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Add example for custom objective
Browse files- docs/examples.md +118 -1
docs/examples.md
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@@ -318,7 +318,124 @@ model.predict(X, -1)
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to make predictions with the most accurate expression.
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## 9.
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For the many other features available in PySR, please
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read the [Options section](options.md).
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to make predictions with the most accurate expression.
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+
## 9. Custom objectives
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You can also pass a custom objectives as a snippet of Julia code,
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which might include symbolic manipulations or custom functional forms.
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These do not even need to be differentiable! First, let's look at the
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default objective used (a simplified version, without weights
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and with mean square error), so that you can see how to write your own:
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```julia
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function default_objective(tree, dataset::Dataset{T,L}, options)::L where {T,L}
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(prediction, completion) = eval_tree_array(tree, dataset.X, options)
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if !completion
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return L(Inf)
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end
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diffs = prediction .- dataset.y
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return sum(diffs .^ 2) / length(diffs)
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end
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```
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Here, the `where {T,L}` syntax defines the function for arbitrary types `T` and `L`.
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If you have `precision=32` (default) and pass in regular floating point data,
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then both `T` and `L` will be equal to `Float32`. If you pass in complex data,
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then `T` will be `ComplexF32` and `L` will be `Float32` (since we need to return
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a real number from the loss function). But, you don't need to worry about this, just
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make sure to return a scalar number of type `L`.
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The `tree` argument is the current expression being evaluated. You can read
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about the `tree` fields [here](https://astroautomata.com/SymbolicRegression.jl/stable/types/).
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For example, let's fix a symbolic form of an expression,
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as a rational function. i.e., $P(X)/Q(X)$ for polynomials $P$ and $Q$.
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```python
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objective = """
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function my_custom_objective(tree, dataset::Dataset{T,L}, options) where {T,L}
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# Require root node to be binary, so we can split it,
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# otherwise return a large loss:
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tree.degree != 2 && return L(Inf)
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P = tree.l
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Q = tree.r
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# Evaluate numerator:
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P_prediction, flag = eval_tree_array(P, dataset.X, options)
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!flag && return L(Inf)
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# Evaluate denominator:
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Q_prediction, flag = eval_tree_array(Q, dataset.X, options)
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!flag && return L(Inf)
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# Impose functional form:
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prediction = P_prediction ./ Q_prediction
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diffs = prediction .- dataset.y
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return sum(diffs .^ 2) / length(diffs)
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end
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"""
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model = PySRRegressor(
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niterations=100,
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binary_operators=["*", "+", "-"],
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full_objective=objective,
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)
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```
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> **Warning**: When using a custom objective like this that performs symbolic
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> manipulations, many functionalities of PySR will not work, such as `.sympy()`,
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> `.predict()`, etc. This is because the SymPy parsing does not know about
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> how you are manipulating the expression, so you will need to do this yourself.
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Note how we did not pass `/` as a binary operator; it will just be implicit
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in the functional form.
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Let's generate an equation of the form $\frac{x_0^2 x_1 - 2}{x_2^2 + 1}$:
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```python
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X = np.random.randn(1000, 3)
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y = (X[:, 0]**2 * X[:, 1] - 2) / (X[:, 2]**2 + 1)
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```
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Finally, let's fit:
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```python
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model.fit(X, y)
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```
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> Note that the printed equation is not the same as the evaluated equation,
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> because the printing functionality does not know about the functional form.
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We can get the string format with:
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```python
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model.get_best().equation
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```
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(or, you could use `model.equations_.iloc[-1].equation`)
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For me, this equation was:
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```text
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(((2.3554819 + -0.3554746) - (x1 * (x0 * x0))) - (-1.0000019 - (x2 * x2)))
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```
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looking at the bracket structure of the equation, we can see that the outermost
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bracket is split at the `-` operator (note that we ignore the root operator in
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the evaluation, as we simply evaluated each argument and divided the result) into
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`((2.3554819 + -0.3554746) - (x1 * (x0 * x0)))` and
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`(-1.0000019 - (x2 * x2))`, meaning that our discovered equation is
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equal to:
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$\frac{x_0^2 x_1 - 2.0000073}{x_2^2 - 1.0000019}$, which
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is nearly the same as the true equation!
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## 10. Additional features
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For the many other features available in PySR, please
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read the [Options section](options.md).
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