' -> [0,0,1].

3. Final Answer:

   [[0,1,0],[0,0,1],[1,0,0],[0,0,1]] -> Option A

4. Quick Check:

   Each word vector matches vocab index [OK]

Hint: Match word order and vocab index for vectors [OK]

Common Mistakes:

• Mixing word order in sentence
• Swapping indices of words
• Using vectors with multiple 1s