16

我正在尝试最小化以下示例函数：

F(x) = f[0]^2(x[0],...,x[n-1]) + ... + f[m-1]^2(x[0],...,x[n-1])

最小化这种函数的正常方法可能是 Levenberg-Marquardt 算法。我想在 C++ 中执行此最小化，并使用 Eigen 进行了一些初始测试，从而产生了预期的解决方案。

我的问题如下：我习惯于使用 ie 在 python 中进行优化scipy.optimize.fmin_powell。这里的输入函数参数是(func, x0, args=(), xtol=0.0001, ftol=0.0001, maxiter=None, maxfun=None, full_output=0, disp=1, retall=0, callback=None, direc=None)。所以我可以定义 a func(x0)，给出x0向量并开始优化。如果需要，我可以更改优化参数。

现在，Eigen Lev-Marq 算法以不同的方式工作。我需要定义一个函数向量（为什么？）此外，我无法设置优化参数。根据：
http
://eigen.tuxfamily.org/dox/unsupported/classEigen_1_1LevenbergMarquardt.html 我应该可以使用setEpsilon()和其他设置功能。

但是当我有以下代码时：

my_functor functor;
Eigen::NumericalDiff<my_functor> numDiff(functor);
Eigen::LevenbergMarquardt<Eigen::NumericalDiff<my_functor>,double> lm(numDiff);
lm.setEpsilon(); //doesn't exist!

所以我有两个问题：

  1. 为什么需要函数向量，为什么函数标量不够？
    我搜索答案的参考资料：
    http ://www.ultimatepp.org/reference$Eigen_demo$en-us.html
    http://www.alglib.net/optimization/levenbergmarquardt.php

  2. 如何使用 set 函数设置优化参数？

4

4 回答 4

24

所以我相信我已经找到了答案。

1) 该函数能够作为函数向量和函数标量工作。
如果存在可m解参数，则需要创建或数值计算 mxm 的雅可比矩阵。为了进行矩阵向量乘法J(x[m]).transpose*f(x[m])，函数向量f(x)应该有m项目。这可以是m不同的功能，但我们也可以给出f1完整的功能并制作其他项目0。

2）参数可以设置和读取使用lm.parameters.maxfev = 2000;

两个答案都已在以下示例代码中进行了测试：

#include <iostream>
#include <Eigen/Dense>

#include <unsupported/Eigen/NonLinearOptimization>
#include <unsupported/Eigen/NumericalDiff>

// Generic functor
template<typename _Scalar, int NX = Eigen::Dynamic, int NY = Eigen::Dynamic>
struct Functor
{
typedef _Scalar Scalar;
enum {
    InputsAtCompileTime = NX,
    ValuesAtCompileTime = NY
};
typedef Eigen::Matrix<Scalar,InputsAtCompileTime,1> InputType;
typedef Eigen::Matrix<Scalar,ValuesAtCompileTime,1> ValueType;
typedef Eigen::Matrix<Scalar,ValuesAtCompileTime,InputsAtCompileTime> JacobianType;

int m_inputs, m_values;

Functor() : m_inputs(InputsAtCompileTime), m_values(ValuesAtCompileTime) {}
Functor(int inputs, int values) : m_inputs(inputs), m_values(values) {}

int inputs() const { return m_inputs; }
int values() const { return m_values; }

};

struct my_functor : Functor<double>
{
my_functor(void): Functor<double>(2,2) {}
int operator()(const Eigen::VectorXd &x, Eigen::VectorXd &fvec) const
{
    // Implement y = 10*(x0+3)^2 + (x1-5)^2
    fvec(0) = 10.0*pow(x(0)+3.0,2) +  pow(x(1)-5.0,2);
    fvec(1) = 0;

    return 0;
}
};


int main(int argc, char *argv[])
{
Eigen::VectorXd x(2);
x(0) = 2.0;
x(1) = 3.0;
std::cout << "x: " << x << std::endl;

my_functor functor;
Eigen::NumericalDiff<my_functor> numDiff(functor);
Eigen::LevenbergMarquardt<Eigen::NumericalDiff<my_functor>,double> lm(numDiff);
lm.parameters.maxfev = 2000;
lm.parameters.xtol = 1.0e-10;
std::cout << lm.parameters.maxfev << std::endl;

int ret = lm.minimize(x);
std::cout << lm.iter << std::endl;
std::cout << ret << std::endl;

std::cout << "x that minimizes the function: " << x << std::endl;

std::cout << "press [ENTER] to continue " << std::endl;
std::cin.get();
return 0;
}
于 2013-08-30T15:15:40.517 回答
9

