Stopword removal in NLP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to remove stopwords from the list of words.

NLP

filtered_words = [word for word in words if word not in [1]]

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

Astopwords

Bpunctuation

Cvocabulary

Dtokens

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'tokens' instead of 'stopwords' causes wrong filtering.

Using 'punctuation' removes symbols, not stopwords.

The variable stopwords contains common words to remove. We keep words not in stopwords.

2fill in blank

medium

Complete the code to import the stopwords list from NLTK.

NLP

from nltk.corpus import [1]

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

Awords

Bstopwords

Ctokenize

Dcorpus

Attempts:

3 left

💡 Hint

Common Mistakes

Importing 'words' instead of 'stopwords' causes attribute errors.

Importing 'tokenize' is unrelated to stopword lists.

We import stopwords from nltk.corpus to access common stopwords.

3fill in blank

hard

Fix the error in the code to get English stopwords from NLTK.

NLP

stop_words = set(stopwords.words([1]))

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

A'eng'

B'English'

C'en'

D'english'

Attempts:

3 left

💡 Hint

Common Mistakes

Using capitalized 'English' causes a KeyError.

Using abbreviations like 'eng' or 'en' are invalid.

The correct language code for English stopwords in NLTK is 'english' all lowercase.

4fill in blank

hard

Fill both blanks to create a list of words without stopwords from a sentence.

NLP

filtered = [[1] for [2] in sentence.split() if [1] not in stop_words]

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

Aword

Bsentence

Cstop_words

Dwords

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the loop and condition causes errors.

Using 'sentence' or 'words' instead of 'word' is incorrect.

We iterate over each word in the sentence split by spaces, and keep words not in stop_words.

5fill in blank

hard

Fill all three blanks to create a dictionary of word counts excluding stopwords.

NLP

word_counts = [1](word for word in words if word not in [2])
filtered_counts = {word: count for word, count in word_counts.[3]()}

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

ACounter

Bstop_words

Citems

Ddict

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'dict' instead of 'Counter' loses counting functionality.

Using 'keys()' instead of 'items()' misses counts.

We use Counter to count words excluding stop_words. Then we create a dictionary from word_counts.items().

Practice

1. What is the main purpose of stopword removal in natural language processing?

easy

A. To correct spelling mistakes in text

B. To translate text into another language

C. To count the number of words in a sentence

D. To remove common words that do not add much meaning

Solution

1. Step 1: Understand what stopwords are

   Stopwords are common words like 'the', 'is', 'and' that usually don't add important meaning.

2. Step 2: Identify the purpose of removing stopwords

   Removing these words helps focus on meaningful words for better analysis.

3. Final Answer:

   To remove common words that do not add much meaning -> Option D

4. Quick Check:

   Stopword removal = Remove common meaningless words [OK]

Hint: Stopwords are common filler words removed to focus on meaning [OK]

Common Mistakes:

• Thinking stopword removal translates text
• Confusing stopword removal with spell checking
• Believing it counts words instead of removing them

2. Which of the following Python code snippets correctly removes stopwords from a list of words using NLTK?

easy

A. filtered_words = [w for w in words if w not in stopwords.words('english')]

B. filtered_words = [w for w in words if w in stopwords.words('english')]

C. filtered_words = stopwords.remove(words)

D. filtered_words = words.remove(stopwords.words('english'))

Solution

1. Step 1: Understand NLTK stopword removal syntax

   We keep words that are NOT in the stopwords list using a list comprehension.

2. Step 2: Check each option

   filtered_words = [w for w in words if w not in stopwords.words('english')] correctly filters out stopwords. filtered_words = [w for w in words if w in stopwords.words('english')] keeps only stopwords, which is wrong. Options C and D use invalid methods.

3. Final Answer:

   filtered_words = [w for w in words if w not in stopwords.words('english')] -> Option A

4. Quick Check:

   Keep words not in stopwords list = filtered_words = [w for w in words if w not in stopwords.words('english')] [OK]

Hint: Filter words not in stopwords list using list comprehension [OK]

Common Mistakes:

• Using 'in' instead of 'not in' to filter stopwords
• Calling non-existent methods like stopwords.remove()
• Confusing filtering logic to keep stopwords instead of removing

3. Given the code below, what is the output?

import nltk
from nltk.corpus import stopwords
nltk.download('stopwords')
words = ['this', 'is', 'a', 'test']
filtered = [w for w in words if w not in stopwords.words('english')]
print(filtered)

medium

A. ['this', 'test']

B. ['this', 'is', 'a', 'test']

C. ['test']

D. []

Solution

1. Step 1: Identify stopwords in the list

   Stopwords in English include 'this', 'is', 'a'. 'test' is not a stopword.

2. Step 2: Filter out stopwords

   The list comprehension removes 'this', 'is', 'a', leaving only 'test'.

3. Final Answer:

   ['test'] -> Option C

4. Quick Check:

   Only non-stopword 'test' remains [OK]

Hint: Remove common words; only meaningful words remain [OK]

Common Mistakes:

• Assuming all words remain after removal
• Forgetting to download stopwords corpus
• Confusing which words are stopwords

4. The following code is intended to remove stopwords from a list of words, but it raises an error. What is the problem?

from nltk.corpus import stopwords
words = ['hello', 'world']
filtered = [w for w in words if w not in stopwords('english')]
print(filtered)

medium

A. stopwords is not a function; should use stopwords.words('english')

B. The list comprehension syntax is incorrect

C. The variable 'words' is not defined

D. The print statement is missing parentheses

Solution

1. Step 1: Check how stopwords are accessed

   stopwords is a module, and stopwords.words('english') returns the list of stopwords.

2. Step 2: Identify the error in code

   The code calls stopwords('english'), which is invalid and causes an error.

3. Final Answer:

   stopwords is not a function; should use stopwords.words('english') -> Option A

4. Quick Check:

   Use stopwords.words('english') to get stopwords list [OK]

Hint: Use stopwords.words('english'), not stopwords('english') [OK]

Common Mistakes:

• Calling stopwords as a function instead of accessing .words()
• Misunderstanding list comprehension syntax
• Assuming print needs no parentheses in Python 3

5. You want to remove stopwords from a text but keep the word 'not' because it changes meaning. How can you modify the stopword list in NLTK to do this?

hard

A. Add 'not' to the stopwords list before filtering

B. Remove 'not' from the stopwords list before filtering

C. Replace 'not' with a synonym before filtering

D. Ignore stopword removal and keep all words

Solution

1. Step 1: Understand default stopwords list

   NLTK's stopwords list includes 'not', which would be removed by default.

2. Step 2: Modify stopwords list to keep 'not'

   Remove 'not' from the stopwords list before filtering to keep it in the text.

3. Final Answer:

   Remove 'not' from the stopwords list before filtering -> Option B

4. Quick Check:

   Modify stopwords list to keep important words [OK]

Hint: Delete 'not' from stopwords list to keep it in text [OK]

Common Mistakes:

• Adding 'not' to stopwords instead of removing
• Replacing words instead of modifying stopwords
• Skipping stopword removal entirely