ceres solver 05 例子：曲线拟合

作者: book_02 | 来源:发表于2019-12-06 21:11 被阅读0次

给出有噪声的曲线数据，用Ceres Solver拟合出曲线方程

1. 曲线方程

假定曲线方程
$y = e^{mx + c}$
我们取 $m=0.3, c=0.1$ 即 $y = e^{0.3x + 0.1}$ ，然后根据这个方程取67个点，然后给这67个点加上标准差为0.2的高斯噪声。

2. Ceres 求解过程

2.1 定义代价函数

把 $y = e^{mx + c}$ 录入到代价函数中去，不过这时有了数据项x和y

struct ExponentialResidual {
  ExponentialResidual(double x, double y)
      : x_(x), y_(y) {}

  template <typename T> bool operator()(const T* const m,
                                        const T* const c,
                                        T* residual) const {
    residual[0] = y_ - exp(m[0] * x_ + c[0]);
    return true;
  }

 private:
  const double x_;
  const double y_;
};

2.2 构建问题，添加残差项

  double m = 0.0;
  double c = 0.0;

  Problem problem;
  for (int i = 0; i < kNumObservations; ++i) {
    problem.AddResidualBlock(
        new AutoDiffCostFunction<ExponentialResidual, 1, 1, 1>(
            new ExponentialResidual(data[2 * i], data[2 * i + 1])),
        NULL,
        &m, &c);
  }

data数组里存的是上面我们生成的带有噪声的曲线上点的数据

2.3 定义求解器的选项参数，并求解

  Solver::Options options;
  options.max_num_iterations = 25;
  options.linear_solver_type = ceres::DENSE_QR;
  options.minimizer_progress_to_stdout = true;

  Solver::Summary summary;
  Solve(options, &problem, &summary);

3. 完整代码

#include "ceres/ceres.h"
#include "glog/logging.h"

using ceres::AutoDiffCostFunction;
using ceres::CostFunction;
using ceres::Problem;
using ceres::Solver;
using ceres::Solve;

// Data generated using the following octave code.
//   randn('seed', 23497);
//   m = 0.3;
//   c = 0.1;
//   x=[0:0.075:5];
//   y = exp(m * x + c);
//   noise = randn(size(x)) * 0.2;
//   y_observed = y + noise;
//   data = [x', y_observed'];

const int kNumObservations = 67;
const double data[] = {
  0.000000e+00, 1.133898e+00,
  7.500000e-02, 1.334902e+00,
  1.500000e-01, 1.213546e+00,
  2.250000e-01, 1.252016e+00,
  3.000000e-01, 1.392265e+00,
  3.750000e-01, 1.314458e+00,
  4.500000e-01, 1.472541e+00,
  5.250000e-01, 1.536218e+00,
  6.000000e-01, 1.355679e+00,
  6.750000e-01, 1.463566e+00,
  7.500000e-01, 1.490201e+00,
  8.250000e-01, 1.658699e+00,
  9.000000e-01, 1.067574e+00,
  9.750000e-01, 1.464629e+00,
  1.050000e+00, 1.402653e+00,
  1.125000e+00, 1.713141e+00,
  1.200000e+00, 1.527021e+00,
  1.275000e+00, 1.702632e+00,
  1.350000e+00, 1.423899e+00,
  1.425000e+00, 1.543078e+00,
  1.500000e+00, 1.664015e+00,
  1.575000e+00, 1.732484e+00,
  1.650000e+00, 1.543296e+00,
  1.725000e+00, 1.959523e+00,
  1.800000e+00, 1.685132e+00,
  1.875000e+00, 1.951791e+00,
  1.950000e+00, 2.095346e+00,
  2.025000e+00, 2.361460e+00,
  2.100000e+00, 2.169119e+00,
  2.175000e+00, 2.061745e+00,
  2.250000e+00, 2.178641e+00,
  2.325000e+00, 2.104346e+00,
  2.400000e+00, 2.584470e+00,
  2.475000e+00, 1.914158e+00,
  2.550000e+00, 2.368375e+00,
  2.625000e+00, 2.686125e+00,

  2.700000e+00, 2.712395e+00,
  2.775000e+00, 2.499511e+00,
  2.850000e+00, 2.558897e+00,
  2.925000e+00, 2.309154e+00,
  3.000000e+00, 2.869503e+00,
  3.075000e+00, 3.116645e+00,
  3.150000e+00, 3.094907e+00,
  3.225000e+00, 2.471759e+00,
  3.300000e+00, 3.017131e+00,
  3.375000e+00, 3.232381e+00,
  3.450000e+00, 2.944596e+00,
  3.525000e+00, 3.385343e+00,
  3.600000e+00, 3.199826e+00,
  3.675000e+00, 3.423039e+00,
  3.750000e+00, 3.621552e+00,
  3.825000e+00, 3.559255e+00,
  3.900000e+00, 3.530713e+00,
  3.975000e+00, 3.561766e+00,
  4.050000e+00, 3.544574e+00,
  4.125000e+00, 3.867945e+00,
  4.200000e+00, 4.049776e+00,
  4.275000e+00, 3.885601e+00,
  4.350000e+00, 4.110505e+00,
  4.425000e+00, 4.345320e+00,
  4.500000e+00, 4.161241e+00,
  4.575000e+00, 4.363407e+00,
  4.650000e+00, 4.161576e+00,
  4.725000e+00, 4.619728e+00,
  4.800000e+00, 4.737410e+00,
  4.875000e+00, 4.727863e+00,
  4.950000e+00, 4.669206e+00,
};