这个答案是两个现有答案的扩展：1）我改编了@Deepfreeze 提供的源代码，以包含额外的注释和两个不同的测试功能。2）我使用@user3361661 的建议以正确的形式重写目标函数。正如他所建议的，它将我的第一个测试问题的迭代次数从 67 次减少到 4 次。

#include <iostream>
#include <Eigen/Dense>

#include <unsupported/Eigen/NonLinearOptimization>
#include <unsupported/Eigen/NumericalDiff>

/***********************************************************************************************/

// Generic functor
// See http://eigen.tuxfamily.org/index.php?title=Functors
// C++ version of a function pointer that stores meta-data about the function
template<typename _Scalar, int NX = Eigen::Dynamic, int NY = Eigen::Dynamic>
struct Functor
{

  // Information that tells the caller the numeric type (eg. double) and size (input / output dim)
  typedef _Scalar Scalar;
  enum { // Required by numerical differentiation module
      InputsAtCompileTime = NX,
      ValuesAtCompileTime = NY
  };

  // Tell the caller the matrix sizes associated with the input, output, and jacobian
  typedef Eigen::Matrix<Scalar,InputsAtCompileTime,1> InputType;
  typedef Eigen::Matrix<Scalar,ValuesAtCompileTime,1> ValueType;
  typedef Eigen::Matrix<Scalar,ValuesAtCompileTime,InputsAtCompileTime> JacobianType;

  // Local copy of the number of inputs
  int m_inputs, m_values;

  // Two constructors:
  Functor() : m_inputs(InputsAtCompileTime), m_values(ValuesAtCompileTime) {}
  Functor(int inputs, int values) : m_inputs(inputs), m_values(values) {}

  // Get methods for users to determine function input and output dimensions
  int inputs() const { return m_inputs; }
  int values() const { return m_values; }

};

/***********************************************************************************************/

// https://en.wikipedia.org/wiki/Test_functions_for_optimization
// Booth Function
// Implement f(x,y) = (x + 2*y -7)^2 + (2*x + y - 5)^2
struct BoothFunctor : Functor<double>
{
  // Simple constructor
  BoothFunctor(): Functor<double>(2,2) {}

  // Implementation of the objective function
  int operator()(const Eigen::VectorXd &z, Eigen::VectorXd &fvec) const {
    double x = z(0);   double y = z(1);
    /*
     * Evaluate the Booth function.
     * Important: LevenbergMarquardt is designed to work with objective functions that are a sum
     * of squared terms. The algorithm takes this into account: do not do it yourself.
     * In other words: objFun = sum(fvec(i)^2)
     */
    fvec(0) = x + 2*y - 7;
    fvec(1) = 2*x + y - 5;
    return 0;
  }
};

/***********************************************************************************************/

// https://en.wikipedia.org/wiki/Test_functions_for_optimization
// Himmelblau's Function
// Implement f(x,y) = (x^2 + y - 11)^2 + (x + y^2 - 7)^2
struct HimmelblauFunctor : Functor<double>
{
  // Simple constructor
  HimmelblauFunctor(): Functor<double>(2,2) {}

  // Implementation of the objective function
  int operator()(const Eigen::VectorXd &z, Eigen::VectorXd &fvec) const {
    double x = z(0);   double y = z(1);
    /*
     * Evaluate Himmelblau's function.
     * Important: LevenbergMarquardt is designed to work with objective functions that are a sum
     * of squared terms. The algorithm takes this into account: do not do it yourself.
     * In other words: objFun = sum(fvec(i)^2)
     */
    fvec(0) = x * x + y - 11;
    fvec(1) = x + y * y - 7;
    return 0;
  }
};

/***********************************************************************************************/

void testBoothFun() {
  std::cout << "Testing the Booth function..." << std::endl;
  Eigen::VectorXd zInit(2); zInit << 1.87, 2.032;
  std::cout << "zInit: " << zInit.transpose() << std::endl;
  Eigen::VectorXd zSoln(2); zSoln << 1.0, 3.0;
  std::cout << "zSoln: " << zSoln.transpose() << std::endl;

  BoothFunctor functor;
  Eigen::NumericalDiff<BoothFunctor> numDiff(functor);
  Eigen::LevenbergMarquardt<Eigen::NumericalDiff<BoothFunctor>,double> lm(numDiff);
  lm.parameters.maxfev = 1000;
  lm.parameters.xtol = 1.0e-10;
  std::cout << "max fun eval: " << lm.parameters.maxfev << std::endl;
  std::cout << "x tol: " << lm.parameters.xtol << std::endl;

