Find offset and scaling of 2 curves (solve linear problem)

Question

I've got 2 curves that have an offset and scaling, which I would like to know. Now finding a factor without offset would not be to hard, as solving Curve1*a = Curve2 is not to hard. Also the offset, solving Curve1+b = Curve2 is not the problem.

It just a linear problem, however, i get stuck when looking for Curve1*a +b = Curve2.

I looked into scipy.minimize, and np.solve, but I'm having a hard time with it.

So I'm trying to minimize f(a,b) = (Curve1*a + b) - Curve2. I can't figure out how.

So imagine craeting two curves;

t = np.arange(0, 10, 0.1);
c1 = np.sin(t)
c2 = c1*2.5 + 4

I want to get a=2.5 and b=4. I tried creating a function

def func(x, c1,c2):
    x1=x[0]
    x2=x[1]
    y= np.mean(c1 - (c2*x1 + x2))
    return y

Then minimizing it

x_start = [2, 10]
result = scipy.optimize.minimize(func, c1, c2, x_start)

But I don't think this is really the correct way, plus it errors ttributeError: 'list' object has no attribute 'lower'

If anyone could help me out, that would be really appreciated! :)

Answer 1

You were not far from the solution but I have found at least three problems in your code.

1. You are not giving the arguments to the minimize() method correctly :

   scipy.optimize.minimize(func,x_start,args=(c1,c2))

2. You have inverted c1 and c2 in the error function func()

3. Given the residuals y= c2 - (c1*x1 + x2) can be either positive or negative, calculate their sum or their mean is not a good marker for the discrepancy between c2 and the optimization. You should better calculate the sum of the square of the residual return np.sum(y*y). This way you will perform a least square minimisation.

Finally I have this code working :

import scipy
import scipy.optimize
import numpy as np

t = np.arange(0, 10, 0.1);
c1 = np.sin(t)
c2 = c1*2.5 + 4

    def func(x,c1,c2):

    x1=x[0]
    x2=x[1]
    y= c2 - (c1*x1 + x2)
    return np.sum(y*y)


x_start = [2, 10]
result = scipy.optimize.minimize(func,x_start,args=(c1,c2))
print(result)

This gives :

fun: 6.604621880500689e-15
 hess_inv: array([[ 0.01133786, -0.00211422],
       [-0.00211422,  0.00539425]])
      jac: array([-9.06677028e-13, -3.57935903e-13])
  message: 'Optimization terminated successfully.'
     nfev: 32
      nit: 5
     njev: 8
   status: 0
  success: True
        x: array([2.5       , 3.99999999])