Records, dictionaries and tuples

Records (spec 8.1.2)

In the Programming Language Subset a record is a list holding fields of mixed types, with the same number of fields in every record. A table of records is a 2D list.

pupils = [["Ava", 16, 72.5, True],
          ["Raj", 17, 64.0, False]]
for p in pupils:
    if p[3]:
        print(p[0], "has paid; mark", p[2])
NAME = 0
print(pupils[1][NAME])

Output:

Ava has paid; mark 72.5
Raj

Named constants for field positions (NAME = 0) make record code readable.

Dictionaries

A dictionary stores key: value pairs. Keys are unique; values are found by key, not by position. Dictionaries are mutable and keep insertion order.

stock = {"pen": 40, "pad": 12}
stock["ruler"] = 8
stock["pen"] = stock["pen"] - 5
print(stock["pen"], len(stock))
print(stock.get("glue", 0))
for item, qty in stock.items():
    print(item, qty)
print(list(stock.keys()))

Output:

35 3
0
pen 35
pad 12
ruler 8
['pen', 'pad', 'ruler']

d[key] raises KeyError for a missing key; d.get(key, default) returns the default instead.

Tuples

A tuple is ordered like a list but immutable: it cannot be changed after creation. Use one for fixed data such as coordinates or the days of the week. A one-item tuple needs a trailing comma.

point = (3, 7)
x, y = point
print(x + y, point[1])
single = (5,)
print(type(single), type((5)))

Output:

10 7
<class 'tuple'> <class 'int'>

Converting between structures

list(), tuple(), set() and dict() convert between structures. Converting to a set removes duplicates (and loses order).

letters = list("noon")
print(letters)
print(tuple(letters))
print(sorted(set(letters)))
pairs = [("a", 1), ("b", 2)]
print(dict(pairs))
print(list({"x": 1, "y": 2}.values()))

Output:

['n', 'o', 'o', 'n']
('n', 'o', 'o', 'n')
['n', 'o']
{'a': 1, 'b': 2}
[1, 2]

Worked example

Choose a structure for: (a) a fixed set of RGB values (b) a phone book looking up numbers by name (c) the names of students in a queue that changes.

1. (a) fixed and must not change: tuple
2. (b) look up by a unique key: dictionary
3. (c) ordered and changing: list

Answer: tuple, dictionary, list

Check yourself

1. What is displayed when this code runs? ```python ages = {"Ali": 15, "Bo": 17} ages["Ali"] = ages["Ali"] + 1 ages["Cy"] = 14 print(ages["Ali"], len(ages)) ```
   Show answer

   16 3. Ali's age becomes 16 and a third key is added: 16 3.

2. What is displayed when this code runs? ```python prices = {"tea": 1.5, "cake": 2.25} print(prices.get("coffee", "n/a")) ```
   • KeyError
   • n/a
   • None
   • coffee
   Show answer

   n/a. get() returns the default when the key is missing: n/a.

3. Which exception is raised when this code runs? ```python colour = (255, 128, 0) colour[0] = 200 ```
   • IndexError
   • ValueError
   • KeyError
   • TypeError
   Show answer

   TypeError. Tuples are immutable, so assigning to an item raises TypeError.

4. What is displayed when this code runs? ```python books = [["Dune", 1965, 4.5], ["Emma", 1815, 4.1], ["Kim", 1901, 3.9]] best = books[0] for b in books: if b[2] > best[2]: best = b print(best[0]) ```
   Show answer

   Dune. The loop keeps the record with the highest rating (field 2): Dune.

5. What is displayed when this code runs? ```python tally = {"a": 3, "b": 5, "c": 2} total = 0 for key, value in tally.items(): if key != "b": total = total + value print(total) ```
   Show answer

   5. Values for a and c are added: 3 + 2 = 5.

6. What is displayed when this code runs? ```python words = ["red", "blue", "red", "green", "blue"] print(len(set(words))) ```
   Show answer

   3. Converting to a set removes duplicates, leaving 3 different words: 3.

7. A program stores each country's capital so it can be looked up quickly by country name. Which structure is most suitable?
   • Tuple
   • Set
   • Dictionary
   • Two-dimensional list
   Show answer

   Dictionary. Each country is a unique key and the capital is its value.

8. This code should display ['pear'] but it does not. Which line number contains the error? ``` 1 inventory = {"apple": 3, "pear": 0, "plum": 5} 2 empty = [] 3 for fruit in inventory: 4 if inventory[fruit] == 0: 5 empty.append(inventory[fruit]) 6 print(empty) ```
   Show answer

   5. Line 5 appends the value (0) instead of the key. It should be empty.append(fruit).

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Computer Science specification (XCP01/YCP01), first teaching 2026
• Pearson IAL Computer Science Unit 2 Programming Language Subset (PLS), Python 3
• Pearson IAL Computer Science IAS sample assessment materials (Unit 2 paper and mark scheme)

Last updated 8 November 2026. Spotted a mistake? Tell us.