SciPy的非線性最小平方

我試圖通過python進行包調整。所以我是測試非線性最小二乘模塊。然後我寫下如下代碼。我想得到正確的Pmat表示三臺攝像機的攝像機投影矩陣。但是我有一個錯誤，「ValueError：對象太深，無法使用所需數組」。SciPy的非線性最小平方

任何人都可以提供線索來解決這個問題？

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]

來源

2010-10-19 Jinho Yoo

我第一次的猜測：leastsq不喜歡矩陣，

使用數組和np.dot，或返回之前轉換np.asarray（ERR），也許轉換p來裏面的殘餘基質功能。

混合矩陣和數組可能是一個難以跟蹤的問題。

來源

2010-10-19 14:39:14 user333700

一對夫婦的小東西：

使用np.array如果你能
不導入*

我已經改變使用np.array證明什麼user333700代碼手段。此外，我將投影矩陣轉換爲12維矢量，因爲大多數優化器都希望您的變量以矢量形式進行優化。

您將運行下面編輯的代碼的錯誤是TypeError：輸入參數不正確。我相信這是因爲您正在嘗試執行線性最小二乘查找12個參數，但您只有8個約束。

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]

來源

2010-10-19 15:08:46

SciPy的非線性最小平方

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