Toy Examples with Code

using SymbolicRegression
using MLJ

1. Simple search

Here's a simple example where we find the expression 2 cos(x4) + x1^2 - 2.

X = 2randn(1000, 5)
y = @. 2*cos(X[:, 4]) + X[:, 1]^2 - 2

model = SRRegressor(
    binary_operators=[+, -, *, /],
    unary_operators=[cos],
    niterations=30
)
mach = machine(model, X, y)
fit!(mach)

Let's look at the returned table:

r = report(mach)
r

We can get the selected best tradeoff expression with:

r.equations[r.best_idx]

2. Custom operator

Here, we define a custom operator and use it to find an expression:

X = 2randn(1000, 5)
y = @. 1/X[:, 1]

my_inv(x) = 1/x

model = SRRegressor(
    binary_operators=[+, *],
    unary_operators=[my_inv],
)
mach = machine(model, X, y)
fit!(mach)
r = report(mach)
println(r.equations[r.best_idx])

3. Multiple outputs

Here, we do the same thing, but with multiple expressions at once, each requiring a different feature. This means that we need to use MultitargetSRRegressor instead of SRRegressor:

X = 2rand(1000, 5) .+ 0.1
y = @. 1/X[:, 1:3]

my_inv(x) = 1/x

model = MultitargetSRRegressor(; binary_operators=[+, *], unary_operators=[my_inv])
mach = machine(model, X, y)
fit!(mach)

The report gives us lists of expressions instead:

r = report(mach)
for i in 1:3
    println("y[$(i)] = ", r.equations[i][r.best_idx[i]])
end

4. Plotting an expression

For now, let's consider the expressions for output 1 from the previous example: We can get a SymbolicUtils version with:

using SymbolicUtils

eqn = node_to_symbolic(r.equations[1][r.best_idx[1]], model)

We can get the LaTeX version with Latexify:

using Latexify

latexify(string(eqn))

We can also plot the prediction against the truth:

using Plots

ypred = predict(mach, X)
scatter(y[1, :], ypred[1, :], xlabel="Truth", ylabel="Prediction")

5. Other types

SymbolicRegression.jl can handle most numeric types you wish to use. For example, passing a Float32 array will result in the search using 32-bit precision everywhere in the codebase:

X = 2randn(Float32, 1000, 5)
y = @. 2*cos(X[:, 4]) + X[:, 1]^2 - 2

model = SRRegressor(binary_operators=[+, -, *, /], unary_operators=[cos], niterations=30)
mach = machine(model, X, y)
fit!(mach)

we can see that the output types are Float32:

r = report(mach)
best = r.equations[r.best_idx]
println(typeof(best))
# Node{Float32}

We can also use Complex numbers (ignore the warning from MLJ):

cos_re(x::Complex{T}) where {T} = cos(abs(x)) + 0im

X = 15 .* rand(ComplexF64, 1000, 5) .- 7.5
y = @. 2*cos_re((2+1im) * X[:, 4]) + 0.1 * X[:, 1]^2 - 2

model = SRRegressor(
    binary_operators=[+, -, *, /],
    unary_operators=[cos_re],
    maxsize=30,
    niterations=100
)
mach = machine(model, X, y)
fit!(mach)

6. Dimensional constraints

One other feature we can exploit is dimensional analysis. Say that we know the physical units of each feature and output, and we want to find an expression that is dimensionally consistent.

We can do this as follows, using DynamicQuantities to assign units. First, let's make some data on Newton's law of gravitation:

using DynamicQuantities
using SymbolicRegression

M = (rand(100) .+ 0.1) .* Constants.M_sun
m = 100 .* (rand(100) .+ 0.1) .* u"kg"
r = (rand(100) .+ 0.1) .* Constants.R_earth

G = Constants.G

F = @. (G * M * m / r^2)

(Note that the u macro from DynamicQuantities will automatically convert to SI units. To avoid this, use the us macro.)

Now, let's ready the data for MLJ:

X = (; M=M, m=m, r=r)
y = F

Since this data has such a large dynamic range, let's also create a custom loss function that looks at the error in log-space:

function loss_fnc(prediction, target)
    # Useful loss for large dynamic range
    scatter_loss = abs(log((abs(prediction)+1e-20) / (abs(target)+1e-20)))
    sign_loss = 10 * (sign(prediction) - sign(target))^2
    return scatter_loss + sign_loss
end

Now let's define and fit our model:

model = SRRegressor(
    binary_operators=[+, -, *, /],
    unary_operators=[square],
    elementwise_loss=loss_fnc,
    complexity_of_constants=2,
    maxsize=25,
    niterations=100,
    populations=50,
    dimensional_constraint_penalty=10^5,
)
mach = machine(model, X, y)
fit!(mach)

You can observe that all expressions with a loss under our penalty are dimensionally consistent! (The "[?]" indicates free units in a constant, which can cancel out other units in the expression.) For example,

"y[m s⁻² kg] = (M[kg] * 2.6353e-22[?])"

would indicate that the expression is dimensionally consistent, with a constant "2.6353e-22[m s⁻²]".

Note that you can also search for dimensionless units by settings dimensionless_constants_only to true.

7. Additional features

For the many other features available in SymbolicRegression.jl, check out the API page for Options. You might also find it useful to browse the documentation for the Python frontend PySR, which has additional documentation. In particular, the tuning page is useful for improving search performance.