# Seven Lines of Julia (examples sought)

I’m loving all the examples being presented here. I would like to encourage folks to submit as many different kinds of examples as possible, and I’ll work on collecting them and getting them onto julialang.org and generally on to social media with something like a #WhyJulia or #WorldOfJulia tag. I think @jiahao used to maintain such a thing a while ago for presentations and such.

It would greatly help to have a 3-5 line description of what it is doing, and why it is interesting, to go along with the code snippet.

-viral

26 Likes

The examples are nice and interesting, but IMO <10 LOC is not where Julia really shines. When not relying on packages, this domain is mostly about surface syntax tricks baked into the language.

The amazing powers of the language start to show around 100 LOC, and come into play fully from 1-5 kLOC, replicating functionality and speed of much larger libraries in other languages.

12 Likes

Agree with Tamas, but it is also nice to have a couple of very short examples. Here a Gibbs sampler sampling the posterior parameters of normal observations with mean distributed a priori as N(μ0, σ0) and precision 1/σ^2 a priori as Gamma(α, β):

``````using Distributions, Statistics
function mc(x, iterations, μ0 = 0., τ0 = 1e-7, α = 0.0001, β = 0.0001)
sumx, n = sum(x), length(x)
μ, τ = sumx/n, 1/var(x)
samples = [(μ, τ)]
for i in 1:iterations
μ = rand(Normal((τ0*μ0 + τ*sumx)/(τ0 + n*τ), 1/sqrt(τ0 + n*τ)))
τ = rand(Gamma(α + n/2, 1/(β + 0.5*sum((xᵢ-μ)^2 for xᵢ in x))))
push!(samples, (μ, τ))
end
samples
end
``````

Usage

``````samples = mc(rand(Normal(0.2, 1.7^(-0.5)), 1_000), 10_000)
μs, τs = first.(samples), last.(samples)
println("mean μ = ", mean(μs), " ± ", std(μs))
println("precision τ = ", mean(τs), " ± ", std(τs))
``````

Output:

`mean μ = 0.2076960556449643 ± 0.023693641548604788`
`precision τ = 1.8301418451485263 ± 0.08225999887566306`

9 Likes

Couldn’t this be written without the need for a loop and `push!`? I think this would get you to seven lines while also being to my eye quite readable:

``````using Distributions, Statistics

function mc(x, iterations, μ₀ = 0., τ₀ = 1e-7, α = 0.0001, β = 0.0001)
∑x, n = sum(x), length(x)
μ, τ = ∑x/n, 1/var(x)
μₛ = rand(Normal((τ₀*μ₀ + τ*∑x)/(τ₀ + n*τ), 1/√(τ₀ + n*τ)), iterations)
τₛ = rand(Gamma(α + n/2, 1/(β + 0.5*sum((xᵢ-μ)^2 for xᵢ in x))), iterations)
samples = [(μ, τ); collect(zip(μₛ, τₛ))]
end
``````
3 Likes

There is interdependence - the next value of `μ` depends on `τ` and vice versa. This is typical for a Gibbs sampler that you cannot parallelise it easily and really need a loop here (and this is exactly what brought me to try Julia in the first place )

8 Likes

Ah yes, I somehow thought the `μ` and `τ` would always come from the outer scope, and the results only differed on the fifth decimal or so when I checked so it seemed equivalent!

1 Like

More than 7 lines of code, but still a quite compact pipeline for geospatial machine learning:

``````using GeoStats, MLJ, CSV, Plots

df.crop = categorical(df.crop)

Ω = georef(df, (:x,:y))

Ωs, Ωt = split(Ω, 0.2, (1.0,-1.0))

𝒫 = LearningProblem(Ωs, Ωt, 𝒯)
𝔏 = PointwiseLearn(model)

Ω̂t = solve(𝒫, 𝔏)

plot(Ω̂t, ms=0.2, c=:cividis, title="crop (prediction)")
``````

Commented example is here: https://juliaearth.github.io/GeoStats.jl/stable/workflow/#Defining-problems

16 Likes

Nice observation that the “wrong” solution is so good… Bernstein von Mises theorem strikes again.

