我试图使用scipy.optimize.minimize来拟合多元函数的参数，然而，无论我向优化器提供多少无噪声数据点，优化器都无法收敛到正确(或接近)的答案。
我想知道我使用优化器的方式是否有错误，但我一直在挠头寻找错误。我将感谢任何建议或猜测，谢谢!
import numpy as np
from scipy.optimize import minimize
import math
def get_transform(ai,aj,ak,x,y,z):
i,j,k = 0, 1, 2
si, sj, sk = math.sin(ai), math.sin(aj), math.sin(ak)
ci, cj, ck = math.cos(ai), math.cos(aj), math.cos(ak)
cc, cs = ci*ck, ci*sk
sc, ss = si*ck, si*sk
M = np.identity(4)
M[i, i] = cj*ck
M[i, j] = sj*sc-cs
M[i, k] = sj*cc+ss
M[j, i] = cj*sk
M[j, j] = sj*ss+cc
M[j, k] = sj*cs-sc
M[k, i] = -sj
M[k, j] = cj*si
M[k, k] = cj*ci
M[0, 3] = x
M[1, 3] = y
M[2, 3] = z
return M
def camera_intrinsic(fx, ppx, fy, ppy):
K = np.zeros((3, 3), dtype='float64')
K[0, 0], K[0, 2] = fx, ppx
K[1, 1], K[1, 2] = fy, ppy
K[2, 2] = 1
return K
def apply_transform(p, matrix):
rotation = matrix[0:3,0:3]
T = np.array([matrix[0][3],matrix[1][3],matrix[2][3]])
transformed = (np.dot(rotation, p.T).T)+T
return transformed
def project(points_3D,internal_calibration):
points_3D = points_3D.T
projections_2d = np.zeros((2, points_3D.shape[1]), dtype='float32')
camera_projection = (internal_calibration).dot(points_3D)
projections_2d[0, :] = camera_projection[0, :]/camera_projection[2, :]
projections_2d[1, :] = camera_projection[1, :]/camera_projection[2, :]
return projections_2d.T
def error(x):
global points,pixels
transform = get_transform(x[0],x[1],x[2],x[3],x[4],x[5])
points_transfered = apply_transform(points, transform)
internal_calibration = camera_intrinsic(x[6],x[7],x[8],x[9])
projected = project(points_transfered,internal_calibration)
# print(((projected-pixels)**2).mean())
return ((projected-pixels)**2).mean()
def generate(points, x):
transform = get_transform(x[0],x[1],x[2],x[3],x[4],x[5])
points_transfered = apply_transform(points, transform)
internal_calibration = camera_intrinsic(x[6],x[7],x[8],x[9])
projected = project(points_transfered,internal_calibration)
return projected
points = np.random.rand(100,3)
x_initial = np.random.rand(10)
pixels = generate(points,x_initial)
x_guess = np.random.rand(10)
results = minimize(error,x_guess, method='nelder-mead', tol = 1e-15)
x = results.x
print(x_initial)
print(x)
您正在解决最小二乘问题，但试图使用最小化标量函数的求解器来优化它。虽然它可能解决问题，但效率非常低。它可能需要更多的迭代，或者根本无法收敛。
更好的方法是使用least_squares而不是minimize。
为使其正常工作，您应该通过返回1D numpy数组而不是标量来修改error函数:
def error(x):
...
return (projected-pixels).flatten()
则调用least_squares:
results = least_squares(error, x_guess)
x = results.x
print(x_initial)
print(x)
print('error:', np.linalg.norm(error(x)))
同样，error(x)目前返回float32的数组，因为project中创建了一个float32的数组。它应该替换为float64，否则最小化无法收敛，因为当使用32位精度时，大多数梯度都会变为零。
def project(points_3D,internal_calibration):
...
projections_2d = np.zeros((2, points_3D.shape[1]), dtype='float64')
通过这些修改，求解器在大多数情况下收敛于解，但有时会失败。这是因为你随机生成了这个问题，所以在某些情况下，这个问题可能是退化的，或者没有物理意义。此类案件应自行调查。
还可以使用鲁棒损耗，如'arctan'，而不是线性损耗:
results = least_squares(error, x_guess, loss='arctan')
结果:
[0.68589904 0.68782115 0.83299068 0.02360941 0.19367124 0.54715374
0.37609235 0.62190714 0.98824796 0.88385802]
[0.68589904 0.68782115 0.83299068 0.02360941 0.19367124 0.54715374
0.37609235 0.62190714 0.98824796 0.88385802]
error: 1.2269443642313758e-12