100 Julia Exercises with Solutions

A set of introductory exercises for Julia. Based on 100 NumPy Exercises.

In order to generate this file:

Clone the repository (Or download).
Activate and instantiate the project.
Run:

using Literate;
Literate.markdown("Julia100Exercises.jl", name = "README", execute = true, flavor = Literate.CommonMarkFlavor());

Remark: Tested with Julia 1.8.2.

To Do:

Reevaluate the difficulty level of each question.

using Literate;
using LinearAlgebra;
using Statistics;
using Dates;
using DelimitedFiles;
using UnicodePlots;
using Random;
using Tullio;
using StaticKernels;

Question 001

Import the LinearAlgebra package under the name LA. (★☆☆)

import LinearAlgebra as LA;

Question 002

Print the version of Julia. (★☆☆)

println(VERSION);

1.8.2

Question 003

Create a non initialized vector of size 10 of Float64. (★☆☆)

vA = Vector{Float64}(undef, 10)

10-element Vector{Float64}:
 0.0
 1.314224183e-315
 1.314224183e-315
 0.0
 1.314248214e-315
 1.31427351e-315
 0.0
 1.314248214e-315
 1.314106556e-315
 0.0

Which is equivalent of

vA = Array{Float64, 1}(undef, 10)

10-element Vector{Float64}:
 0.0
 1.314248214e-315
 1.314106556e-315
 0.0
 1.314248214e-315
 1.314106556e-315
 0.0
 1.314248214e-315
 1.314106556e-315
 0.0

Question 004

Find the memory size of any array. (★☆☆)

sizeof(vA)

Question 005

Show the documentation of the + (Add) method. (★☆☆)

@doc +

+(x, y...)

Addition operator. x+y+z+... calls this function with all arguments, i.e. +(x, y, z, ...).

Examples

julia> 1 + 20 + 4
25

julia> +(1, 20, 4)
25

dt::Date + t::Time -> DateTime

The addition of a Date with a Time produces a DateTime. The hour, minute, second, and millisecond parts of the Time are used along with the year, month, and day of the Date to create the new DateTime. Non-zero microseconds or nanoseconds in the Time type will result in an InexactError being thrown.

Question 006

Create a vector of zeros of size 10 but the fifth value which is 1. (★☆☆)

vA = zeros(10);
vA[5] = 1.0;
vA

10-element Vector{Float64}:
 0.0
 0.0
 0.0
 0.0
 1.0
 0.0
 0.0
 0.0
 0.0
 0.0

Question 007

Create a vector with values ranging from 7 to 12. (★☆☆)

vA = 7:12

7:12

The above is efficient type. In order to explicitly create a vector:

vA = collect(7:12)

6-element Vector{Int64}:
  7
  8
  9
 10
 11
 12

Question 008

Reverse a vector (first element becomes last). (★☆☆)

vA = collect(1:3);
vB = vA[end:-1:1];
vB

3-element Vector{Int64}:
 3
 2
 1

Alternative 001:

vB = reverse(vA);

Alternative 002 (In place):

reverse!(vA);
vA

3-element Vector{Int64}:
 3
 2
 1

Question 009

Create a 3x3 matrix with values ranging from 0 to 8. (★☆☆)

mA = reshape(0:8, 3, 3)

3×3 reshape(::UnitRange{Int64}, 3, 3) with eltype Int64:
 0  3  6
 1  4  7
 2  5  8

Another way would be:

mA = Matrix{Float64}(undef, 3, 3);
mA[:] = 0:8;

Question 010

Find indices of non zero elements from [1, 2, 0, 0, 4, 0]. (★☆☆)

findall(!iszero, [1, 2, 0, 0, 4, 0])

3-element Vector{Int64}:
 1
 2
 5

Question 011

Create a 3x3 identity matrix. (★☆☆)

mA = I(3)

3×3 LinearAlgebra.Diagonal{Bool, Vector{Bool}}:
 1  ⋅  ⋅
 ⋅  1  ⋅
 ⋅  ⋅  1

An alternative method (Explicit matrix) would be:

mA = Matrix(I, 3, 3) #<! For Float64: Matrix{Float64}(I, 3, 3)

3×3 Matrix{Bool}:
 1  0  0
 0  1  0
 0  0  1

Question 012

Create a 2x2x2 array with random values. (★☆☆)

mA = randn(2, 2, 2)

2×2×2 Array{Float64, 3}:
[:, :, 1] =
 -1.34961   -0.225955
  0.116267   1.03802

[:, :, 2] =
  0.737672  2.2642
 -0.839961  0.218133

Question 013

Create a 5x5 array with random values and find the minimum and maximum values. (★☆☆)

mA = rand(5, 5);
minVal = minimum(mA)

0.02305480092860468

maxVal = maximum(mA)

0.9708814560256962

Using extrema() one could get both values at once:

minVal, maxVal = extrema(mA);

Question 014

Create a random vector of size 30 and find the mean value. (★☆☆)

meanVal = mean(randn(30))

-0.04218053798839749

Question 015

Create a 2d array with 1 on the border and 0 inside. (★☆☆)

mA = zeros(4, 4);
mA[:, [1, end]] .= 1;
mA[[1, end], :] .= 1;
mA

4×4 Matrix{Float64}:
 1.0  1.0  1.0  1.0
 1.0  0.0  0.0  1.0
 1.0  0.0  0.0  1.0
 1.0  1.0  1.0  1.0

An alternative way (Different dimensions):

mA = ones(4, 5);
mA[2:(end - 1), 2:(end - 1)] .= 0;

Using one line code:

mA = zeros(4, 5);
mA[[LinearIndices(mA)[cartIdx] for cartIdx in CartesianIndices(mA) if (any(cartIdx.I .== 1) || cartIdx.I[1] == size(mA, 1) || cartIdx.I[2] == size(mA, 2))]] .= 1;

By Tomer Arnon:

numRows = 5;
numCols = 4;
mA = Int[ii ∈ (1, numRows) || jj ∈ (1, numCols) for ii in 1:numRows, jj in 1:numCols];

Question 016

Add a border of zeros around the array. (★☆☆)

mB = zeros(size(mA) .+ 2);
mB[2:(end - 1), 2:(end - 1)] = mA;
mB

7×6 Matrix{Float64}:
 0.0  0.0  0.0  0.0  0.0  0.0
 0.0  1.0  1.0  1.0  1.0  0.0
 0.0  1.0  0.0  0.0  1.0  0.0
 0.0  1.0  0.0  0.0  1.0  0.0
 0.0  1.0  0.0  0.0  1.0  0.0
 0.0  1.0  1.0  1.0  1.0  0.0
 0.0  0.0  0.0  0.0  0.0  0.0

Question 017

Evaluate the following expressions. (★☆☆)

0 * NaN

NaN

NaN == NaN

false

Inf > NaN

false

NaN - NaN

NaN

NaN in [NaN]

false

0.3 == 3 * 0.1

false

Question 018

Create a 5x5 matrix with values [1, 2, 3, 4] just below the diagonal. (★☆☆)

mA = diagm(5, 5, -1 => 1:4)

5×5 Matrix{Int64}:
 0  0  0  0  0
 1  0  0  0  0
 0  2  0  0  0
 0  0  3  0  0
 0  0  0  4  0

