A3.1 Arrays, records and tuples answers

Arrays in one, two and three dimensions, records and fields, tuples and lists, and static structures.

Questions 6 marks in all

[1 mark]Which of these describes an array?

AA fixed-size, indexed collection of elements of the same data type
BA collection of named fields that can have different data types
CAn immutable ordered collection of values of any types
DA collection of key-value pairs

Answer: A. An array has a fixed size, one data type and an index for each element. The others describe a record, a tuple and a dictionary.

[1 mark]What does this program print?

scans = [[[1, 2], [3, 4]], [[5, 6], [7, 8]], [[9, 10], [11, 12]]]
print(len(scans), len(scans[0]), len(scans[0][0]))
print(scans[2][0][1])

Answer:

3 2 2
10

The array is 3 by 2 by 2. scans[2] is the third block, [0] its first row, and [1] the second element of that row: 10.

[1 mark]A two-dimensional array with 8 columns is stored in row-major order, starting at position 0. At what position is element [3][5]?

Answer: 29. Row-major order stores whole rows one after another, so [r][c] is at r x columns + c = 3 x 8 + 5 = 29.

[1 mark]BugBot's position() returns a tuple rather than a list. Which is the best reason?

AA tuple is immutable, so the x and y cannot be changed separately by mistake
BA tuple can hold more values than a list
CA tuple is sorted automatically
DA tuple can only hold numbers

Answer: A. A tuple cannot be changed once made, which suits a pair of values that belong together.

[1 mark]What does this program print?

grid = [[0] * 2] * 3
grid[1][0] = 7
print(grid)

Answer:

[[7, 0], [7, 0], [7, 0]]

Multiplying the outer list makes three references to the same row, so changing one row changes all three.

[1 mark]Which of these are true of a record?

Tick every answer that is true.

AIt groups fields that can have different data types
BEach field has its own name
CAll of its fields must have the same data type
DA file is often a collection of records with the same fields

Answer: A, B, D. A record describes one thing with named fields of any types, and a file of records keeps many of them on storage.

The task: the depth cube

scans is a 3D array of three saved 4 by 4 depth scans, indexed scans[scan][row][col], each value a whole number of centimetres from 1 to 100. Using nested loops and len(), print: 1. size: 3 x 4 x 4, working the three sizes out with len(). 2. For each scan s from 0 to 2, scan <s> average: <mean>, the mean of its 16 values rounded to 1 decimal place with round(value, 1), for example scan 0 average: 56.0. 3. nearest: <cm> cm at [<scan>][<row>][<col>], the smallest value and its three indexes. The smallest value appears only once. The robot does not move.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

# scans[scan][row][col]: three saved 4 by 4 depth scans, in cm
scans = [
    [[45, 44, 43, 50], [40, 38, 39, 41], [60, 58, 57, 59], [80, 79, 81, 82]],
    [[44, 42, 41, 49], [39, 36, 12, 40], [61, 57, 56, 60], [78, 80, 80, 83]],
    [[46, 43, 42, 51], [41, 37, 38, 42], [59, 59, 55, 58], [81, 78, 82, 80]],
]

The hint students can ask for: Three nested loops reach every reading: the outer one picks the scan, the middle one the row, the inner one the column. Keep a running total for each scan, and keep the smallest value seen so far together with the three indexes where you saw it.

A solution

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

# scans[scan][row][col]: three saved 4 by 4 depth scans, in cm
scans = [
    [[45, 44, 43, 50], [40, 38, 39, 41], [60, 58, 57, 59], [80, 79, 81, 82]],
    [[44, 42, 41, 49], [39, 36, 12, 40], [61, 57, 56, 60], [78, 80, 80, 83]],
    [[46, 43, 42, 51], [41, 37, 38, 42], [59, 59, 55, 58], [81, 78, 82, 80]],
]

print("size:", len(scans), "x", len(scans[0]), "x", len(scans[0][0]))

best = scans[0][0][0]
where = [0, 0, 0]
for s in range(len(scans)):
    total = 0
    count = 0
    for r in range(len(scans[s])):
        for c in range(len(scans[s][r])):
            value = scans[s][r][c]
            total = total + value
            count = count + 1
            if value < best:
                best = value
                where = [s, r, c]
    print(f"scan {s} average: {round(total / count, 1)}")
print(f"nearest: {best} cm at [{where[0]}][{where[1]}][{where[2]}]")

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.

What this lesson is about

Questions 6 marks in all

The task: the depth cube

A solution