Non-Linear Least Square Fitting Using Python

Question

I am trying to find the optimal parameter values based on 2 dimensional data.
I searched for other Q&A threads and one of the questions looked similar to mine and the answer seemed to be duable link for me to replicate but I get the following error:

TypeError: only size-1 arrays can be converted to Python scalars

I rarely changed the codes but customized a bit only but seems like my equation does not accept Numpy arrays. Is there a way to address this issue and derive the parameter values?
Below is the code:

from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt

t_data = np.array([0,5,10,15,20,25,27])
y_data = np.array([1771,8109,22571,30008,40862,56684,59101])

def func_nl_lsq(t, *args):
    K, A, L, b = args
    return math.exp(L*x)/((K+A*(b**x)))

popt, pcov = curve_fit(func_nl_lsq, t_data, y_data, p0=[1, 1, 1, 1])
plt.plot(t_data, y_data, 'o')
plt.plot(t_data, func_nl_lsq(t_data, *popt), '-')
plt.show()
print(popt[0], popt[1], popt[2], popt[3])

For more clarifications, I also attach a screenshot of the (1) equation that I want to run NLS fitting and the data observations from the journal paper that I've read.

... (1)

year(x_data)	value(y_data)
1960	1771
1965	8109
1970	22571
1975	30008
1980	40862
1985	56684
1987	59101

Hey Todd, thanks for the clear question, just one thing, what is the `x` in `return math.exp(L*x)/((K+A*(b**x)))`? I cannot run the code without its value (also you forgot `import math` at the top) — Caridorc, Feb 08 '23 at 01:01
Also the function `func_nl_lsq` ignores the argument `t`, this looks like a bug — Caridorc, Feb 08 '23 at 01:02
The code runs if I set `x = t` and use `np.exp` instead of `math.exp` but the result is still not great — Caridorc, Feb 08 '23 at 01:05
Ok I think I fixed your code, please take a look at my answer — Caridorc, Feb 08 '23 at 01:20
They are four parameters in your equation but only seven experimental points. It is not sufficient to expect a reliable result. More over the curve is quite linear. — JJacquelin, Feb 08 '23 at 13:35

Caridorc · Accepted Answer · 2023-02-08T01:26:06.570

Ok so there was quite some work to do:

Set x = t (probably a typo)
Use np.exp instead of math.exp. np.exp is also used in the docs: https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html
Change the initial guesses to 0.1 as nonlinear optimization is sensible to initial guesses. Take care to choose guesses as near as possible to your ballpark estimate of the parameters order of magnitude. If you cannot estimate, run the fitting algorithm many times for many initial guesses, then you can select the (non-failed) fit with the smallest sum of residuals squared, see here learn how to calculate those residuals: Getting the r-squared value using curve_fit

Here is the final result:

from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
import math

t_data = np.array([0,5,10,15,20,25,27])
y_data = np.array([1771,8109,22571,30008,40862,56684,59101])

def func_nl_lsq(t, *args):
    K, A, L, b = args
    x=t # typo on your part?
    exponential = np.exp(L*x) # np.exp rather than math.exp
    denominator = K+A*(b**x)
    return exponential/denominator

popt, pcov = curve_fit(func_nl_lsq, t_data, y_data, p0=[0.1, 0.1, 0.1, 0.1]) # changed initial conditions
plt.plot(t_data, y_data, 'o')
plt.plot(t_data, func_nl_lsq(t_data, *popt), '-')
plt.show()
print(popt[0], popt[1], popt[2], popt[3])

With output:

7.36654485198494e-05 0.0011449352764853055 0.05547299686499709 0.5902083262954019

And plot:

Thank you very much, Caridorc. You rescued me from where I was struggling. I learned a lot, especially choosing good initial guess set seems very important. My guess such as 1,1,1,1 was not even fitting to the data points at all. Thanks, again. — Todd, Feb 09 '23 at 17:14
You are welcome @Todd , good luck in your data analysis journey! — Caridorc, Feb 09 '23 at 17:41

Non-Linear Least Square Fitting Using Python

1 Answers1