Question 019

Create a 8x8 matrix and fill it with a checkerboard pattern. (★☆☆)

mA = zeros(8, 8);
mA[2:2:end, 1:2:end] .= 1;
mA[1:2:end, 2:2:end] .= 1;
mA

8×8 Matrix{Float64}:
 0.0  1.0  0.0  1.0  0.0  1.0  0.0  1.0
 1.0  0.0  1.0  0.0  1.0  0.0  1.0  0.0
 0.0  1.0  0.0  1.0  0.0  1.0  0.0  1.0
 1.0  0.0  1.0  0.0  1.0  0.0  1.0  0.0
 0.0  1.0  0.0  1.0  0.0  1.0  0.0  1.0
 1.0  0.0  1.0  0.0  1.0  0.0  1.0  0.0
 0.0  1.0  0.0  1.0  0.0  1.0  0.0  1.0
 1.0  0.0  1.0  0.0  1.0  0.0  1.0  0.0

By Tomer Arnon (https://github.com/tomerarnon):

mA = Int[isodd(ii + jj) for ii in 1:8, jj in 1:8];

Question 020

Convert the linear index 100 to a Cartesian Index of a size (6,7,8). (★☆☆)

mA = rand(6, 7, 8);
cartIdx = CartesianIndices(mA)[100]; #<! See https://discourse.julialang.org/t/14666
mA[cartIdx] == mA[100]

true

Question 021

Create a checkerboard 8x8 matrix using the repeat() function. (★☆☆)

mA = repeat([0 1; 1 0], 4, 4)

8×8 Matrix{Int64}:
 0  1  0  1  0  1  0  1
 1  0  1  0  1  0  1  0
 0  1  0  1  0  1  0  1
 1  0  1  0  1  0  1  0
 0  1  0  1  0  1  0  1
 1  0  1  0  1  0  1  0
 0  1  0  1  0  1  0  1
 1  0  1  0  1  0  1  0

Question 022

Normalize a 4x4 random matrix. (★☆☆)

mA = rand(4, 4);
mA .= (mA .- mean(mA)) ./ std(mA) #<! Pay attention that `@.` will yield error (`std()` and `mean()`)

4×4 Matrix{Float64}:
 -1.0589     0.696722    0.474169   0.0807478
  1.33922    1.23329    -0.559528  -1.24305
 -0.652489   0.919517    1.16875   -1.49035
 -0.0565761  0.0550669   0.711533  -1.61812

Question 023

Create a custom type that describes a color as four unsigned bytes (RGBA). (★☆☆)

struct sColor
    R::UInt8;
    G::UInt8;
    B::UInt8;
    A::UInt8;
end

sMyColor = sColor(rand(UInt8, 4)...)

Main.var"##312".sColor(0xb0, 0xae, 0x35, 0x81)

Question 024

Multiply a 2x4 matrix by a 4x3 matrix. (★☆☆)

mA = rand(2, 4) * randn(4, 3)

2×3 Matrix{Float64}:
 -1.47402   -1.10919   -1.22996
 -0.745242   0.188422  -3.19746

Question 025

Given a 1D array, negate all elements which are between 3 and 8, in place. (★☆☆)

vA = rand(1:10, 8);
map!(x -> ((x > 3) && (x < 8)) ? -x : x, vA, vA)

8-element Vector{Int64}:
 -6
  3
  1
 -4
  3
 -4
 -5
 -4

Julia allows Math like notation as well (See Q0027):

vA = rand(1:10, 8);
map!(x -> 3 < x < 8 ? -x : x, vA, vA)

8-element Vector{Int64}:
 -7
 -6
 -4
  1
 -6
 10
  8
 -5

Using logical indices one could use:

vA[3 .< vA .< 8] .*= -1;

Question 026

Sum the array 1:4 with initial value of -10. (★☆☆)

sum(1:4, init = -10)

Question 027

Consider an integer vector vZ validate the following expressions. (★☆☆)

vZ .^ vZ
2 << vZ >> 2
vZ <- vZ
1im * vZ
vZ / 1 / 1
vZ < Z > Z

vZ = rand(1:10, 3);

vZ .^ vZ

3-element Vector{Int64}:
 387420489
 387420489
     46656

try
    2 << vZ >> 2
catch e
    println(e)
end

MethodError(<<, (2, [9, 9, 6]), 0x0000000000007f1f)

vZ <- vZ

false

1im * vZ

3-element Vector{Complex{Int64}}:
 0 + 9im
 0 + 9im
 0 + 6im

vZ / 1 / 1

3-element Vector{Float64}:
 9.0
 9.0
 6.0

vZ < vZ > vZ

false

Question 028

Evaluate the following expressions. (★☆☆)

[0] ./ [0]

1-element Vector{Float64}:
 NaN

try
    [0] .÷ [0]
catch e
    println(e)
end

DivideError()

try
    convert(Float, convert(Int, NaN))
catch e
    println(e)
end

InexactError(:Int64, Int64, NaN)

Question 029

Round away from zero a float array. (★☆☆)

vA = randn(10);
map(x -> x > 0 ? ceil(x) : floor(x), vA)

10-element Vector{Float64}:
  1.0
 -1.0
 -1.0
 -1.0
  2.0
 -1.0
  2.0
  2.0
  2.0
  1.0

Question 030

Find common values between two arrays. (★☆☆)

vA = rand(1:10, 6);
vB = rand(1:10, 6);

vA[findall(in(vB), vA)]

2-element Vector{Int64}:
 7
 5

Question 031

Suppress Julia's warnings. (★☆☆)

One could use Suppressor.jl.

Question 032

Compare sqrt(-1) and sqrt(-1 + 0im). (★☆☆)

try
    sqrt(-1)
catch e
    println(e)
end

DomainError(-1.0, "sqrt will only return a complex result if called with a complex argument. Try sqrt(Complex(x)).")

sqrt(-1 + 0im)

0.0 + 1.0im

Question 033

Display yesterday, today and tomorrow's date. (★☆☆)

println("Yesterday: $(today() - Day(1))");
println("Today: $(today())");
println("Tomorrow: $(today() + Day(1))");

Yesterday: 2022-10-26
Today: 2022-10-27
Tomorrow: 2022-10-28

Question 034

Display all the dates corresponding to the month of July 2016. (★★☆)

collect(Date(2016,7,1):Day(1):Date(2016,7,31))

31-element Vector{Dates.Date}:
 2016-07-01
 2016-07-02
 2016-07-03
 2016-07-04
 2016-07-05
 2016-07-06
 2016-07-07
 2016-07-08
 2016-07-09
 2016-07-10
 2016-07-11
 2016-07-12
 2016-07-13
 2016-07-14
 2016-07-15
 2016-07-16
 2016-07-17
 2016-07-18
 2016-07-19
 2016-07-20
 2016-07-21
 2016-07-22
 2016-07-23
 2016-07-24
 2016-07-25
 2016-07-26
 2016-07-27
 2016-07-28
 2016-07-29
 2016-07-30
 2016-07-31

Question 035

Compute ((mA + mB) * (-mA / 2)) in place. (★★☆)

mA = rand(2, 2);
mB = rand(2, 2);
mA .= ((mA .+ mB) .* (.-mA ./ 2))

2×2 Matrix{Float64}:
 -0.35038   -0.685683
 -0.510546  -0.444504

Using the dot macro:

@. mA = ((mA + mB) * (-mA / 2));

