Word frequency analysis in Data Analysis Python - Time & Space Complexity

Time Complexity: Word frequency analysis

O(n)

Understanding Time Complexity

We want to know how the time to count words grows as the text gets longer.

How does the work change when we have more words to analyze?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

text = "this is a sample text with several words this is a test"
words = text.split()
word_counts = {}
for word in words:
    if word in word_counts:
        word_counts[word] += 1
    else:
        word_counts[word] = 1

This code counts how many times each word appears in the text.

Identify Repeating Operations

• Primary operation: Looping through each word in the list.
• How many times: Once for every word in the text.

How Execution Grows With Input

As the number of words grows, the loop runs more times, increasing work linearly.

Input Size (n)	Approx. Operations
10	About 10 checks and updates
100	About 100 checks and updates
1000	About 1000 checks and updates

Pattern observation: The work grows directly with the number of words.

Final Time Complexity

Time Complexity: O(n)

This means the time to count words grows in direct proportion to the number of words.

Common Mistake

[X] Wrong: "Checking if a word is in the dictionary takes time proportional to the number of words counted so far."

[OK] Correct: Dictionary lookups are very fast and do not grow with the number of words counted, so each check is almost constant time.

Interview Connect

Understanding how counting words scales helps you explain how your code handles larger texts efficiently.

Self-Check

"What if we used a list instead of a dictionary to store counts? How would the time complexity change?"

Practice

1.

What does word frequency analysis help us understand about a text?

easy

A. How many times each word appears in the text

B. The total number of characters in the text

C. The number of sentences in the text

D. The length of the longest word in the text

Solution

1. Step 1: Understand the purpose of word frequency analysis

   Word frequency analysis counts how often each word appears in a text.

2. Step 2: Compare options with the purpose

   Only How many times each word appears in the text matches this purpose exactly.

3. Final Answer:

   How many times each word appears in the text -> Option A

4. Quick Check:

   Word frequency = count of each word [OK]

Hint: Focus on counting words, not characters or sentences [OK]

Common Mistakes:

• Confusing word count with character count
• Thinking it counts sentences
• Assuming it finds longest word length

2.

Which Python code correctly imports the tool to count word frequencies?

?

easy

A. import wordcount from collections

B. import Counter from collections

C. from collections import wordfreq

D. from collections import Counter

Solution

1. Step 1: Recall the correct import syntax for Counter

   The correct way is 'from collections import Counter'.

2. Step 2: Check each option's syntax

   Only from collections import Counter uses valid Python import syntax and correct name.

3. Final Answer:

   from collections import Counter -> Option D

4. Quick Check:

   Correct import = from collections import Counter [OK]

Hint: Remember 'from collections import Counter' exactly [OK]

Common Mistakes:

• Using wrong import order
• Trying to import non-existent modules
• Using invalid syntax like 'import Counter from'

3.

What is the output of this code?

from collections import Counter
text = "apple banana apple orange banana apple"
words = text.split()
count = Counter(words)
print(count)

medium

A. Counter({'apple': 1, 'banana': 1, 'orange': 1})

B. {'apple': 3, 'banana': 2, 'orange': 1}

C. Counter({'apple': 3, 'banana': 2, 'orange': 1})

D. {'apple': 1, 'banana': 2, 'orange': 3}

Solution

1. Step 1: Split the text into words

   The text splits into ['apple', 'banana', 'apple', 'orange', 'banana', 'apple'].

2. Step 2: Count each word using Counter

   'apple' appears 3 times, 'banana' 2 times, 'orange' 1 time.

3. Final Answer:

   Counter({'apple': 3, 'banana': 2, 'orange': 1}) -> Option C

4. Quick Check:

   Count matches word occurrences [OK]

Hint: Count words carefully; Counter shows counts with 'Counter()' [OK]

Common Mistakes:

• Ignoring Counter output format
• Miscounting word frequencies
• Confusing dictionary and Counter print styles

4.

Find the error in this code that tries to count words:

from collections import Counter
text = "Hello world! Hello"
words = text.split()
count = Counter(words.lower())
print(count)

medium

A. Cannot call lower() on a list; should call on each word

B. Missing import statement for Counter

C. split() should be split(' ')

D. Counter cannot count strings

Solution

1. Step 1: Identify the type of 'words'

   words is a list of strings after split().

2. Step 2: Check the use of lower()

   Calling words.lower() is invalid because lists have no lower() method; lower() applies to strings.

3. Final Answer:

   Cannot call lower() on a list; should call on each word -> Option A

4. Quick Check:

   lower() works on strings, not lists [OK]

Hint: Apply lower() to each word, not the whole list [OK]

Common Mistakes:

• Calling lower() on list instead of string
• Ignoring error messages
• Thinking split() needs arguments

5.

You want to count words ignoring case and punctuation in this text:

text = "Hello, hello! How are you? Are you fine?"

Which code correctly cleans and counts words?

hard

A. import re
from collections import Counter
words = re.findall(r'\w+', text)
count = Counter(words)
print(count)

B. import string
from collections import Counter
clean_text = text.lower().translate(str.maketrans('', '', string.punctuation))
words = clean_text.split()
count = Counter(words)
print(count)

C. from collections import Counter
words = text.split()
count = Counter(words)
print(count)

D. from collections import Counter
words = text.lower().split()
count = Counter(words)
print(count)

Solution

1. Step 1: Understand the need to remove punctuation and lowercase

   To count words ignoring case and punctuation, text must be lowercased and punctuation removed.

2. Step 2: Check each option's approach

   import string from collections import Counter clean_text = text.lower().translate(str.maketrans('', '', string.punctuation)) words = clean_text.split() count = Counter(words) print(count) uses string.punctuation with translate() to remove punctuation and lowercases text before splitting, which is correct.

3. Final Answer:

   Code that lowercases and removes punctuation before counting -> Option B

4. Quick Check:

   Clean text + Counter = accurate word frequency [OK]

Hint: Remove punctuation and lowercase before splitting words [OK]

Common Mistakes:

• Not removing punctuation
• Counting words with mixed cases separately
• Using split() without cleaning text