如何表征最小二乘估计的适应度

Question

我正在进行一个本地化项目，并使用最小二乘估计来确定发射器的位置。我需要一种方法来统计地描述我的解决方案在我的程序中的“适合度”，这可以用来告诉我我是否有一个好的答案，或者我需要额外的测量，或者有坏的数据。我已经阅读了一些有关使用“确定系数”或 R 平方的信息，但找不到任何好的示例。任何关于如何表征我是否有一个好的解决方案或需要额外测量的想法将不胜感激。

谢谢！

我的代码给了我以下输出，

grid_lat 和 grid_lon 对应于可能目标位置网格的经纬度坐标

grid_lat = [[ 38.16755799  38.16755799  38.16755799  38.16755799  38.16755799
  38.16755799]
  [ 38.17717199  38.17717199  38.17717199  38.17717199  38.17717199
    38.17717199]
  [ 38.186786    38.186786    38.186786    38.186786    38.186786    38.186786  ]
  [ 38.1964      38.1964      38.1964      38.1964      38.1964      38.1964    ]
  [ 38.20601401  38.20601401  38.20601401  38.20601401  38.20601401
    38.20601401]
  [ 38.21562801  38.21562801  38.21562801  38.21562801  38.21562801
    38.21562801]
  [ 38.22524202  38.22524202  38.22524202  38.22524202  38.22524202
    38.22524202]]

grid_lon = [[-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]
  [-75.83805812 -75.83006167 -75.82206522 -75.81406878 -75.80607233
    -75.79807588]]

grid_error 对应于每个点的解决方案的“好”程度。如果我们有 0.0 的误差，我们就有一个完美的解决方案。网格误差针对网格上的每个点计算到每个测量位置（下面测量中的轨迹）。每个测量位置都有一个到发射器的估计范围。 “误差”对应于从测量到发射机的估计范围，减去在测量范围位置和网格点之间计算的实际范围。误差越小，我们越接近实际发射器位置

# Calculate distance between every grid point and every measurement in meters 
measured_distance = spatial.distance.cdist(grid_ecef_array, measurement_ecef_array,    'euclidean')

measurement_error = [pow((measurement - estimated_distance),2) for measurement in measured_distance]

mean_squared_error = [numpy.sqrt(numpy.mean(measurement)) for measurement in measurement_error]

# Find minimum solution 
# Convert array of mean_squared_errors to 2D grid for graphing
N3, N4 = numpy.array(grid_lon).shape
grid_error = numpy.array(mean_squared_error).reshape((N3, N4))

grid_error = [[ 2.33608445  2.02805063  1.85638288  1.84620283  2.02757163  2.38035108]
  [ 1.73675429  1.40649524  1.21799211  1.06503271  1.27373554  1.74265406]
  [ 1.44967789  0.96835022  0.62667257  0.52804942  0.91189678  1.50067864]
  [ 1.70155286  1.24024402  0.9642869   1.00517531  1.32606411  1.81754752]
  [ 2.40218247  2.07449106  1.91044903  1.94272889  2.15511638  2.51683715]
  [ 3.29679348  3.05353929  2.93662134  2.95839307  3.11583615  3.39320682]
  [ 4.27303679  4.08195869  3.99203754  4.00926823  4.13247105  4.35378011]]

# Generate the 3D plot with the Z coordinate being the mean squared error estimate
plot3Dcoordinates(grid_lon, grid_lat, grid_error)

# Generic function using matplotlib to plot coordinates
def plot3Dcoordinates(X, Y, Z):
    fig = plt.figure()
    ax = Axes3D(fig)

    surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.jet,
          linewidth=0, antialiased=False)

    fig.colorbar(surf, shrink=0.5, aspect=5)

这是在更大的网格上处理算法的示例图像。我可以直观地看出我有一个很好的解决方案，因为形状平滑地收敛在一个最小点（解决方案）上，看起来有点像倒置的女巫帽。

第二张图片显示了所有测量值和位置，解决方案绘制在顶部，最小点作为解决方案（红色 x）。

Answer 1

R 平方越接近 1.0，您的拟合度就越好。选择你自己的“足够好”的阈值，我认为典型的阈值在 .92-.98 范围内。