Handling out-of-vocabulary words in NLP - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

What is the main purpose of using an UNK token in NLP models?

In natural language processing, when a word is not found in the model's vocabulary, an UNK token is often used. What is the main purpose of this token?

A To remove unknown words from the input text before processing to avoid errors.

B To replace unknown words with their synonyms from a dictionary.

C To increase the vocabulary size dynamically during model training.

D To represent all out-of-vocabulary words with a single placeholder so the model can handle unknown words during inference.

💡 Hint

Think about how the model deals with words it has never seen before.

The UNK token acts as a catch-all for any word not in the vocabulary, allowing the model to process inputs without crashing or ignoring unknown words.

❓ Predict Output

intermediate

What is the output of this tokenization with OOV handling?

Given the vocabulary {'hello': 1, 'world': 2} and the sentence 'hello there world', what is the tokenized output using 0 as the UNK token index?

NLP

vocab = {'hello': 1, 'world': 2}
sentence = 'hello there world'
tokens = [vocab.get(word, 0) for word in sentence.split()]
print(tokens)

A [1, 1, 2]

B [1, 0, 2]

C [0, 1, 2]

D [1, 2, 0]

💡 Hint

Words not in the vocabulary get the UNK token index 0.

The word 'there' is not in the vocabulary, so it is replaced by 0. 'hello' maps to 1 and 'world' maps to 2.

❓ Model Choice

advanced

Which model architecture is best suited to handle out-of-vocabulary words using subword units?

You want to build an NLP model that can handle out-of-vocabulary words effectively by breaking words into smaller parts. Which model architecture or technique is best for this?

A A model using Byte Pair Encoding (BPE) or WordPiece tokenization to split words into subword units.

B A model that uses a fixed vocabulary of whole words only.

C A model that replaces all unknown words with a single UNK token without subword splitting.

D A model that ignores any word not in the vocabulary during training and inference.

💡 Hint

Think about how breaking words into smaller parts helps with unknown words.

Subword tokenization methods like BPE or WordPiece split rare or unknown words into known smaller units, allowing the model to understand and generate words it has never seen before.

❓ Hyperparameter

advanced

Which hyperparameter affects the size of the vocabulary and thus the handling of OOV words in subword tokenization?

When using Byte Pair Encoding (BPE) for tokenization, which hyperparameter directly controls the vocabulary size and impacts how many out-of-vocabulary words appear?

A The batch size used during model training.

B The learning rate of the model optimizer.

C The number of merge operations performed during BPE training.

D The dropout rate applied in the model layers.

💡 Hint

Think about what controls how many subword units are created.

The number of merge operations in BPE determines how many subword units are merged, controlling vocabulary size and granularity. More merges mean larger vocabulary and fewer OOV tokens.

❓ Metrics

expert

How does the presence of many out-of-vocabulary words affect the model's perplexity on a test set?

You evaluate a language model on a test set containing many out-of-vocabulary (OOV) words. How does this typically affect the model's perplexity metric?

A Perplexity increases because the model struggles to predict unknown words, indicating worse performance.

B Perplexity becomes zero because the model replaces all OOV words with UNK tokens.

C Perplexity remains unchanged because OOV words are ignored during evaluation.

D Perplexity decreases because unknown words are easier to predict.

💡 Hint

Think about how unknown words affect prediction confidence.

When many OOV words appear, the model cannot predict them well, leading to higher uncertainty and thus higher perplexity values, which means worse model performance.

Practice

1. What is the main purpose of using an <UNK> token in natural language processing?

easy

A. To separate words in a sentence

B. To mark the end of a sentence

C. To represent words not seen during training

D. To highlight important keywords

Solution

1. Step 1: Understand the role of <UNK> token

   The <UNK> token is used to replace words that the model has not seen during training, known as out-of-vocabulary words.

2. Step 2: Identify the correct purpose

   Since <UNK> stands for unknown words, it helps the model handle new or rare words by treating them as a single token.

3. Final Answer:

   To represent words not seen during training -> Option C

4. Quick Check:

   <UNK> = unknown words [OK]

Hint: Think of <UNK> as a placeholder for unknown words [OK]

Common Mistakes:

• Confusing <UNK> with sentence delimiters
• Using <UNK> for common words
• Thinking <UNK> highlights keywords

2. Which of the following is the correct way to replace out-of-vocabulary words with <UNK> in a Python list of tokens named tokens given a vocabulary set vocab?

easy

A. tokens = [word if word in vocab else '<UNK>' for word in tokens]

B. tokens = [word for word in tokens if word in vocab else '<UNK>']

C. tokens = [word in vocab ? word : '<UNK>' for word in tokens]

D. tokens = [word if word not in vocab else '<UNK>' for word in tokens]

Solution

1. Step 1: Understand list comprehension syntax

   The correct Python syntax for conditional expressions inside a list comprehension is: [x if condition else y for x in list].