  Eigen::VectorXd z = zInit;
  int ret = lm.minimize(z);
  std::cout << "iter count: " << lm.iter << std::endl;
  std::cout << "return status: " << ret << std::endl;
  std::cout << "zSolver: " << z.transpose() << std::endl;
  std::cout << "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~" << std::endl;
}

/***********************************************************************************************/

void testHimmelblauFun() {
  std::cout << "Testing the Himmelblau function..." << std::endl;
  // Eigen::VectorXd zInit(2); zInit << 0.0, 0.0;  // soln 1
  // Eigen::VectorXd zInit(2); zInit << -1, 1;  // soln 2
  // Eigen::VectorXd zInit(2); zInit << -1, -1;  // soln 3
  Eigen::VectorXd zInit(2); zInit << 1, -1;  // soln 4
  std::cout << "zInit: " << zInit.transpose() << std::endl;
  std::cout << "soln 1: [3.0, 2.0]" << std::endl;
  std::cout << "soln 2: [-2.805118, 3.131312]" << std::endl;
  std::cout << "soln 3: [-3.77931, -3.28316]" << std::endl;
  std::cout << "soln 4: [3.584428, -1.848126]" << std::endl;

  HimmelblauFunctor functor;
  Eigen::NumericalDiff<HimmelblauFunctor> numDiff(functor);
  Eigen::LevenbergMarquardt<Eigen::NumericalDiff<HimmelblauFunctor>,double> lm(numDiff);
  lm.parameters.maxfev = 1000;
  lm.parameters.xtol = 1.0e-10;
  std::cout << "max fun eval: " << lm.parameters.maxfev << std::endl;
  std::cout << "x tol: " << lm.parameters.xtol << std::endl;

  Eigen::VectorXd z = zInit;
  int ret = lm.minimize(z);
  std::cout << "iter count: " << lm.iter << std::endl;
  std::cout << "return status: " << ret << std::endl;
  std::cout << "zSolver: " << z.transpose() << std::endl;
  std::cout << "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~" << std::endl;
}

/***********************************************************************************************/

int main(int argc, char *argv[])
{

std::cout << "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~" << std::endl;
testBoothFun();
testHimmelblauFun();
return 0;
}

运行此测试脚本的命令行输出为：

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Testing the Booth function...
zInit:  1.87 2.032
zSoln: 1 3
max fun eval: 1000
x tol: 1e-10
iter count: 4
return status: 2
zSolver: 1 3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Testing the Himmelblau function...
zInit:  1 -1
soln 1: [3.0, 2.0]
soln 2: [-2.805118, 3.131312]
soln 3: [-3.77931, -3.28316]
soln 4: [3.584428, -1.848126]
max fun eval: 1000
x tol: 1e-10
iter count: 8
return status: 2
zSolver:  3.58443 -1.84813
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
于 2018-03-02T21:58:05.943 回答
6

作为替代方案，您可以像这样简单地创建一个新的仿函数，

struct my_functor_w_df : Eigen::NumericalDiff<my_functor> {};

然后像这样初始化 LevenbergMarquardt 实例，

my_functor_w_df functor;
Eigen::LevenbergMarquardt<my_functor_w_df> lm(functor);

就个人而言，我觉得这种方法更干净一些。

于 2014-03-26T17:30:48.073 回答
4

似乎功能更通用：

  1. 假设您有一个 m 参数模型。
  2. 您有 n 个观测值，希望以最小二乘的方式拟合 m 参数模型。
  3. 如果提供雅可比矩阵，它将是 n 乘以 m。

您需要在 fvec 中提供 n 个错误值。此外，不需要对f 值进行平方，因为隐含地假设整体误差函数由 fvec 分量的平方和组成。

因此，如果您遵循这些准则并将代码更改为：

fvec(0) = sqrt(10.0)*(x(0)+3.0);
fvec(1) = x(1)-5.0;

它将以极少的迭代次数收敛——比如少于 5 次。我还在一个更复杂的例子上进行了尝试——http: //www.itl.nist.gov/div898/strd/nls/data/上的 Hahn1 基准测试hahn1.shtml具有 m=7 参数和 n=236 个观察值，它仅在 11 次迭代中收敛到已知的正确解，并使用数值计算的雅可比行列式。

于 2014-02-27T17:34:55.633 回答