Question 036

Extract the integer part of a random array of positive numbers using 4 different methods. (★★☆)

mA = 5 * rand(3, 3);

Option 1:

floor.(mA)

3×3 Matrix{Float64}:
 0.0  0.0  4.0
 2.0  3.0  4.0
 0.0  3.0  3.0

Option 2:

round.(mA .- 0.5) #<! Generates -0.0 for numbers smaller than 0.5

3×3 Matrix{Float64}:
  0.0  0.0  4.0
  2.0  3.0  4.0
 -0.0  3.0  3.0

Option 3:

mA .÷ 1

3×3 Matrix{Float64}:
 0.0  0.0  4.0
 2.0  3.0  4.0
 0.0  3.0  3.0

Option 4:

mA .- rem.(mA, 1)

3×3 Matrix{Float64}:
 0.0  0.0  4.0
 2.0  3.0  4.0
 0.0  3.0  3.0

Question 037

Create a 5x5 matrix with row values ranging from 0 to 4. (★★☆)

mA = repeat(reshape(0:4, 1, 5), 5, 1)

5×5 Matrix{Int64}:
 0  1  2  3  4
 0  1  2  3  4
 0  1  2  3  4
 0  1  2  3  4
 0  1  2  3  4

One could also generate row like range using tranpose:

mA = repeat((0:4)', 5, 1);

Question 038

Generate an array using a generator of 10 numbers. (★☆☆)

vA = collect(x for x in 1:10)

10-element Vector{Int64}:
  1
  2
  3
  4
  5
  6
  7
  8
  9
 10

In Julia the result of collect can be achieved directly using Array Comprehension:

vA = [x for x in 1:10];

Question 039

Create a vector of size 10 with values ranging from 0 to 1, both excluded. (★★☆)

vA = LinRange(0, 1, 12)[2:(end - 1)]

10-element LinRange{Float64, Int64}:
 0.0909091,0.181818,0.272727,0.363636,…,0.636364,0.727273,0.818182,0.909091

Question 040

Create a random vector of size 10 and sort it. (★★☆)

vA = rand(1:10, 10);
sort(vA) #<! Use `sort!()` for inplace sorting

10-element Vector{Int64}:
  1
  3
  3
  4
  4
  4
  7
  8
  9
 10

Question 041

Implement the sum() function manually. (★★☆)

vA = rand(100);

function MySum(vA::Vector{T}) where{T <: Number}
sumVal = vA[1];
for ii in 2:length(vA)
    sumVal += vA[ii];
end

return sumVal;

end

MySum(vA)

51.34991384741698

Question 042

Check for equality of 2 arrays. (★★☆)

vA = rand(10);
vB = rand(10);

all(vA .== vB)

false

Question 043

Make an array immutable (Read only). (★★☆)

This is a work in progress for Julia at in Issue 31630.

Question 044

Consider a random 10x2 matrix representing cartesian coordinates, convert them to polar coordinates. (★★☆)

mA = rand(10, 2);

ConvToPolar = vX -> [hypot(vX[1], vX[2]), atan(vX[2], vX[1])]

mB = [ConvToPolar(vX) for vX in eachrow(mA)]

10-element Vector{Vector{Float64}}:
 [0.9505364569557584, 0.031648622825633764]
 [0.704392126760974, 0.5147716023358581]
 [0.29679995764188105, 1.201649642883764]
 [0.637435149299262, 0.02987122275134911]
 [0.933745980010277, 0.7244055260446777]
 [0.9187482092583683, 0.36651502794834323]
 [1.0164226188075605, 0.9581226557842126]
 [0.5946171056720952, 1.4392243895565575]
 [0.7889136757232136, 0.2964469880257975]
 [1.044698619009772, 0.8754548962664869]

In order to have the same output size:

mC = reduce(hcat, mB)';

Question 045

Create random vector of size 10 and replace the maximum value by 0. (★★☆)

vA = randn(10);

In case of a single maximum or all different values:

vA[argmax(vA)] = 0;
vA

10-element Vector{Float64}:
  0.15593661212531384
  0.8177229521444311
 -1.4850097210409072
  0.0
  0.8552383451908633
  1.476127811761641
  0.2407698166125247
 -0.6363509339561092
 -0.5514594178114751
  0.8756142494364301

General solution:

maxVal = maximum(vA);
vA .= (valA == maxVal ? 0 : valA for valA in vA); #<! Non allocating generator by using `.=`

Question 046

Create a a grid of x and y coordinates covering the [0, 1] x [0, 1] area. (★★☆)

numGridPts = 5;
vX = LinRange(0, 1, numGridPts);
vY = LinRange(0, 1, numGridPts);
MeshGrid = (vX, vY) -> ([x for _ in vY, x in vX], [y for y in vY, _ in vX]);

mX, mY = MeshGrid(vX, vY); #<! See https://discourse.julialang.org/t/48679
@show mX

5×5 Matrix{Float64}:
 0.0  0.25  0.5  0.75  1.0
 0.0  0.25  0.5  0.75  1.0
 0.0  0.25  0.5  0.75  1.0
 0.0  0.25  0.5  0.75  1.0
 0.0  0.25  0.5  0.75  1.0

@show mY

5×5 Matrix{Float64}:
 0.0   0.0   0.0   0.0   0.0
 0.25  0.25  0.25  0.25  0.25
 0.5   0.5   0.5   0.5   0.5
 0.75  0.75  0.75  0.75  0.75
 1.0   1.0   1.0   1.0   1.0

By Tomer Arnon:

mXY = [(ii, jj) for ii in 0:0.25:1, jj in 0:0.25:1]; #<! Also `tuple.(0:0.25:1, (0:0.25:1)')`

Question 047

Given two vectors, vX and vY, construct the Cauchy matrix mC: (Cij = 1 / (xi - yj)). (★★☆)

vX = rand(5);
vY = rand(5);

mC = 1 ./ (vX .- vY')

5×5 Matrix{Float64}:
 21.646    3.23455  34.1025   3.79522  -3.10044
 12.1532   2.89647  15.2886   3.33807  -3.49101
  5.11271  2.18076   5.59545  2.422    -5.77562
 27.6005   3.34229  51.6618   3.94442  -3.00751
  2.36374  1.45768   2.46193  1.56164  18.4072

Question 048

Print the minimum and maximum representable value for each Julia scalar type. (★★☆)

vT = [UInt8 UInt16 UInt32 UInt64 Int8 Int16 Int32 Int64 Float16 Float32 Float64]

for juliaType in vT
    println(typemin(juliaType));
    println(typemax(juliaType));
end

0
255
0
65535
0
4294967295
0
18446744073709551615
-128
127
-32768
32767
-2147483648
2147483647
-9223372036854775808
9223372036854775807
-Inf
Inf
-Inf
Inf
-Inf
Inf

Question 049

Print all the values of an array. (★★☆)

mA = rand(3, 3);
print(mA);

[0.6807535316053605 0.849791747148149 0.32690265736495716; 0.9773227682466125 0.5276250185081583 0.9071015474523568; 0.3192975731085147 0.009594822627075117 0.21726819354916904]

Question 050

Find the closest value to a given scalar in a vector. (★★☆)

inputVal = 0.5;
vA = rand(10);

vA[argmin(abs.(vA .- inputVal))]

0.5023858297186229

By Tomer Arnon:

function ClosestValue(vA::Vector{T}, inputVal::T) where{T <: Number}
    return argmin(y -> abs(y - inputVal), vA);
end

