SymbolicRegression.jl searches for symbolic expressions which optimize a particular objective.

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Install in Julia with:

using Pkg

The heart of this package is the EquationSearch function, which takes a 2D array (shape [features, rows]) and attempts to model a 1D array (shape [rows]) using analytic functional forms.

Run with:

using SymbolicRegression

X = randn(Float32, 5, 100)
y = 2 * cos.(X[4, :]) + X[1, :] .^ 2 .- 2

options = SymbolicRegression.Options(
    binary_operators=[+, *, /, -],
    unary_operators=[cos, exp],

hall_of_fame = EquationSearch(
    X, y, niterations=40, options=options,

You can view the resultant equations in the dominating Pareto front (best expression seen at each complexity) with:

dominating = calculate_pareto_frontier(X, y, hall_of_fame, options)

This is a vector of PopMember type - which contains the expression along with the score. We can get the expressions with:

trees = [member.tree for member in dominating]

Each of these equations is a Node{T} type for some constant type T (like Float32).

You can evaluate a given tree with:

tree = trees[end]
output, did_succeed = eval_tree_array(tree, X, options)

The output array will contain the result of the tree at each of the 100 rows. This did_succeed flag detects whether an evaluation was successful, or whether encountered any NaNs or Infs during calculation (such as, e.g., sqrt(-1)).

Constructing trees

You can also manipulate and construct trees directly. For example:

using SymbolicRegression

options = Options(;
    binary_operators=[+, -, *, ^, /], unary_operators=[cos, exp, sin]
x1, x2, x3 = [Node(; feature=i) for i=1:3]
tree = cos(x1 - 3.2 * x2) - x1^3.2

This tree has Float64 constants, so the type of the entire tree will be promoted to Node{Float64}.

We can convert all constants (recursively) to Float32:

float32_tree = convert(Node{Float32}, tree)

We can then evaluate this tree on a dataset:

X = rand(Float32, 3, 100)
output, did_succeed = eval_tree_array(tree, X, options)

Exporting to SymbolicUtils.jl

We can view the equations in the dominating Pareto frontier with:

dominating = calculate_pareto_frontier(X, y, hall_of_fame, options)

We can convert the best equation to SymbolicUtils.jl with the following function:

eqn = node_to_symbolic(dominating[end].tree, options)
println(simplify(eqn*5 + 3))

We can also print out the full pareto frontier like so:


for member in dominating
    complexity = compute_complexity(member.tree, options)
    loss = member.loss
    string = string_tree(member.tree, options)


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