← VCE Software Development U3O1

Choosing a data structure

Pick the structure by how you use the data: a list for an ordered collection, a dict for lookup by key, and a tuple for a fixed group of values that should not change.

Part of: VCE Software Development - U3O1 - Programming — Module 102, VCE Software Development - U4O1 - development and evaluation — Module 46, VCE Data Analytics - U3O2 - Analysis and Design — Module 55, 9-10 Digital Technologies: Data structures in Python — Module 10, VCE Applied Computing U1O2 Programming: Data structures — Module 10, QCE Digital Solutions Unit 2 — Data-Driven Problems and Solution Requirements — Module 16, QCE Digital Solutions Unit 2 — Data and Programming Techniques — Module 27, QCE Digital Solutions Unit 3 — Interactions Between Users, Data and Digital Systems — Module 52, QCE Digital Solutions Unit 3 — Real-World Problems and Solution Requirements — Module 44, QCE Digital Solutions Unit 4 — Digital Methods for Exchanging Data — Module 122, WA Computer Science ATAR Year 11 — Control Structures and Data Types — Module 62, WA Computer Science ATAR Year 11 — Data Management — Module 11

Understanding Choosing a data structure in VCE Software Development U3O1

What it does: Round brackets make a tuple. An x, y point is a fixed pair of values that should not be changed. Each dictionary is one record with named fields, and the list keeps the records in order. A list is an ordered collection that can grow or change, which suits a series of scores. A dictionary lets you look up a value by its key, so phones["Mia"] gives the number straight away. Tuples are fixed, so there is no append(). A list is better when items are added or removed. Looking up a word by name and updating its count is a key lookup job, which dictionaries do best.

Common mistakes: Scores is an ordered collection that grows, so it needs a list. A tuple is fixed and has no append(), so line 2 would fail. It should be scores = [85, 92, 78]. A dictionary is looked up by key, not by position number. There is no key 0, so this causes a KeyError. It should be print(ages["Mia"]).

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This module's key terms are also in a printable crossword (Modules 98–113) and word search (Modules 72–113).

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Learning
Duo
You need to store the scores of 30 students in the order they were entered. Which structure fits best?
A list: scores = [72, 85, 90]A tuple of student names stored as keys
Matches
colours = ("red", "blue") colours.append("green")
↓
A tuple cannot be changed, so append() causes an error. Use a list if the values need to grow.
Applying
Meanings
Term
list, dict, tuple
Meaning
Pick the structure by how you use the data: a list for an ordered collection, a dict for lookup by key, and a tuple for a fixed group of values that should not change.
Practice
point = (3, 7)
Testing
Gaps
print(["Mia"])
Check Answer
Spots
1scores = (85, 92, 78)
2scores.append(90)

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Quick Questions

1. You need to store the scores of 30 students in the order they were entered. Which structure fits best?
A. A list: scores = [72, 85, 90] ✓
B. A tuple of student names stored as keys
A list is an ordered collection that can grow or change, which suits a series of scores.
2. You want to look up a student's phone number using their name. Which structure fits best?
A. A dictionary: phones = {"Mia": "0412"} ✓
B. A list: phones = ["Mia", "0412"]
A dictionary lets you look up a value by its key, so phones["Mia"] gives the number straight away.
3. Which is the best choice to store the fixed latitude and longitude of Melbourne, which should never change?
A. A list: location = [-37.81, 144.96]
B. A tuple: location = (-37.81, 144.96) ✓
A tuple is a fixed group of values that cannot be changed after it is created, which suits coordinates.
4. True or false: A tuple is a good choice for a shopping list that you keep adding items to.
A. False: use a list, because tuples cannot be changed ✓
B. True: tuples are the best for growing collections
Tuples are fixed, so there is no append(). A list is better when items are added or removed.
5. You are storing 3 students, each with a name, a year level and a house. Which structure is best for the whole collection?
A. A single tuple holding every name, year and house mixed together
B. A list of dictionaries, one dictionary per student ✓
A list of dictionaries stores records: the list keeps the order of students and each dictionary labels the fields with keys.

Match the Pairs

scores = [85, 92, 78] → A list: an ordered collection of values that you can loop through and change
ages = {"Mia": 15, "Leo": 14} → A dictionary: look up a value quickly using its key
A tuple is best for a fixed group of values that should not change, like a map point → point = (10, 25)
A list of dictionaries is best for storing records, where each dictionary holds one item's details → students = [{"name": "Mia", "year": 9}]
colours = ("red", "blue") colours.append("green") → A tuple cannot be changed, so append() causes an error. Use a list if the values need to grow.
phone = ["Mia", "0412 345 678"] print(phone["Mia"]) → A list is indexed by position, not by name. To look up by a key, use a dictionary instead.

Key Term

list, dict, tuple
Pick the structure by how you use the data: a list for an ordered collection, a dict for lookup by key, and a tuple for a fixed group of values that should not change.

Fill in the Blanks

1. ages = {"Mia": 15, "Leo": 16} print(ages["Mia"])
Hint: Type the name of the variable holding the data you look up by key.
2. rgb = tuple([255, 128, 0])
Hint: Type the function that makes a fixed group of values that cannot change.
3. kids = [{"age": 9}, {"age": 12}] print(kids[0]["age"])
Hint: Type the index of the first record in the list of dictionaries.
4. stock = {"apple": 4, "pear": 7} print(stock.get("pear"))
Hint: Type the dictionary method that looks up a value using its key.
5. names = ["Zoe", "Sam", "Ali"] for name in names: print(name)
Hint: Type the variable for the ordered collection to loop through.

Spot the Error

Question 1
A scores = (85, 92, 78)
B scores.append(90)
C print(scores[0])
Line A — Scores is an ordered collection that grows, so it needs a list. A tuple is fixed and has no append(), so line 2 would fail. It should be scores = [85, 92, 78].
Question 2
A ages = {"Mia": 15, "Leo": 16}
B print(ages[0])
C print(len(ages))
Line B — A dictionary is looked up by key, not by position number. There is no key 0, so this causes a KeyError. It should be print(ages["Mia"]).
Question 3
A RGB = (255, 128, 0)
B RGB[0] = 200
C print(red)
Line B — A tuple holds a fixed group of values and cannot be changed after it is created. Assigning to RGB[0] causes a TypeError. The line should read the value: red = RGB[0].
Question 4
A student = ["name": "Ava"]
B student["year"] = 10
C print(student["name"])
Line A — Key: value pairs belong in a dictionary, which uses curly brackets. Square brackets make a list, which cannot hold key: value pairs. It should be student = {"name": "Ava"}.
Question 5
A dog = {"name": "Rex"}
B cat = {"name", "Kit"}
C pets = [dog, cat]
D print(pets[1]["name"])
Line B — A dictionary needs a colon between the key and the value. A comma makes a set of two strings instead of a record. It should be cat = {"name": "Kit"}.

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