ClosestValue(vA, inputVal)

0.5023858297186229

Question 051

Create a structured array representing a position (x, y) and a color (r, g, b). (★★☆)

struct sPosColor
    x::Int
    y::Int
    R::UInt8;
    G::UInt8;
    B::UInt8;
    A::UInt8;
end

numPixels   = 10;
maxVal      = typemax(UInt32);
vMyColor    = [sPosColor(rand(1:maxVal, 2)..., rand(UInt8, 4)...) for _ in 1:numPixels];

Question 052

Consider a random vector with shape (5, 2) representing coordinates, find the distances matrix mD: $ {D}{i, j} = {\left| {x}{i} - {x}{j} \right|}{2} $. (★★☆)

mX = rand(5, 2);
vSumSqr = sum(vX -> vX .^ 2, mX, dims = 2);
mD = vSumSqr .+ vSumSqr' - 2 * (mX * mX');
mD #<! Apply `sqrt.()` for the actual norm

5×5 Matrix{Float64}:
 -4.44089e-16  0.135171   0.729876     0.724544   0.525945
  0.135171     0.0        0.539868     0.255352   0.136451
  0.729876     0.539868  -1.38778e-17  0.403845   0.814356
  0.724544     0.255352   0.403845     0.0        0.135001
  0.525945     0.136451   0.814356     0.135001  -4.44089e-16

Question 053

Convert a float (32 bits) array into an integer (32 bits) in place. (★★☆)

vA = 9999 .* rand(Float32, 5);
vB = reinterpret(Int32, vA); #<! Creates a view
@. vB = trunc(Int32, vA) #<! Updates the byes in th view (Inplace for `vA`)

5-element reinterpret(Int32, ::Vector{Float32}):
 8898
 1815
 6821
 6523
 9953

The above is equivalent of:

for ii in eachindex(vB)
    vB[ii] = trunc(Int32, vA[ii]);
end
vB

Question 054

Read the following file (Q0054.txt). (★★☆)

1, 2, 3, 4, 5
6,  ,  , 7, 8
 ,  , 9,10,11

mA = readdlm("Q0054.txt", ',')

3×5 Matrix{Any}:
 1     2      3       4   5
 6      "  "   "  "   7   8
  " "   "  "  9      10  11

Question 055

Enumerate array in a loop. (★★☆)

mA = rand(3, 3);

for (elmIdx, elmVal) in enumerate(mA) #<! See https://discourse.julialang.org/t/48877
    println(elmIdx);
    println(elmVal);
end

1
0.7028850425818378
2
0.9169645026836734
3
0.887675432103485
4
0.6988267487719038
5
0.7143993871851885
6
0.8303140449034555
7
0.43387559304673373
8
0.13163225199507878
9
0.6590727215475687

Question 056

Generate a generic 2D Gaussian like array with μ = 0, σ = 1 and indices over {-5, -4, ..., 0, 1, ..., 5}. (★★☆)

vA = -5:5;
μ = 0;
σ = 1;
mG = [(1 / (2 * pi * σ)) * exp(-0.5 * ((([x, y] .- μ)' * ([x, y] .- μ)) / (σ * σ))) for x in vA, y in vA];

heatmap(mG)

      ┌───────────┐              
   11 │▄▄▄▄▄▄▄▄▄▄▄│  ┌──┐ 0.2    
      │▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│        
      │▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│        
      │▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│        
      │▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│        
    1 │▄▄▄▄▄▄▄▄▄▄▄│  └──┘ 2.0e-12
      └───────────┘              
       1        11

Using the separability of the Gaussian function:

vG = (1 / (sqrt(2 * pi) * σ)) .* exp.(-0.5 .* (((vA .- μ) .^ 2) / (σ * σ)));
mG = vG * vG';

Question 057

Place 5 elements in a 5x5 array randomly. (★★☆)

mA = rand(5, 5);
mA[rand(1:25, 5)] = rand(5);

Another option which avoids setting into the same indices:

mA[randperm(25)[1:5]] = rand(5);

Question 058

Subtract the mean of each row of a matrix. (★★☆)

mA = rand(3, 3);
mA .-= mean(mA, dims = 2);
mean(mA, dims = 1)

1×3 Matrix{Float64}:
 0.261517  -0.0218978  -0.23962

Question 059

Sort an array by a column. (★★☆)

colIdx = 2;

mA = rand(3, 3);
mA[sortperm(mA[:, colIdx]), :]

3×3 Matrix{Float64}:
 0.870197  0.229455  0.462788
 0.812345  0.61344   0.0990058
 0.271644  0.722126  0.182457

Using sortslices():

sortslices(mA, dims = 1, by = x -> x[colIdx]);

Question 060

Tell if a given 2D array has null (All zeros) columns. (★★☆)

mA = rand(0:1, 3, 9);
any(all(iszero.(mA), dims = 1))

false

Question 061

Find the 2nd nearest value from a given value in an array. (★★☆)

inputVal = 0.5;
vA = rand(10);

vA[sortperm(abs.(vA .- inputVal))[2]]

0.5947029703004421

Alternative way (More efficient)

closeFirst  = Inf;
closeSecond = Inf;
closeFirstIdx  = 0;
closeSecondIdx = 0;

# Using `global` for scope in Literate
for (elmIdx, elmVal) in enumerate(abs.(vA .- inputVal))
    if (elmVal < closeFirst)
        global closeSecond = closeFirst;
        global closeFirst = elmVal;
        global closeSecondIdx  = closeFirstIdx;
        global closeFirstIdx   = elmIdx;
    elseif (elmVal < closeSecond)
        global closeSecond = elmVal;
        global closeSecondIdx = elmIdx;
    end
end

vA[closeSecondIdx] == vA[sortperm(abs.(vA .- inputVal))[2]]

true

By Tomer Arnon:

vA[partialsortperm(abs.(vA .- inputVal), 2)]

0.5947029703004421

Question 062

Considering two arrays with shape (1, 3) and (3, 1), Compute their sum using an iterator. (★★☆)

vA = rand(1, 3);
vB = rand(3, 1);

sum(aVal + bVal for aVal in vA, bVal in vB)

11.149909948305584

Question 063

Create an array class that has a name attribute. (★★☆)

One could use NamedArrays.jl or AxisArrays.jl.

Question 064

Given a vector, add 1 to each element indexed by a second vector (Be careful with repeated indices). (★★★)

vA = rand(1:10, 5);
vB = rand(1:5, 3);

println(vA);

# Julia is very efficient with loops
for bIdx in vB
    vA[bIdx] += 1;
end

println(vA);

[1, 6, 2, 6, 2]
[3, 7, 2, 6, 2]

Question 065

Accumulate elements of a vector X to an array F based on an index list I. (★★★)

vX = rand(1:5, 10);
vI = rand(1:15, 10);

numElements = maximum(vI);
vF = zeros(numElements);

for (ii, iIdx) in enumerate(vI)
    vF[iIdx] += vX[ii];
end

println("vX: $vX");
println("vI: $vI");
println("vF: $vF");

vX: [5, 4, 3, 5, 2, 5, 2, 5, 5, 1]
vI: [3, 2, 6, 11, 12, 10, 5, 11, 12, 2]
vF: [0.0, 5.0, 5.0, 0.0, 2.0, 3.0, 0.0, 0.0, 0.0, 5.0, 10.0, 7.0]

One could also use counts() from StatsBase.jl.

