One-hot encoding for text in NLP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a one-hot encoded vector for the word 'cat' using the vocabulary.

NLP

vocab = ['cat', 'dog', 'bird']
word = 'cat'
one_hot = [1 if w == [1] else 0 for w in vocab]

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

Aword

B'cat'

C'dog'

D'bird'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable 'word' instead of the string 'cat' inside the list comprehension.

Using a wrong string like 'dog' or 'bird' for comparison.

We compare each word in the vocabulary to the string 'cat' to create the one-hot vector.

2fill in blank

medium

Complete the code to build a one-hot encoding dictionary for all words in the vocabulary.

NLP

vocab = ['apple', 'banana', 'cherry']
one_hot_dict = {word: [1 if w == [1] else 0 for w in vocab] for word in vocab}

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

Aword

B'apple'

C'banana'

D'cherry'

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed string like 'apple' instead of the variable 'word'.

Confusing the roles of 'word' and 'w' in the comprehension.

We use the variable 'word' to create one-hot vectors for each word in the vocabulary.

3fill in blank

hard

Fix the error in the code to correctly create a one-hot vector for the word 'dog'.

NLP

vocab = ['cat', 'dog', 'fish']
word = 'dog'
one_hot = [1 if w == [1] else 0 for w in vocab]

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

A'dog'

Bword

Cdog

D'cat'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable dog without quotes causing a NameError.

Using the variable 'word' without quotes when the code expects a string.

The comparison must be with the string 'dog' (in quotes) to match vocabulary words correctly.

4fill in blank

hard

Fill both blanks to create a one-hot encoding dictionary for the vocabulary.

NLP

vocab = ['red', 'green', 'blue']
one_hot_dict = {word: [1 if w == [1] else 0 for w in [2]] for word in vocab}

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

Aword

Bvocab

Cwords

Dcolors

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variables like 'words' or 'colors' instead of 'vocab'.

Mixing up the blanks by putting 'vocab' in the first blank.

We compare each word 'w' to the current 'word' and iterate over the vocabulary list 'vocab'.

5fill in blank

hard

Fill all three blanks to create a one-hot encoding dictionary and print the vector for 'blue'.

NLP

vocab = ['red', 'green', 'blue']
one_hot_dict = { [1]: [1 if w == [2] else 0 for w in [3] ] for word in vocab }
print(one_hot_dict['blue'])

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

Aword

Cvocab

Dw

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'w' as the dictionary key instead of 'word'.

Using 'w' in the comparison instead of 'word'.

Iterating over an undefined variable instead of 'vocab'.

The dictionary key is 'word', the comparison is with 'word', and the iteration is over 'vocab'.

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