Python 最小二乘scipy.optimize.leastsq()的残差,最优解
我知道scipy.optimize中有一个最小二乘法的例子,但我在残差函数方面遇到了实际问题,这已经有三天多了。我决定全面描述这个问题。我在网站上发现了一些其他类似的问题,但我无法真正通过编程解决,在我的案例中我真的很困惑。我无法正确创建适合scipy.optimize.leastsq残差的残差函数。我真的很累 这个问题就是A*x=b问题。让我简单解释一下: 输入:Python 最小二乘scipy.optimize.leastsq()的残差,最优解,python,scipy,mathematical-optimization,Python,Scipy,Mathematical Optimization,我知道scipy.optimize中有一个最小二乘法的例子,但我在残差函数方面遇到了实际问题,这已经有三天多了。我决定全面描述这个问题。我在网站上发现了一些其他类似的问题,但我无法真正通过编程解决,在我的案例中我真的很困惑。我无法正确创建适合scipy.optimize.leastsq残差的残差函数。我真的很累 这个问题就是A*x=b问题。让我简单解释一下: 输入: import numpy as np from scipy.optimize import leastsq def funct
import numpy as np
from scipy.optimize import leastsq
def function(x,delta):
return dot(delta, x)
def residual(x, delta, y):
error = y - dot(delta, x)
return sum(error**2)
def main():
# INPUTS
# Unknown values
x = [x2, x4, x5, x6, x7, x10, x13, x16]
delta = np.array([1.76762035, 2.04349174, 1.25674742],
[0.94873891, 2.01859342, 1.46348023],
[0.83678402, 1.12030343, 0.92516861],
[1.43, 2., 2., 1.57])
y = np.array([0.8353410485015903, 0.73620941924970962,
0.45428639186344633, 1.6180418445100002]
x_init = np.zeros(len(x))
result = leastsq(residual, x, args=(delta,y) )
print result[0]
if __name__ == '__main__':
main()
- y-例如:y1=点(δ_1,数组([x2,x5,x6]))
- x-8个未知数[x2,x4,x5,x6,x7,x10,x13,x16]
- 增量长度
import numpy as np
from scipy.optimize import leastsq
def function(x,delta):
return dot(delta, x)
def residual(x, delta, y):
error = y - dot(delta, x)
return sum(error**2)
def main():
# INPUTS
# Unknown values
x = [x2, x4, x5, x6, x7, x10, x13, x16]
delta = np.array([1.76762035, 2.04349174, 1.25674742],
[0.94873891, 2.01859342, 1.46348023],
[0.83678402, 1.12030343, 0.92516861],
[1.43, 2., 2., 1.57])
y = np.array([0.8353410485015903, 0.73620941924970962,
0.45428639186344633, 1.6180418445100002]
x_init = np.zeros(len(x))
result = leastsq(residual, x, args=(delta,y) )
print result[0]
if __name__ == '__main__':
main()
- 函数f(x_m,delta_n_m),n行,m列
- 残差| b|n-f(x,delta)^2
- 使用scipy.optimize.leastsq(残差,x0,arg(增量))最小化残差
import numpy as np
from scipy.optimize import leastsq
def function(x,delta):
return dot(delta, x)
def residual(x, delta, y):
error = y - dot(delta, x)
return sum(error**2)
def main():
# INPUTS
# Unknown values
x = [x2, x4, x5, x6, x7, x10, x13, x16]
delta = np.array([1.76762035, 2.04349174, 1.25674742],
[0.94873891, 2.01859342, 1.46348023],
[0.83678402, 1.12030343, 0.92516861],
[1.43, 2., 2., 1.57])
y = np.array([0.8353410485015903, 0.73620941924970962,
0.45428639186344633, 1.6180418445100002]
x_init = np.zeros(len(x))
result = leastsq(residual, x, args=(delta,y) )
print result[0]
if __name__ == '__main__':
main()
leastsq期望残差函数只返回残差,即:
def residual(x, delta, y):
return y - dot(delta, x)
你好,克里斯蒂安,谢谢!我对此进行了测试,但是,如果不更改delta的形状,您会如何修复
ValueError:对象未对齐??顺便说一下,在代码的开头,应该禁用x factor,因为它是未知的。请修复您的代码,使其可运行,然后我将能够查看错误消息。