Question 066

Considering an image of size w x h x 3 image of type UInt8, compute the number of unique colors. (★★☆)

mI = rand(UInt8, 1000, 1000, 3);

numColors = length(unique([reinterpret(UInt32, [iPx[1], iPx[2], iPx[3], 0x00])[1] for iPx in eachrow(reshape(mI, :, 3))])); #<! Reshaping as at the moment `eachslice()` doesn't support multiple `dims`.
print("Number of Unique Colors: $numColors");

Number of Unique Colors: 970711

Another option:

numColors = length(unique([UInt32(iPx[1]) + UInt32(iPx[2]) << 8 + UInt32(iPx[3]) << 16 for iPx in eachrow(reshape(mI, :, 3))]));
print("Number of Unique Colors: $numColors");

Number of Unique Colors: 970711

Simpler way to slice a pixel:

numColors = length(unique([UInt32(mI[ii, jj, 1]) + UInt32(mI[ii, jj, 2]) << 8 + UInt32(mI[ii, jj, 3]) << 16 for ii in 1:size(mI, 1), jj in 1:size(mI, 2)]));
print("Number of Unique Colors: $numColors");

Number of Unique Colors: 970711

Question 067

Considering a four dimensions array, get sum over the last two axis at once. (★★★)

mA = rand(2, 2, 2, 2);
sum(reshape(mA, (2, 2, :)), dims = 3)

2×2×1 Array{Float64, 3}:
[:, :, 1] =
 1.817    1.16234
 2.44291  2.32796

Question 068

Considering a one dimensional vector vA, how to compute means of subsets of vA using a vector vS of same size describing subset indices. (★★★)

# Bascially extending `Q0065` with another vector of number of additions.

vX = rand(1:5, 10);
vI = rand(1:15, 10);

numElements = maximum(vI);
vF = zeros(numElements);
vN = zeros(Int, numElements);

for (ii, iIdx) in enumerate(vI)
    vF[iIdx] += vX[ii];
    vN[iIdx] += 1;
end

# We only divide the mean if the number of elements accumulated is bigger than 1
for ii in 1:numElements
    vF[ii] = ifelse(vN[ii] > 1, vF[ii] / vN[ii], vF[ii]);
end

println("vX: $vX");
println("vI: $vI");
println("vF: $vF");

vX: [5, 5, 4, 2, 1, 1, 3, 5, 3, 2]
vI: [14, 7, 8, 10, 5, 10, 14, 2, 7, 13]
vF: [0.0, 5.0, 0.0, 0.0, 1.0, 0.0, 4.0, 4.0, 0.0, 1.5, 0.0, 0.0, 2.0, 4.0]

Question 069

Get the diagonal of a matrix product. (★★★)

mA = rand(5, 7);
mB = rand(7, 4);

numDiagElements = min(size(mA, 1), size(mB, 2));
vD = [dot(mA[ii, :], mB[:, ii]) for ii in 1:numDiagElements]

4-element Vector{Float64}:
 1.8937792321469207
 1.035584608236753
 2.0251852803210024
 2.2065505485118653

Alternative way:

vD = reshape(sum(mA[1:numDiagElements, :]' .* mB[:, 1:numDiagElements], dims = 1), numDiagElements)

4-element Vector{Float64}:
 1.8937792321469207
 1.035584608236753
 2.0251852803210024
 2.2065505485118653

Question 070

Consider the vector [1, 2, 3, 4, 5], build a new vector with 3 consecutive zeros interleaved between each value. (★★★)

vA = 1:5;

# Since Julia is fast with loops, it would be the easiest choice

numElements = (4 * length(vA)) - 3;
vB = zeros(Int, numElements);

for (ii, bIdx) in enumerate(1:4:numElements)
    vB[bIdx] = vA[ii];
end
println(vB);

# Alternative (MATLAB style) way:

mB = [reshape(collect(vA), 1, :); zeros(Int, 3, length(vA))];
vB = reshape(mB[1:(end - 3)], :);
println(vB);

[1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 5]
[1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 5]

Question 071

Consider an array of dimension 5 x 5 x 3, mulitply it by an array with dimensions 5 x 5 using broadcasting. (★★★)

mA = rand(5, 5, 3);
mB = rand(5, 5);

mA .* mB #<! Very easy in Julia

5×5×3 Array{Float64, 3}:
[:, :, 1] =
 0.492335  0.301617  0.0158155   0.411494  0.0331824
 0.555254  0.148368  0.111992    0.366332  0.192798
 0.104442  0.249306  0.0138448   0.018588  0.721434
 0.665075  0.136419  0.00305929  0.430373  0.208216
 0.290075  0.642668  0.519127    0.511527  0.151719

[:, :, 2] =
 0.234647   0.447608  0.000903852  0.00299349  0.0626502
 0.0159529  0.164814  0.0807125    0.27415     0.180858
 0.103613   0.291711  0.00759939   0.013771    0.11713
 0.314552   0.210101  0.00966737   0.343302    0.377711
 0.129541   0.642804  0.833245     0.372239    0.816422

[:, :, 3] =
 0.774094  0.181579   0.0151706   0.26963    0.0364216
 0.167574  0.0016081  0.0130672   0.823491   0.398857
 0.062678  0.40945    0.00790564  0.0141378  0.468662
 0.344903  0.18419    0.00697708  0.487335   0.504131
 0.118821  0.0130137  0.803091    0.754898   0.0470187

Question 072

Swap two rows of a 2D array. (★★★)

mA = rand(UInt8, 3, 2);
println(mA);
mA[[1, 2], :] .= mA[[2, 1], :];
println(mA);

UInt8[0xc7 0x2f; 0xbf 0x8f; 0xac 0x2a]
UInt8[0xbf 0x8f; 0xc7 0x2f; 0xac 0x2a]

Question 073

Consider a set of 10 triplets describing 10 triangles (with shared vertices), find the set of unique line segments composing all the triangles. (★★★)

mA = rand(0:100, 10, 3); #<! Each row composes 3 veritces ([1] -> [2], [2] -> [3], [3] -> [1])
mC = [sort!([vC[mod1(ii, end)], vC[mod1(ii + 1, end)]]) for ii in 1:(size(mA, 2) + 1), vC in eachrow(mA)][:] #<! Sorted combinations of vertices
mC = unique(mC)

30-element Vector{Vector{Int64}}:
 [52, 86]
 [23, 86]
 [23, 52]
 [53, 88]
 [53, 95]
 [88, 95]
 [4, 31]
 [4, 24]
 [24, 31]
 [34, 80]
 [34, 99]
 [80, 99]
 [19, 69]
 [19, 21]
 [21, 69]
 [38, 74]
 [36, 74]
 [36, 38]
 [18, 20]
 [20, 48]
 [18, 48]
 [0, 23]
 [23, 77]
 [0, 77]
 [23, 42]
 [42, 75]
 [23, 75]
 [20, 69]
 [7, 69]
 [7, 20]

Question 074

Given a sorted array vC that corresponds to a bincount, produce an array vA such that bincount(vA) == vC. (★★★)

vC = rand(0:7, 5);
numElements = sum(vC);
vA = zeros(Int, numElements);

elmIdx = 1;
# Using `global` for scope in Literate
for (ii, binCount) in enumerate(vC)
    for jj in 1:binCount
        vA[elmIdx] = ii;
        global elmIdx += 1;
    end
end

