Python exponential curve fitting

Question

I have added excel plot from which I get the exponential equation, I am trying to curve fit this in Python.

My fitted equation is not as close to the empirical data i have provided when i use it to predict the y data, the prediction gives f(-25)= 5.30e-11, while the empirical data f(-25) gives = 5.3e-13

How can i improve the code to be predicting close to empirical data, or i have made mistakes in my code??

python fitted plot

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

pd.set_option('precision', 14)

def f(x,A,B):
    return A * np.exp((-B) * (x))

y_data= [2.156e-05,  1.85e-07,  1.02e-10 , 1.268e-11,  5.352e-13]
x=  [-28.8,  -27.4,  -26 , -25.5,  -25]
p, pcov = optimize.curve_fit(f,  x,   y_data,   p0=[10**(-59),4],    maxfev=5000)

plt.figure()
plt.plot(x, y_data, 'ko', label="Empirical BER")
plt.plot(x, f(x, *p ), 'g-', label="Fitted BER" )
plt.title(" BER  ")
plt.xlabel('Power Rx (dB)')
plt.ylabel('')
plt.legend()
plt.grid()
plt.yscale("log")
plt.show()

When I compare the graph and the code, it seems that in the code you are confusing `x` and `y`. — John Coleman, Jun 14 '21 at 09:44
@mkrieger1, thanks for formatting the question, the problem is that the fitting is not predicting correct y_data set data. I am wondering where i have gone wrong ? fresh eyes might catch my mistake — martin, Jun 14 '21 at 13:16
@John Coleman , thanks , can you explain further where the confusion is — martin, Jun 14 '21 at 13:18
I have no idea what it is predicting and what you expected it to predict instead. — mkrieger1, Jun 14 '21 at 13:27
I think the problem is simply that the model function is not suitable at all for this data. — mkrieger1, Jun 14 '21 at 13:33
When adding a constant term to `f` and using less extreme initial values, it kind of works. — mkrieger1, Jun 14 '21 at 13:40
@John Coleman, I have corrected the x , y data sets, thanks but this mistake was just on posting, — martin, Jun 14 '21 at 13:55
@stackoverflow, kindly re open the question, I have addressed issues raised , Thanks — martin, Jun 14 '21 at 14:16

M Newville · Answer 1 · 2021-06-22T00:14:05.470

Since you are plotting the data with a log-plot, your view of the data and fit is emphasizing the "tiny" compared to the "small". Fitting uses the sum of the squares of the misfit to determine the best fit. A misfit of a few percent of the data with a y-value of ~2e-5 would completely swamp a misfit of a factor of 10 or even 100 for the data with a y-value of 1.e-11. Your plot is consistent with that.

There are two possible routes to a better fit:

a) if you have uncertainties in the y-values, use those. It's quite possible that the uncertainty in the data with y~2e-5 is much larger than the uncertainty in the date with y~1.e-11, and scaling by the uncertainty so that the minimization is of the sum-of-squares of (data-model)/uncertainty will help fit the low-value data better. OTOH, if the errors are constant, plotting those uncertainties might show that the fit you have is actually not that bad -- the misfit where y~1.e-11 is only 1.e-10.

b) realize that you are assessing the fit quality by plotting the log of the data, and embrace that observation so that you fit the log(data) to log(model). Conveniently for a simple exponential function, the log of that model is linear, so you could do linear regression of the log of your data.

Bonus round: recognize that options a) and b) are related. Since a fit minimizes Sum[ ((data-model)/uncertainty)**2], not providing values for uncertainty is effectively saying that the has same uncertainty (=1.0 in fact) for all values of x and y. If you fit the log of the model to the log of the data, as withSum[ (log(data) - log(model))**2] is effectively saying that the uncertainty in the log(data) is the same for all values of x and y.

@M Newville thanks a lot for the clarification and interpretation the data — martin, Jul 07 '21 at 12:54

Python exponential curve fitting

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