Phase 2: Programming Fundamentals II · ~35 minutes · Python · uv

Strings and Text Processing

s[0] = "H" fails because strings never change in place. Every string method you call returns a brand-new string.

Hiring signal: Knows strings are immutable and reasons correctly about which string method call actually does what's needed

What you will learn

Use core string methods: .split(), .strip(), .join(), .replace(), and f-strings
Explain why strings are immutable and why every 'modification' actually creates a new string
Diagnose a real TypeError from attempting in-place string mutation
Parse unstructured text into structured pieces using .split()

Introduction

Type: Learn Languages: Python Prerequisites: Lesson 02 (Dictionaries and Sets) Time: ~35 minutes

Objective

Learning objectives

Use core string methods: .split(), .strip(), .join(), .replace(), and f-strings
Explain why strings are immutable and why every "modification" actually creates a new string
Diagnose a real TypeError from attempting in-place string mutation
Parse unstructured text into structured pieces using .split()

What you're building

A script (log_parser.py) that:

Takes at least 3 raw log lines in the format TIMESTAMP LEVEL MESSAGE (space-separated, message may contain spaces), hardcoded as strings
Parses each into a dictionary with keys "timestamp", "level", and "message", using .split(" ", N) with the correct limit so multi-word messages stay intact
Prints each parsed line using an f-string in a clean, readable format
Contains a comment demonstrating the s[0] = "X" TypeError and explaining, in your own words, why .replace() is the correct tool instead

What does " data science ".strip().replace(" ", "_") return?

.strip() runs first, removing only the leading and trailing whitespace: "data science" (the single space between the words stays, since .strip() only touches the outer edges). .replace(" ", "_") then runs on that result, converting every remaining space — just the one in the middle — to an underscore: "data_science". Order matters here: running .replace() before .strip() would still work in this case, but chaining methods left-to-right, each operating on the previous result, is the general pattern worth internalizing.

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