Question 075

Compute averages using a sliding window over an array. (★★★)

numElements = 10;
winRadius   = 1;
winReach    = 2 * winRadius;
winLength   = 1 + winReach;

vA = rand(0:3, numElements);
vB = zeros(numElements - (2 * winRadius));

aIdx = 1 + winRadius;
# Using `global` for scope in Literate
for ii in 1:length(vB)
    vB[ii] = mean(vA[(aIdx - winRadius):(aIdx + winRadius)]); #<! Using integral / running sum it would be faster.
    global aIdx += 1;
end

Another method using running sum:

vC = zeros(numElements - winReach);

jj = 1;
sumVal = sum(vA[1:winLength]);
vC[jj] = sumVal / winLength;
jj += 1;

# Using `global` for scope in Literate
for ii in 2:(numElements - winReach)
    global sumVal += vA[ii + winReach] - vA[ii - 1];
    vC[jj] = sumVal / winLength;
    global jj += 1;
end

maximum(abs.(vC - vB)) < 1e-8

true

Question 076

Consider a one dimensional array vA, build a two dimensional array whose first row is [ vA[0], vA[1], vA[2] ] and each subsequent row is shifted by 1. (★★★)

vA = rand(10);
numCols = 3;

numRows = length(vA) - numCols + 1;
mA = zeros(numRows, numCols);

for ii in 1:numRows
    mA[ii, :] = vA[ii:(ii + numCols - 1)]; #<! One could optimize the `-1` out
end

Question 077

Negate a boolean or to change the sign of a float inplace. (★★★)

vA = rand(Bool, 10);
vA .= .!vA;

vA = randn(10);
vA .*= -1;

Question 078

Consider 2 sets of points mP1, mP2 describing lines (2d) and a point vP, how to compute distance from the point vP to each line i: [mP1[i, :], mP2[i, :]. (★★★)

# See distance of a point from a line in Wikipedia (https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line).
# Specifically _Line Defined by Two Points_.

numLines = 10;
mP1 = randn(numLines, 2);
mP2 = randn(numLines, 2);
vP  = randn(2);

vD = [(abs(((vP2[1] - vP1[1]) * (vP1[2] - vP[2])) - ((vP1[1] - vP[1]) * (vP2[2] - vP1[2]))) / hypot((vP2 - vP1)...)) for (vP1, vP2) in zip(eachrow(mP1), eachrow(mP2))];
minDist = minimum(vD);

println("Min Distance: $minDist");

Min Distance: 0.42811311783896533

Question 079

Consider 2 sets of points mP1, mP2 describing lines (2d) and a set of points mP, how to compute distance from the point vP = mP[j, :] to each line i: [mP1[i, :], mP2[i, :]. (★★★)

numLines = 5;
mP1 = randn(numLines, 2);
mP2 = randn(numLines, 2);
mP  = randn(numLines, 2);

mD = [(abs(((vP2[1] - vP1[1]) * (vP1[2] - vP[2])) - ((vP1[1] - vP[1]) * (vP2[2] - vP1[2]))) / hypot((vP2 - vP1)...)) for (vP1, vP2) in zip(eachrow(mP1), eachrow(mP2)), vP in eachrow(mP)];

for jj in 1:numLines
    minDist = minimum(mD[jj, :]);
    println("The minimum distance from the $jj -th point: $minDist");
end

┌ Warning: Assignment to `minDist` in soft scope is ambiguous because a global variable by the same name exists: `minDist` will be treated as a new local. Disambiguate by using `local minDist` to suppress this warning or `global minDist` to assign to the existing global variable.
└ @ string:9
The minimum distance from the 1 -th point: 0.33333028326949365
The minimum distance from the 2 -th point: 0.09596286876613971
The minimum distance from the 3 -th point: 0.15987404532559377
The minimum distance from the 4 -th point: 0.09922388894592074
The minimum distance from the 5 -th point: 0.0701461047753776

Question 080

Consider an arbitrary 2D array, write a function that extract a subpart with a fixed shape and centered on a given element (Handel out of bounds). (★★★)

# One could use `PaddedViews.jl` to easily solve this.

arrayLength = 10;
winRadius   = 3;
vWinCenter  = [7, 9];

mA = rand(arrayLength, arrayLength);
winLength = (2 * winRadius) + 1;
mB = zeros(winLength, winLength);

verShift = -winRadius;
# Using `global` for scope in Literate
for ii in 1:winLength
    horShift = -winRadius;
    for jj in 1:winLength
        mB[ii, jj] = mA[min(max(vWinCenter[1] + verShift, 1), arrayLength), min(max(vWinCenter[2] + horShift, 1), arrayLength)]; #<! Nearest neighbor extrapolation
        horShift += 1;
    end
    global verShift += 1;
end

Question 081

Consider an array vA = [1, 2, 3, ..., 13, 14], generate an array vB = [[1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6], ..., [11, 12, 13, 14]]. (★★★)

vA = collect(1:14);

winNumElements  = 4;
winReach        = winNumElements - 1;

vB = [vA[ii:(ii + winReach)] for ii in 1:(length(vA) - winReach)]

11-element Vector{Vector{Int64}}:
 [1, 2, 3, 4]
 [2, 3, 4, 5]
 [3, 4, 5, 6]
 [4, 5, 6, 7]
 [5, 6, 7, 8]
 [6, 7, 8, 9]
 [7, 8, 9, 10]
 [8, 9, 10, 11]
 [9, 10, 11, 12]
 [10, 11, 12, 13]
 [11, 12, 13, 14]

Question 082

Compute a matrix rank. (★★★)

numRows = 5;
numCols = 4;
mA = randn(numRows, numCols);
rank(mA)

Question 083

Find the most frequent value in an array. (★★★)

vA = rand(1:5, 15);

MATLAB Style (Manual loop might be faster)

vB = unique(vA);
# vB[argmax(sum(vA .== vB', dims = 1)[:])] #<! The input to `argmax()` is a `1 x n` vector, hence squeezed so `argmax()` won't return Cartesian Index.
vB[argmax(dropdims(sum(vA .== vB', dims = 1), dims = 1))] #<! The input to `argmax()` is a `1 x n` vector, hence squeezed so `argmax()` won't return Cartesian Index.

