#0003

Count Word Frequency

EasyPython12 min10 XP

Problem

Given a sentence, return a dictionary counting lowercase words. Words are separated by spaces. Strip basic punctuation from the beginning and end of each word.

Why This Matters

Frequency maps are everywhere: text analysis, feature engineering, logs, surveys, and classic interview problems.

Function Signature

def word_counts(text):

Examples

Example 1

Inputtext = "ML is fun, ml is useful!"

Output{"ml": 2, "is": 2, "fun": 1, "useful": 1}

"ML" and "ml" count as the same word after lowercasing.

Constraints

Treat words case-insensitively.
Strip punctuation characters .,!?:; from word edges.
Return a normal Python dictionary.

CodePython

def word_counts(text):
    """
    Input example: text = "ML is fun, ml is useful!"
    Return: {"ml": 2, "is": 2, "fun": 1, "useful": 1}
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    counts = {}
    # write your code here
    return counts

def word_counts(text):
    """
    Input example: text = "ML is fun, ml is useful!"
    Return: {"ml": 2, "is": 2, "fun": 1, "useful": 1}
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    counts = {}
    # write your code here
    return counts

Visible browser tests run here when available.

Testcases2 visible / 4 hidden categories

Case and punctuation

Input"ML is fun, ml is useful!"

Expected{"ml": 2, "is": 2, "fun": 1, "useful": 1}

Lowercasing and stripping punctuation prevent duplicate-looking keys.

Empty text

Input""

Expected{}

No words means no counts.

Hidden Test Categories

Repeated punctuationSingle wordMixed capitalizationExtra spaces