2. Step 2: Apply correct condition for replacing OOV words

   We want to keep the word if it is in the vocabulary; otherwise, replace it with '<UNK>'. tokens = [word if word in vocab else '<UNK>' for word in tokens] correctly implements this logic.

3. Final Answer:

   tokens = [word if word in vocab else '<UNK>' for word in tokens] -> Option A

4. Quick Check:

   Correct Python conditional list comprehension [OK]

Hint: Remember: x if condition else y inside list comprehensions [OK]

Common Mistakes:

• Using incorrect syntax like 'if-else' outside list comprehension
• Confusing Python with other languages' ternary syntax
• Reversing the condition logic

3. Given the following code snippet, what will be the output?

vocab = {'hello', 'world'}
tokens = ['hello', 'there', 'world']
tokens = [word if word in vocab else '<UNK>' for word in tokens]
print(tokens)

medium

A. ['hello', 'there', 'world']

B. ['hello', 'world', '<UNK>']

C. ['<UNK>', '<UNK>', '<UNK>']

D. ['hello', '<UNK>', 'world']

Solution

1. Step 1: Check each token against the vocabulary

   'hello' is in vocab, so it stays 'hello'. 'there' is not in vocab, so it becomes '<UNK>'. 'world' is in vocab, so it stays 'world'.

2. Step 2: Construct the resulting list

   The new tokens list is ['hello', '<UNK>', 'world'].

3. Final Answer:

   ['hello', '<UNK>', 'world'] -> Option D

4. Quick Check:

   Replace OOV words with <UNK> [OK]

Hint: Replace words not in vocab with <UNK> [OK]

Common Mistakes:

• Not replacing 'there' because of misunderstanding
• Replacing all words regardless of vocab
• Confusing list order in output

4. The following code is intended to replace out-of-vocabulary words with <UNK>. What is the error?

vocab = {'cat', 'dog'}
tokens = ['cat', 'bird', 'dog']
tokens = [word if word not in vocab else '<UNK>' for word in tokens]
print(tokens)

medium

A. The vocabulary should be a list, not a set

B. The condition is reversed; it replaces in-vocab words instead of OOV

C. The list comprehension syntax is invalid

D. The print statement is missing parentheses

Solution

1. Step 1: Analyze the condition in list comprehension

   The condition word if word not in vocab else '<UNK>' means words NOT in vocab stay as they are, and words IN vocab become '<UNK>'. This is the opposite of the intended behavior.

2. Step 2: Identify the correct logic

   We want to keep words in vocab and replace words not in vocab with '<UNK>'. So the condition should be word if word in vocab else '<UNK>'.

3. Final Answer:

   The condition is reversed; it replaces in-vocab words instead of OOV -> Option B

4. Quick Check:

   Correct condition keeps vocab words, replaces others [OK]

Hint: Check if condition matches intended keep-or-replace logic [OK]

Common Mistakes:

• Mixing up 'in' and 'not in' in conditions
• Assuming set vs list affects membership test
• Ignoring Python 3 print syntax

5. You have a pretrained word embedding model that does not include the word 'unicorn'. Which approach best helps your model handle this out-of-vocabulary word during inference?

hard

A. Use subword tokenization to break 'unicorn' into known parts

B. Ignore 'unicorn' and remove it from the input

C. Add 'unicorn' to the vocabulary without retraining

D. Replace 'unicorn' with <UNK> token embedding

Solution

1. Step 1: Understand limitations of <UNK> token

   Replacing with <UNK> loses specific meaning, which may reduce model accuracy.

2. Step 2: Consider subword tokenization benefits

   Subword tokenization breaks unknown words into smaller known units, allowing the model to infer meaning from parts.

3. Step 3: Evaluate other options

   Ignoring the word loses information; adding it without retraining is not feasible; subword tokenization is the best practical approach.

4. Final Answer:

   Use subword tokenization to break 'unicorn' into known parts -> Option A

5. Quick Check:

   Subword methods handle OOV words better than <UNK> [OK]

Hint: Break unknown words into smaller known pieces with subword tokenization [OK]

Common Mistakes:

• Thinking <UNK> always preserves meaning
• Trying to add words without retraining embeddings
• Removing unknown words loses important info