Comparing bits:

One could convert at the bits level to integers and then use something like counts() from StatsBase.jl. Support to 1:4 bytes of data:

numBytes = sizeof(vA[1]);
if (sizeof(vA[1]) == 1)
    vB = reinterpret(UInt8, vA);
elseif (sizeof(vA[1]) == 2)
    vB = reinterpret(UInt16, vA);
elseif (sizeof(vA[1]) == 4)
    vB = reinterpret(UInt32, vA);
elseif (sizeof(vA[1]) == 8)
    vB = reinterpret(UInt64, vA);
end

Question 084

Extract all the contiguous 3x3 blocks from a random 5x5 matrix. (★★★)

numRows = 5;
numCols = 5;

mA = rand(1:9, numRows, numCols);

winRadius   = 1;
winReach    = 2 * winRadius;
winLength   = winReach + 1;

mB = [mA[ii:(ii + winReach), jj:(jj + winReach)] for ii in 1:(numRows - winReach), jj in 1:(numCols - winReach)]

3×3 Matrix{Matrix{Int64}}:
 [9 1 1; 3 1 2; 9 4 6]  [1 1 4; 1 2 2; 4 6 2]  [1 4 7; 2 2 1; 6 2 2]
 [3 1 2; 9 4 6; 4 2 1]  [1 2 2; 4 6 2; 2 1 7]  [2 2 1; 6 2 2; 1 7 1]
 [9 4 6; 4 2 1; 2 9 8]  [4 6 2; 2 1 7; 9 8 1]  [6 2 2; 1 7 1; 8 1 4]

Question 085

Create a 2D array struct such that mA[i, j] == mA[j, i] (Symmetric matrix). (★★★)

struct SymmetricMatrix{T <: Number} <: AbstractArray{T, 2}
    numRows::Int
    data::Matrix{T}

    function SymmetricMatrix(mA::Matrix{T}) where {T <: Number}
        size(mA, 1) == size(mA, 2) || throw(ArgumentError("Input matrix must be square"))
        new{T}(size(mA, 1), Matrix(Symmetric(mA)));
    end

end

function Base.size(mA::SymmetricMatrix)
    (mA.numRows, mA.numRows);
end
function Base.getindex(mA::SymmetricMatrix, ii::Int)
    mA.data[ii];
end
function Base.getindex(mA::SymmetricMatrix, ii::Int, jj::Int)
    mA.data[ii, jj];
end
function Base.setindex!(mA::SymmetricMatrix, v, ii::Int, jj::Int)
    setindex!(mA.data, v, ii, jj);
    setindex!(mA.data, v, jj, ii);
end

mA = SymmetricMatrix(zeros(Int, 2, 2));
mA[1, 2] = 5;
mA

2×2 Main.var"##312".SymmetricMatrix{Int64}:
 0  5
 5  0

Question 086

Consider a set of p matrices of shape nxn and a set of p vectors with length n. Compute the sum of of the p matrix vector products at once (Result is a vector of length n). (★★★)

# One could use `TensorOperations.jl` or `Einsum.jl` for a more elegant solution.

numRows = 5;
numMat  = 3;

tP = [randn(numRows, numRows) for _ in 1:numMat];
mP = [randn(numRows) for _ in 1:numMat];

vA = reduce(+, (mP * vP for (mP, vP) in zip(tP, mP)));

vB = zeros(numRows);
for ii in 1:numMat
    vB .+= tP[ii] * mP[ii];
end

vA == vB

true

Question 087

Consider a 16x16 array, calculate the block sum (Block size is 4x4). (★★★)

We solve a more general case for any size of blocks.

numRows = 16;
numCols = 8;

vBlockSize = [2, 4]; #<! [numRows, numCols] ./ vBlockSize == integer

mA = rand(numRows, numCols);

numBlocksVert   = numRows ÷ vBlockSize[1];
numBlocksHori   = numCols ÷ vBlockSize[2];
numBlocks       = numBlocksVert * numBlocksHori;

mA = reshape(mA, vBlockSize[1], :);

We number the blocks column wise and create their block index per column of the reshaped mA.

vBlockIdx = 1:numBlocks;
mBlockIdx = reshape(vBlockIdx, numBlocksVert, numBlocksHori);
mBlockIdx = repeat(mBlockIdx, 1, 1, vBlockSize[2]);
mBlockIdx = permutedims(mBlockIdx, (1, 3, 2));
vBlockIdx = mBlockIdx[:]; #<! Matches the block index per column of the reshaped `mA`.

vA = dropdims(sum(mA, dims = 1), dims = 1);
vB = zeros(numBlocks);

for ii = 1:(numBlocks * vBlockSize[2])
    vB[vBlockIdx[ii]] += vA[ii];
end
vB

16-element Vector{Float64}:
 4.354231620225193
 3.734856822720345
 3.297606250613403
 4.256409222766379
 3.2941384563138993
 3.776165923653995
 3.600353929303568
 4.518180088610547
 3.6679777425586186
 4.6420262282688345
 4.296040312870272
 5.1547123230902825
 3.0214360614973477
 3.202315920090961
 4.669003591708878
 3.3257368418156616

Question 088

Implement the simulation Game of Life using arrays. (★★★)

numRows = 20;
numCols = 20;

gofKernel           = @kernel w -> w[-1, -1] + w[-1, 0] + w[-1, 1] + w[0, -1] + w[0, 1] + w[1, -1] + w[1, 0] + w[1, 1];
gofNumLives         = round(Int, 0.05 * numRows * numCols);
gofNumGenerations   = 50;

vI = randperm(numRows * numCols)[1:gofNumLives];

mG = zeros(UInt8, numRows, numCols);
mG[vI] .= UInt8(1);
mB = similar(mG);

heatmap(mG) #<! Initialization

      ┌────────────────────┐        
   20 │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  ┌──┐ 1
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
    1 │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  └──┘ 0
      └────────────────────┘        
       1                 20

for ii in 1:numGridPts
    map!(gofKernel, mB, extend(mG, StaticKernels.ExtensionConstant(0))); #<! One may use `ExtensionCircular`
    for ii in eachindex(mB)
        mG[ii] = UInt8((mB[ii] >= 3) || ((mB[ii] > 0) && (mB[ii] == 2))); #<! Could be done with broadcasting
    end
end

heatmap(mG) #<! Final state

      ┌────────────────────┐        
   20 │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  ┌──┐ 1
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
      │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  │▄▄│  
    1 │▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄│  └──┘ 0
      └────────────────────┘        
       1                 20

TODO: Use O(1) implementation for Box Blur.

Question 089

Get the n largest values of an array. (★★★)

vA = rand(10);
numValues = 3;

vA[partialsortperm(vA, 1:numValues, rev = true)]

3-element Vector{Float64}:
 0.8713271882275592
 0.8486847944348448
 0.8445471427655805

Question 090

Given an arbitrary number of vectors, build the Cartesian Product (Every combinations of every item). (★★★)

function CartesianProduct(tupleX)
    return collect(Iterators.product(tupleX...))[:];
end

vA = 1:3;
vB = 8:9;
vC = 4:5;

CartesianProduct((vA, vB, vC))

12-element Vector{Tuple{Int64, Int64, Int64}}:
 (1, 8, 4)
 (2, 8, 4)
 (3, 8, 4)
 (1, 9, 4)
 (2, 9, 4)
 (3, 9, 4)
 (1, 8, 5)
 (2, 8, 5)
 (3, 8, 5)
 (1, 9, 5)
 (2, 9, 5)
 (3, 9, 5)

Question 091

Create an array which can be accessed like a record array in NumPy. (★★★)

One could use StructArrays.jl.

Question 092

Consider a large vector vA, compute vA to the power of 3 using 3 different methods. (★★★)

vA = rand(1000);

Method 001:

vB = vA .^ 3;

Method 002:

vC = [valA ^ 3 for valA in vA];

Method 003:

vD = zeros(length(vA));
for (ii, valA) in enumerate(vA)
    vD[ii] = valA * valA * valA;
end

vB ≈ vC ≈ vD

true

Question 093

Consider two arrays mA and mB of shape 8x3 and 2x2. Find rows of mA that contain elements of each row of mB regardless of the order of the elements in mB. (★★★)

The way I interpret the question is rows in mA which contain at least 1 element from each row of mB.

mA = rand(0:4, 8, 3);
mB = rand(0:4, 2, 2);
mC = [any(vA .== vB') for vB in eachrow(mB), vA in eachrow(mA)]; #<! General solution, will work for any size of `mA` and `mB`
vD = [all(vC) for vC in eachcol(mC)]

8-element Vector{Bool}:
 0
 1
 1
 1
 0
 0
 1
 1

In order to have a solution without the intermediate array mC

function Iterate2(iA; iterState = missing)
    if(ismissing(iterState))
        valA, iterState = iterate(iA);
    else
        valA, iterState = iterate(iA, iterState);
    end
    valB, iterState = iterate(iA, iterState);
    return (valA, valB), iterState
end

tT = (any(vA .== vB') for vB in eachrow(mB), vA in eachrow(mA));

iterState = missing;

vE = zeros(Bool, size(mA, 1));

for ii = 1:length(vD)
    global iterState;
    (valA, valB), iterState = Iterate2(tT; iterState = iterState);
    vE[ii] = valA && valB;
end

vD == vE

true

Question 094

Considering a 10x3 matrix, extract rows with unequal values. (★★★)

mA = rand(1:3, 10, 3);
vD = [maximum(vA) != minimum(vA) for vA in eachrow(mA)]

10-element Vector{Bool}:
 1
 0
 1
 1
 1
 1
 1
 1
 1
 1

Question 095

Convert a vector of ints into a matrix binary representation. (★★★)

vA = rand(UInt8, 10);
mB = zeros(Bool, length(vA), 8);

# See https://discourse.julialang.org/t/26663
for ii in 1:length(vA)
    vS = bitstring(vA[ii]);
    for jj in 1:size(mB, 2)
        mB[ii, jj] = vS[jj] == '1';
    end
end

mB

10×8 Matrix{Bool}:
 0  0  1  0  1  1  1  1
 1  1  0  1  1  1  0  0
 0  0  0  0  1  0  1  0
 1  1  1  0  0  1  0  1
 0  1  0  1  0  1  0  1
 0  0  0  1  0  1  1  1
 1  0  0  0  1  0  1  0
 1  1  0  1  1  1  0  0
 0  0  1  1  1  1  0  0
 1  0  0  0  0  1  0  0

By Tomer Arnon:

mB = reverse!(reduce(hcat, digits.(vA, base = 2, pad = 8))', dims = 2);

Question 096

Given a two dimensional array, extract unique rows. (★★★)

mA = UInt8.(rand(1:3, 10, 3));

vS = [reduce(*, bitstring(valA) for valA in vA) for vA in eachrow(mA)]; #<! Supports any array!
vU = unique(vS);
vI = [findfirst(valU .== vS) for valU in vU];

An alternative way:

vB = indexin(vU, vS);
vB == vI

true

Question 097

Considering 2 vectors vA and vB, write the einsum equivalent (Using Einsum.jl) of inner, outer, sum, and mul function. (★★★)

vA = rand(5);
vB = rand(5);

Inner Product

@tullio tullioVal = vA[ii] * vB[ii];
tullioVal ≈ dot(vA, vB) #<! Inner product

true

Outer Product

@tullio mTullio[ii, jj] := vA[ii] * vB[jj]; #<! Outer product
mTullio ≈ vA * vB'

true

Sum

@tullio tullioVal = vA[ii];
tullioVal ≈ sum(vA) #<! Sum

true

Multiplication

@tullio vTullio[ii] := vA[ii] * vB[ii];
vTullio ≈ vA .* vB #<! Multiplication

true

Question 098

Considering a path described by two vectors vX and vY, sample it using equidistant samples. (★★★)

The way I interpreted the question is to create sub segments of the same length.

numPts      = 100;
numSegments = 1000;

vX = sort(10 * rand(numPts));
vY = sort(10 * rand(numPts));

vR = cumsum(hypot.(diff(vX), diff(vY)));
pushfirst!(vR, 0.0);
vRSegment = LinRange(0.0, vR[end], numSegments);

struct LinearInterpolator1D{T <: Number} <: AbstractArray{T, 1}
    vX::Vector{T};
    vY::Vector{T};

    function LinearInterpolator1D(vX::Vector{T}, vY::Vector{T}) where {T <: Number}
        length(vX) == length(vX) || throw(ArgumentError("Input vectors must have the same length"));
        new{T}(vX, vY);
    end
end

function Base.size(vA::LinearInterpolator1D)
    size(vA.vX);
end
function Base.getindex(vA::LinearInterpolator1D, ii::Number)
    if (ii >= vA.vX[end])
        return vA.vY[end];
    end
    if (ii <= vA.vX[1])
        return vA.vY[1];
    end

    rightIdx = findfirst(vA.vX .>= ii);
    leftIdx = rightIdx - 1;

    tt = (ii - vA.vX[leftIdx]) / (vA.vX[rightIdx] - vA.vX[leftIdx]);

    return ((1 - tt) * vA.vY[leftIdx]) + (tt * vA.vY[rightIdx]);

end
function Base.setindex!(vA::LinearInterpolator1D, valX, valY, ii::Int, jj::Int)
    setindex!(sLinInterp.vX, valX, ii);
    setindex!(sLinInterp.vY, valY, ii);
end

vXInt = LinearInterpolator1D(vR, vX);
vYInt = LinearInterpolator1D(vR, vY);

vXSegment = [vXInt[intIdx] for intIdx in vRSegment];
vYSegment = [vYInt[intIdx] for intIdx in vRSegment];

hP = lineplot(vX, vY, canvas = DotCanvas, name = "Samples");
lineplot!(hP, vXSegment, vYSegment, name = "Interpolated");
hP

      ┌────────────────────────────────────────┐             
   10 │                                     ..:│ Samples     
      │                                 ...''  │ Interpolated
      │                              .:''      │             
      │                             .:         │             
      │                            .:          │             
      │                        .:'''           │             
      │                       :'               │             
      │                    ..:'                │             
      │                  :''                   │             
      │              ....:                     │             
      │            ..:                         │             
      │        :''''                           │             
      │     .:''                               │             
      │   .:'                                  │             
    0 │..:'                                    │             
      └────────────────────────────────────────┘             
       0                                     10

Remark: In order to be mathematically accurate the resolution of vRSegment must be high enough to be an integer factor of each segment. It can be done if the resolution is 1 / (prod(vR)) which is easily infeasible with Float64. So this is a good enough approximation.

Question 099

Given an integer n and a 2D array mA, find the rows which can be interpreted as draws from a multinomial distribution with n (Rows which only contain integers and which sum to n). (★★★)

mA = rand([0, 0.5, 1, 2, 3], 15, 3);
sumVal = 4;
vI = [all(vA .== round.(vA)) && sum(vA) == sumVal for vA in eachrow(mA)];

Question 100

Compute bootstrapped 95% confidence intervals for the mean of a 1D array vA. Namely, resample the elements of an array with replacement N times, compute the mean of each sample and then compute percentiles over the means. (★★★)

numTrials   = 10000;
numSamples  = 1000;
μ           = 0.5;

vA = μ .+ randn(numSamples);
tM = (mean(vA[rand(1:numSamples, numSamples)]) for _ in 1:numTrials);
quantile(tM, [0.025, 0.975])

2-element Vector{Float64}:
 0.4620914889349693
 0.5830619321705541

This page was generated using Literate.jl.

RoyiAvital/Julia100Exercises

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100 Julia Exercises with Solutions

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