4 Likes

Love everyone’s contributions… Hate to say it but I hard a hard time finding something around 7 LOC to submit :D. But, I did think of something I did a while back.

``````mutable struct βLASSO
η::Float32
λ::Float32
β::Float32
end

βLASSO(η = Float32(0.01), λ = Float32(0.009), β = Float32(50.0)) = βLASSO(η, λ, β)

function apply!(o::βLASSO, x, Δ)
Δ = o.η .* ( Δ .+ ( o.λ .* sign.( Δ ) ) )
Δ = Δ .* Float32.( abs.( x.data ) .> ( opt.β * opt.λ ) )
return Δ
end
``````

(full sloppy gist here: https://gist.github.com/caseykneale/b21c4c6cf5119c58d4f933baac16136b)

Summary:
I was reading the following paper: https://arxiv.org/pdf/2007.13657.pdf around when it was first printed. The paper describes a projected gradient descent method for a penalized(β) LASSO regularization. In < 30min I was able to write my own optimizer, implement the majority of the paper and play with the loss function using Flux.jl. Below is a synthetic example of their loss function properly selecting a single variable of interest while minimizing the contributions of the others.

Why Julia:
The tools in the ecosystem and the language facilitate quickly implementing cutting edge stuff. Plus, unicode support for readability!

7 Likes

@juliohm, the nice plot took very long time to display in my laptop (2nd time plotting it was also very slow). Any recommendation on this?

It may be an issue with Plots.jl @rafael.guerra, I had to lock the version to some old version in the documentation for example because plots became too slow. Are you also using the GR backend? We can try to address this in Zulip or in a private message.

Like with Julia, experts say that Shakespearean language can be tricky but that acting it out makes it fun and understandable. On the same vein, the following MIT list of Shakespearean insults might be a good way to learn both languages:

``````using CSV, DataFrames, HTTP
f = CSV.File(HTTP.get("http://web.mit.edu/dryfoo/Funny-pages/shakespeare-insult-kit.html").body);
sk = DataFrame(f[17:66])  # TODO: findfirst "artless" word to set range
r = size(sk,1)
sk = DataFrame([split(sk[i,1]) for i in 1:r])  # output = 3 rows x 50 columns
Shakespeare() = println("Thou "*sk[1,rand(1:r)]*" "*sk[2,rand(1:r)]*" "*sk[3,rand(1:r)])
``````

Then call Shakespeare:

``````julia> Shakespeare()
julia> Shakespeare()
Thou villainous motley-minded giglet
``````
17 Likes

With 15 lines I could write a code to perform a particle simulation with periodic boundary conditions, a langevin thermostat, and a quadratic potential between the particles, and produce an animation:

``````using Plots ; ENV["GKSwstype"]="nul"
const N, τ, Δt, λ, T, k = 100, 1000, 0.01, 1e-3, 0.26, 1e-6
const x, v, f = -0.5 .+ rand(3,N), -0.01 .+ 0.02*randn(3,N), zeros(3,N)
wrap(x,y) = (x-y) > 0.5 ? (x-y)-1 : ( (x-y) < -0.5 ? (x-y)+1 : (x-y) )
anim = @animate for t in 1:τ
f .= 0
for i in 1:N-1, j in i+1:N
f[:,i] .+= wrap.(x[:,i],x[:,j]) .- λ .* v[:,i]
f[:,j] .+= wrap.(x[:,j],x[:,i]) .- λ .* v[:,j]
end
x .= wrap.(x .+ v*Δt .+ (f/2)*Δt^2,zeros(3))
v .= v .+ f*Δt .+ sqrt.(2*λ*k*T*Δt)*randn()
scatter(x[1,:],x[2,:],x[3,:],label="",lims=(-0.5,0.5),aspect_ratio=1,framestyle=:box)
end
gif(anim,"anim.gif",fps=10)
``````

As Tamas mentioned, if we go to a few more lines, this could include an actual Lennard-Jones potential and be parallelized. And that without using any package besides Plots.

54 Likes

I love watching particles dance! Awesome Leandro!

