SciPy的非线性最小平方

我试图通过python进行包调整。所以我是测试非线性最小二乘模块。然后我写下如下代码。我想得到正确的Pmat表示三台摄像机的摄像机投影矩阵。但是我有一个错误，“ValueError：对象太深，无法使用所需数组”。SciPy的非线性最小平方

任何人都可以提供线索来解决这个问题？

Regards， 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]

来源

2010-10-19 Jinho Yoo

我第一次的猜测：leastsq不喜欢矩阵，

使用数组和np.dot，或返回之前转换np.asarray（ERR），也许转换p来里面的残余基质功能。

混合矩阵和数组可能是一个难以跟踪的问题。

来源

2010-10-19 14:39:14 user333700

一对夫妇的小东西：

使用np.array如果你能
不导入*

我已经改变使用np.array证明什么user333700代码手段。此外，我将投影矩阵转换为12维矢量，因为大多数优化器都希望您的变量以矢量形式进行优化。

您将运行下面编辑的代码的错误是TypeError：输入参数不正确。我相信这是因为您正在尝试执行线性最小二乘查找12个参数，但您只有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]

来源

2010-10-19 15:08:46

SciPy的非线性最小平方

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