私はこの質問がかなり古く、i80でこれに答えるのに優れた仕事をしたことを知っていますが、Apache Mathを使用して導関数または偏導関数を計算する非常に簡単な方法があることを(将来のSO-erのために)追加しようと思いました( Jacobian Matrixに対して独自の差別化を行うため)。それはDerivativeStructureです。
DerivativeStructureクラスを使用するためにi80andの回答を拡張します。
//Everything stays the same except for the Jacobian Matrix
import java.util.*;
import org.apache.commons.math3.analysis.ParametricUnivariateFunction;
import org.apache.commons.math3.fitting.AbstractCurveFitter;
import org.apache.commons.math3.fitting.leastsquares.LeastSquaresBuilder;
import org.apache.commons.math3.fitting.leastsquares.LeastSquaresProblem;
import org.apache.commons.math3.fitting.WeightedObservedPoint;
import org.apache.commons.math3.linear.DiagonalMatrix;
import org.apache.commons.math3.analysis.differentiation.DerivativeStructure;
class MyFunc implements ParametricUnivariateFunction {
public double value(double t, double... parameters) {
return parameters[0] * Math.pow(t, parameters[1]) * Math.exp(-parameters[2] * t);
}
// Jacobian matrix of the above. In this case, this is just an array of
// partial derivatives of the above function, with one element for each parameter.
public double[] gradient(double t, double... parameters) {
final double a = parameters[0];
final double b = parameters[1];
final double c = parameters[2];
// Jacobian Matrix Edit
// Using Derivative Structures...
// constructor takes 4 arguments - the number of parameters in your
// equation to be differentiated (3 in this case), the order of
// differentiation for the DerivativeStructure, the index of the
// parameter represented by the DS, and the value of the parameter itself
DerivativeStructure aDev = new DerivativeStructure(3, 1, 0, a);
DerivativeStructure bDev = new DerivativeStructure(3, 1, 1, b);
DerivativeStructure cDev = new DerivativeStructure(3, 1, 2, c);
// define the equation to be differentiated using another DerivativeStructure
DerivativeStructure y = aDev.multiply(DerivativeStructure.pow(t, bDev))
.multiply(cDev.negate().multiply(t).exp());
// then return the partial derivatives required
// notice the format, 3 arguments for the method since 3 parameters were
// specified first order derivative of the first parameter, then the second,
// then the third
return new double[] {
y.getPartialDerivative(1, 0, 0),
y.getPartialDerivative(0, 1, 0),
y.getPartialDerivative(0, 0, 1)
};
}
}
public class MyFuncFitter extends AbstractCurveFitter {
protected LeastSquaresProblem getProblem(Collection<WeightedObservedPoint> points) {
final int len = points.size();
final double[] target = new double[len];
final double[] weights = new double[len];
final double[] initialGuess = { 1.0, 1.0, 1.0 };
int i = 0;
for(WeightedObservedPoint point : points) {
target[i] = point.getY();
weights[i] = point.getWeight();
i += 1;
}
final AbstractCurveFitter.TheoreticalValuesFunction model = new
AbstractCurveFitter.TheoreticalValuesFunction(new MyFunc(), points);
return new LeastSquaresBuilder().
maxEvaluations(Integer.MAX_VALUE).
maxIterations(Integer.MAX_VALUE).
start(initialGuess).
target(target).
weight(new DiagonalMatrix(weights)).
model(model.getModelFunction(), model.getModelFunctionJacobian()).
build();
}
public static void main(String[] args) {
MyFuncFitter fitter = new MyFuncFitter();
ArrayList<WeightedObservedPoint> points = new ArrayList<WeightedObservedPoint>();
// Add points here; for instance,
WeightedObservedPoint point = new WeightedObservedPoint(1.0,
1.0,
1.0);
points.add(point);
final double coeffs[] = fitter.fit(points);
System.out.println(Arrays.toString(coeffs));
}
}
以上です。使用するのはかなり複雑で紛らわしいクラスであることは知っていますが、偏導関数を手作業で取得するのが面倒な非常に複雑な方程式を扱う場合(これはそれほど前に私には起こりませんでした)、または偏導関数を導出したい場合は、2次または3次と言います。
2次、3次などの次数導関数の場合、実行する必要があるのは次のとおりです。
// specify the required order as the second argument, say second order so 2
DerivativeStructure aDev = new DerivativeStructure(3, 2, 0, a);
DerivativeStructure bDev = new DerivativeStructure(3, 2, 1, b);
DerivativeStructure cDev = new DerivativeStructure(3, 2, 2, c);
// and then specify the order again here
y.getPartialDerivative(2, 0, 0),
y.getPartialDerivative(0, 2, 0),
y.getPartialDerivative(0, 0, 2)
うまくいけば、これはいつか誰かを助けるでしょう。