1 Like

Infinizoom! (Background: Simulates a Brownian motion and uses the Brownian self-similarity Wiener process - Wikipedia combined with simple data-augmentation/Brownian bridge sampling to zoom in infinitely.)

``````using Makie
dt = 0.0005; ϵ = 0.1; t = collect(-1.0:dt:1.0); n = length(t)
x = sqrt(dt)*cumsum(randn(n)); x .-= x[end÷2]
T, X = Node(t), Node(x)
p = lines(T, X); display(p)
xlims!(-1.0, 1.0); ylims!(-1.5,1.5)
while true
t .*= (1 + ϵ)
x .*= sqrt(1 + ϵ)
if t[end] > 2.0
t .= collect(-1.0:dt:1.0)
x[1:2:end] = x[n÷4+1:3n÷4+1]
x[2:2:end] = (x[1:2:end-2]+x[3:2:end])/2 + sqrt(dt)/2*randn(n÷2)
x[end÷2+1] = 0
end
T[] = t; X[] = x; sleep(0.02)
end
``````

22 Likes

@FedericoStra, found interesting to perform a zoom-in loop and recompute your fractal and to use animated gif technique shown by @lmiq to create:

``````xc, yc = -0.55, 0.61;
x0, x1 = xc - 2, xc + 2; y0, y1 = yc - 2, yc + 2;
anim = @animate for t in 1:50
x, y = range(x0, x1; length=1000), range(y0, y1; length=1000)
heatmap(x, y, -log.(mandelbrot.(x' .+ y .* im));aspect_ratio=1,border=:none,legend=:none);
x0, x1 = (15x0 + x1)/16, (15x1 + x0)/16;  y0, y1 = (15y0 + y1)/16, (15y1 + y0)/16
end
gif(anim,"mandelbrot_zooms.gif",fps=10)
``````
15 Likes

for reference, what is the extended code that gives your example (a) parallelized (b) with the actual L-J potential (c) both

1 Like

Nice! I was thinking of doing something similar, but I couldn’t find the time, nor fit it in less than 7 lines (you cheated! ).

I think that, to make it better, the number of iterations in `mandelbrot` has to increase the more one zooms in, otherwise the boundary of the fractal becomes too imprecise. Here is a complete version that does so:

``````using Plots
exprange(start, stop, len) = exp.(range(log(start), log(stop), length=len))
function mandelbrot(z; lim=75) w = z
for n = 1:lim;  abs2(w) < 4 ? w = w^2 + z : return n end
lim + 1
end
x₀, y₀ = -0.5626805, 0.6422555
anim = @animate for (r, l) in zip(exprange(2, 1.35e-6, 120), exprange(100, 2500, 120))
x = range(x₀-r, x₀+r; length=600); y = range(y₀-r, y₀+r; length=600);
heatmap(x, y, -log.(log.(mandelbrot.(x' .+ y .* im; lim=round(l))));
legend=:none, border=:none, ticks=:none, size=(600,600), ratio=1)
end
g = gif(anim; fps=12)
``````

Apparently I cannot attach the animated GIF file (which you can find here), so I include only the last frame:

10 Likes

Calculating a cross-correlation in any number of dimensions:

``````for I in CartesianIndices(img)
for J in CartesianIndices(kernel)    # kernel with centered indices
if I+J in CartesianIndices(img)
filtered[I] += img[I+J] * kernel[J]
end
end
end
``````

The code in action:

``````using OffsetArrays
using Makie

# A 3D "image" of random values (either 0 or 1)
img = rand([0.0, 1.0], 20, 60, 40)

# Plot
s1 = volume(img, algorithm=:iso, resolution=(800, 500))

# Function to generate an N-dimensional kernel of uniform values that sum to 1
# The center of the hypercube will have indices (0, 0, ..., 0)
uniform_kernel(n, dim) = OffsetArray(fill(1/n^dim, fill(n, dim)...), fill(-n÷2-1, dim)...)

# 5x5x5 array of uniform values that sum to 1
kernel = uniform_kernel(5, 3) # 3 dimensions

filtered = zero(img)

for I in CartesianIndices(img)
for J in CartesianIndices(kernel)
if I+J in CartesianIndices(img)
filtered[I] += img[I+J] * kernel[J]
end
end
end

s2 = volume(filtered, algorithm=:iso, resolution=(800, 500))
``````

Original image:

Filtered image:

14 Likes

What about a simulation with balls bouncing against each other?
Or a simulation of moving people getting infected with a virus when they meet other people.