struct ExponentialResidual {
  ExponentialResidual(double x, double y)
      : x_(x), y_(y) {}

  template <typename T> bool operator()(const T* const m,
                                        const T* const c,
                                        T* residual) const {
    residual[0] = y_ - exp(m[0] * x_ + c[0]);
    return true;
  }

 private:
  const double x_;
  const double y_;
};

int main(int argc, char** argv) {
  google::InitGoogleLogging(argv[0]);

  double m = 0.0;
  double c = 0.0;

  Problem problem;
  for (int i = 0; i < kNumObservations; ++i) {
    problem.AddResidualBlock(
        new AutoDiffCostFunction<ExponentialResidual, 1, 1, 1>(
            new ExponentialResidual(data[2 * i], data[2 * i + 1])),
        NULL,
        &m, &c);
  }

  Solver::Options options;
  options.max_num_iterations = 25;
  options.linear_solver_type = ceres::DENSE_QR;
  options.minimizer_progress_to_stdout = true;

  Solver::Summary summary;
  Solve(options, &problem, &summary);
  std::cout << summary.BriefReport() << "\n";
  std::cout << "Initial m: " << 0.0 << " c: " << 0.0 << "\n";
  std::cout << "Final   m: " << m << " c: " << c << "\n";
  return 0;
}

运行结果如下：

  iter      cost      cost_change  |gradient|   |step|    tr_ratio  tr_radius  ls_iter  iter_time  total_time
   0  1.211734e+02    0.00e+00    3.61e+02   0.00e+00   0.00e+00  1.00e+04       0    5.34e-04    2.56e-03
   1  1.211734e+02   -2.21e+03    0.00e+00   7.52e-01  -1.87e+01  5.00e+03       1    4.29e-05    3.25e-03
   2  1.211734e+02   -2.21e+03    0.00e+00   7.51e-01  -1.86e+01  1.25e+03       1    1.10e-05    3.28e-03
   3  1.211734e+02   -2.19e+03    0.00e+00   7.48e-01  -1.85e+01  1.56e+02       1    1.41e-05    3.31e-03

   4  1.211734e+02   -2.02e+03    0.00e+00   7.22e-01  -1.70e+01  9.77e+00       1    1.00e-05    3.34e-03
   5  1.211734e+02   -7.34e+02    0.00e+00   5.78e-01  -6.32e+00  3.05e-01       1    1.00e-05    3.36e-03
   6  3.306595e+01    8.81e+01    4.10e+02   3.18e-01   1.37e+00  9.16e-01       1    2.79e-05    3.41e-03
   7  6.426770e+00    2.66e+01    1.81e+02   1.29e-01   1.10e+00  2.75e+00       1    2.10e-05    3.45e-03
   8  3.344546e+00    3.08e+00    5.51e+01   3.05e-02   1.03e+00  8.24e+00       1    2.10e-05    3.48e-03

   9  1.987485e+00    1.36e+00    2.33e+01   8.87e-02   9.94e-01  2.47e+01       1    2.10e-05    3.52e-03
  10  1.211585e+00    7.76e-01    8.22e+00   1.05e-01   9.89e-01  7.42e+01       1    2.10e-05    3.56e-03
  11  1.063265e+00    1.48e-01    1.44e+00   6.06e-02   9.97e-01  2.22e+02       1    2.60e-05    3.61e-03
  12  1.056795e+00    6.47e-03    1.18e-01   1.47e-02   1.00e+00  6.67e+02       1    2.10e-05    3.64e-03
  13  1.056751e+00    4.39e-05    3.79e-03   1.28e-03   1.00e+00  2.00e+03       1    2.10e-05    3.68e-03

Ceres Solver Report: Iterations: 13, Initial cost: 1.211734e+02, Final cost: 1.056751e+00, Termination: CONVERGENCE
Initial m: 0 c: 0
Final   m: 0.291861 c: 0.131439

曲线拟合结果

最后求出来m=0.291861, c=0.131439,与我们设定的m=0.3, c=0.1相近

4. 总结

总结使用步骤如下：

定义代价函数。根据目标函数定义代价代价函数
构建问题。添加残差项。这里要根据有噪声的数据点，不断地添加残差项。
求解问题。定义求解器的选项参数和求解器的结果报告

5. 参考

http://ceres-solver.org/nnls_tutorial.html#curve-fitting

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