Handling out-of-vocabulary words in NLP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to replace unknown words with a special token.

NLP

def replace_oov(word, vocab):
    if word not in vocab:
        return [1]
    return word

Drag options to blanks, or click blank then click option'

A"<UNK>"

B"<PAD>"

C"<EOS>"

D"<SOS>"

Attempts:

3 left

💡 Hint

Common Mistakes

Using padding token '' instead of unknown token.

Using end-of-sequence token '' incorrectly.

The special token '' is commonly used to represent out-of-vocabulary words.

2fill in blank

medium

Complete the code to convert words to indices, using the unknown token index for out-of-vocabulary words.

NLP

def word_to_index(word, word_index, unk_index):
    return word_index.get(word, [1])

Drag options to blanks, or click blank then click option'

Aunk_index

B0

C-1

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 which might be padding index.

Returning None which causes errors in model input.

If the word is not found, we return the index for the unknown token.

3fill in blank

hard

Fix the error in the code that handles out-of-vocabulary words by filling the blank.

NLP

def preprocess_sentence(sentence, vocab, unk_token):
    return [word if word in vocab else [1] for word in sentence]

Drag options to blanks, or click blank then click option'

Aword

Bunk_token

C"<PAD>"

Dvocab[word]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the original word even if it's not in vocab.

Using padding token instead of unknown token.

We replace words not in vocab with the unknown token variable 'unk_token'.

4fill in blank

hard

Fill both blanks to create a dictionary mapping words to indices, assigning the unknown token index for out-of-vocabulary words.

NLP

def create_index(sentence, vocab, unk_index):
    return {word: vocab.get(word, [1]) for word in sentence if word [2] vocab}

Drag options to blanks, or click blank then click option'

Aunk_index

Bin

Cnot in

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' instead of 'not in' in the condition.

Using wrong index for unknown words.

We assign 'unk_index' for words not in vocab and filter words that are not in vocab.

5fill in blank

hard

Fill all three blanks to build a list of indices for a sentence, replacing out-of-vocabulary words with the unknown token index.

NLP

def sentence_to_indices(sentence, vocab, unk_index):
    return [vocab.get([1], [2]) for [3] in sentence]

Drag options to blanks, or click blank then click option'

Aword

Bunk_index

Dtoken

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name inconsistently.

Not providing the default index for missing words.

We iterate over 'word' in sentence and get its index or 'unk_index' if missing.

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