Python least squares leastsq function for normal solution and x-axis vertical problem

lines perpendicular to the x-axis cannot be calculated when y=kx+b is used. Therefore, the normal ysin (theta) +xcos (theta) = dist is used. It seems that the use of a bit complicated, direct use of ax+by=1 do not know can calculate, not tested.

  
 

   
  

  
# 修改自 http://www.cnblogs.com/NanShan2016/p/5493429.html
   
  

  

   
  

  

   
  

  
### 最小二乘法 python leastsq###
   
  

  
import numpy as np
   
  

  
import math
   
  

  
from scipy.optimize import leastsq
   
  

  

   
  

  
###采样点(Xi,Yi)###
   
  

  
Xi=np.array([-1,-1])
   
  

  
Yi=np.array([0,10])
   
  

  

   
  

  

   
  

  

   
  

  

   
  

  
# p是个数组，表示所有参数!!!
   
  

  
### 定义误差函数，拟合y=kx+b，p[0]表示k，p[1]表示b
   
  

  
### 法线式 y*sin(theta)+x*cos(theta) = dist
   
  

  
def   error   ( p   x   y    
   
  

  
 return y*math.sin(p[0])+x*math.cos(p[0])-p[1] #x、y都是列表，故返回值也是个列表
   
  

  

   
  

  

   
  

  

   
  

  
###主函数从此开始###
   
  

  
# 可能是使用梯度下降法而非矩阵运算，因此需要给定初始参数p0
   
  

  
p0=[0,1]
   
  

  
Para=leastsq(error,p0,args=(Xi,Yi)) #把error函数中除了p以外的参数打包到args中
   
  

  
theta = Para[0][0]
   
  

  
dist = Para[0][1]
   
  

  
print("theta=",theta,‘\n‘,"dist=",dist)
   
  

  

   
  

  
###绘图，看拟合效果###
   
  

  
import matplotlib.pyplot as plt
   
  

  
plt.axis([-10,10,-10,10])
   
  

  

   
  

  

   
  

  
plt.scatter(Xi,Yi,color="red",label="Sample Point",linewidth=3) #画样本点
   
  

  

   
  

  

   
  

  

   
  

  
if theta != 0: 
   
  

  
 x=np.linspace(-10,10,10)
   
  

  
 y=dist/math.sin(theta)-x/math.tan(theta)
   
  

  
else:
   
  

  
 x = dist
   
  

  
 y = np.linspace(-10,10,10)
   
  

  

   
  

  

   
  

  
plt.plot(x,y,color="orange",label="Fitting Line",linewidth=2) #画拟合直线
   
  

  
plt.legend()
   
  

  
plt.show()
   
  

  

  
 

From Wiznote

Python least squares leastsq function for normal solution and x-axis vertical problem