Question

我试图通过python进行bundle调整。所以我测试了非线性最小二乘模块。然后我编写了如下代码。我想要正确Pmat代表三个相机的相机投影矩阵。但是我有一个错误，“ValueError：对象太深了所需的数组”。

任何可以提供解决此问题的线索的人？

此致 Jinho Yoo。

from math import* from numpy import *

import pylab as p from scipy.optimize
import leastsq  

Projected_x = \ mat([[ -69.69  , 255.3825,    1.    ],
        [ -69.69  ,  224.6175,    1.    ],
        [-110.71  ,  224.6175,    1.    ],
        [-110.71  ,  255.3825,    1.    ],
        [ 709.69  ,  224.6175,    1.    ],
        [ 709.69  ,  255.3825,    1.    ],
        [ 750.71  ,  255.3825,    1.    ],
        [ 750.71  ,  224.6175,    1.    ]])

Projected_x = Projected_x.transpose()

Pmat = \ mat(   [[  5.79746167e+02,  0.00000000e+00,   3.20000000e+02,    0.00000000e+00],
        [  0.00000000e+00,   4.34809625e+02,   2.40000000e+02, 0.00000000e+00],
        [  0.00000000e+00,   0.00000000e+00,   1.00000000e+00, 0.00000000e+00] ]  )

reconst_X = \ mat([[-0.95238194, -0.58146697,  0.61506506,  0.00539229],
        [-0.99566105, -0.76178453,  0.72451719,  0.00502341],
        [-1.15401215, -0.81736486,  0.79417098,  0.00546999],
        [-1.11073304, -0.6370473 ,  0.68471885,  0.00583888],
        [ 2.71283058,  2.34190758, -1.80448545, -0.00612243],
        [ 2.7561097 ,  2.52222514, -1.91393758, -0.00575354],
        [ 2.9144608 ,  2.57780547, -1.98359137, -0.00620013],
        [ 2.87118168,  2.39748791, -1.87413925, -0.00656901]])

def residuals(p, y, x):
    err = y - p*x.transpose()

    err = err * err.transpose()

    return err

p0 = Pmat  

plsq = leastsq(residuals, p0, args=(Projected_x, reconst_X  )  )

print plsq[0]

Answer 1

我的第一个猜测：minimalsq不喜欢矩阵，

使用数组和np.dot，或者在返回之前转换np.asarray（err），并且可能在剩余函数内将p转换为矩阵。

混合矩阵和数组可能很难跟踪。

Answer 2

一些小事：

如果可以
不要导入*

我已将代码更改为使用np.array来演示user333700的含义。此外，我将投影矩阵转换为12维向量，因为大多数优化器都希望您的变量以向量形式进行优化。

您将运行下面编辑过的代码的错误是TypeError：输入参数不正确。我相信这是因为你试图执行线性最小二乘法来找到12个参数，但你只有8个约束。

import numpy as np

import pylab as p
from scipy.optimize import leastsq

Projected_x = np.array([[ -69.69  , 255.3825,    1.    ],
        [ -69.69  ,  224.6175,    1.    ],
        [-110.71  ,  224.6175,    1.    ],
        [-110.71  ,  255.3825,    1.    ],
        [ 709.69  ,  224.6175,    1.    ],
        [ 709.69  ,  255.3825,    1.    ],
        [ 750.71  ,  255.3825,    1.    ],
        [ 750.71  ,  224.6175,    1.    ]])

Projected_x = Projected_x.transpose()

Pmat = np.array(   [  5.79746167e+02,  0.00000000e+00,   3.20000000e+02,    0.00000000e+00,
          0.00000000e+00,   4.34809625e+02,   2.40000000e+02, 0.00000000e+00,
          0.00000000e+00,   0.00000000e+00,   1.00000000e+00, 0.00000000e+00]   )

reconst_X = np.array([[-0.95238194, -0.58146697,  0.61506506,  0.00539229],
        [-0.99566105, -0.76178453,  0.72451719,  0.00502341],
        [-1.15401215, -0.81736486,  0.79417098,  0.00546999],
        [-1.11073304, -0.6370473 ,  0.68471885,  0.00583888],
        [ 2.71283058,  2.34190758, -1.80448545, -0.00612243],
        [ 2.7561097 ,  2.52222514, -1.91393758, -0.00575354],
        [ 2.9144608 ,  2.57780547, -1.98359137, -0.00620013],
        [ 2.87118168,  2.39748791, -1.87413925, -0.00656901]])

def residuals(p, y, x):
    err = y - np.dot(p.reshape(3,4),x.T)

    print p

    return np.sum(err**2, axis=0)

p0 = Pmat

plsq = leastsq(residuals, p0, args=(Projected_x, reconst_X  )  )

print plsq[0]

SciPy的非线性最小二乘法

2 个答案: