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Lesson 148 of 1455

Python Lists & Dicts — The Two Collections You Can't Live Without

Lists are ordered rows; dicts are labeled lookups. You'll use both to solve a real problem, and catch the mistakes autocomplete makes.

Builders · AI-Assisted Coding · ~18 min read

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Two containers, two jobs

A list holds items in order, accessed by index. A dict holds values under keys, like a labeled drawer system. Pick the one that matches how you will look things up later.

Always use .get() when a key might be missing — it returns None (or a default) instead of crashing.

python

# Lists: ordered, indexed by number scores = [87, 92, 76, 95] print(scores[0]) # 87 print(scores[-1]) # 95 (last item) scores.append(88) # add to the end print(len(scores)) # 5 # Dicts: accessed by key student = { "name": "Ava", "grade": 8, "subjects": ["math", "art"], } print(student["name"]) # Ava student["age"] = 13 # add a new key print(student.get("age", 0))# safe access with default

List comprehension — Python's superpower

Comprehensions read left-to-right: 'the result, for each thing, if some condition.'

python

# Without comprehension doubled = [] for n in [1, 2, 3, 4]: doubled.append(n * 2) # With comprehension — one line doubled = [n * 2 for n in [1, 2, 3, 4]] # With a filter evens = [n for n in range(10) if n % 2 == 0] # Dict comprehension squares = {n: n*n for n in range(1, 6)} # {1: 1, 2: 4, 3: 9, 4: 16, 5: 25}

Mini-exercise: class roster

1Build a list of 4 student dicts, each with name and score
2Write a comprehension that returns just the names of students with score >= 80
3Write another that returns a dict mapping name -> grade letter (A/B/C/F)
4Have your AI assistant draft both — then check if it handled ties and missing keys

Compare the options

Use a list when	Use a dict when
Order matters	You look up by key, not position
You iterate in order	You need O(1) lookup
Example: queue of tasks	Example: user -> profile

A pattern you'll use forever: dict as counter, then sorted by value.

python

# Real mini-project: word frequency counter text = "the cat sat on the mat and the dog sat too" words = text.split() freq = {} for word in words: freq[word] = freq.get(word, 0) + 1 # Sort by count, descending top = sorted(freq.items(), key=lambda pair: pair[1], reverse=True) print(top[:3]) # [('the', 3), ('sat', 2), ('cat', 1)]

Key terms in this lesson

Big idea: most real programs are lists of dicts or dicts of lists. Master these two and 80% of Python feels natural.

End-of-lesson quiz

Check what stuck

8 questions · Score saves to your progress.

Lesson help

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Python Lists & Dicts — The Two Collections You Can't Live Without

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Questions are best handled with a grown-up here.

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