python nonlinear least squares fitting

Question

I am a little out of my depth in terms of the math involved in my problem, so I apologise for any incorrect nomenclature.

I was looking at using the scipy function leastsq, but am not sure if it is the correct function. I have the following equation:

eq = lambda PLP,p0,l0,kd : 0.5*(-1-((p0+l0)/kd) + np.sqrt(4*(l0/kd)+(((l0-p0)/kd)-1)**2))

I have data (8 sets) for all the terms except for kd (PLP,p0,l0). I need to find the value of kd by non-linear regression of the above equation. From the examples I have read, leastsq seems to not allow for the inputting of the data, to get the output I need.

Thank you for your help

Answer 1

Another option is to use lmfit.

They provide a great example to get you started:.

#!/usr/bin/env python
#<examples/doc_basic.py>
from lmfit import minimize, Minimizer, Parameters, Parameter, report_fit
import numpy as np

# create data to be fitted
x = np.linspace(0, 15, 301)
data = (5. * np.sin(2 * x - 0.1) * np.exp(-x*x*0.025) +
        np.random.normal(size=len(x), scale=0.2) )

# define objective function: returns the array to be minimized
def fcn2min(params, x, data):
    """ model decaying sine wave, subtract data"""
    amp = params['amp']
    shift = params['shift']
    omega = params['omega']
    decay = params['decay']
    model = amp * np.sin(x * omega + shift) * np.exp(-x*x*decay)
    return model - data

# create a set of Parameters
params = Parameters()
params.add('amp',   value= 10,  min=0)
params.add('decay', value= 0.1)
params.add('shift', value= 0.0, min=-np.pi/2., max=np.pi/2)
params.add('omega', value= 3.0)


# do fit, here with leastsq model
minner = Minimizer(fcn2min, params, fcn_args=(x, data))
kws  = {'options': {'maxiter':10}}
result = minner.minimize()


# calculate final result
final = data + result.residual

# write error report
report_fit(result)

# try to plot results
try:
    import pylab
    pylab.plot(x, data, 'k+')
    pylab.plot(x, final, 'r')
    pylab.show()
except:
    pass

#<end of examples/doc_basic.py>

Answer 2

This is a bare-bones example of how to use scipy.optimize.leastsq:

import numpy as np
import scipy.optimize as optimize
import matplotlib.pylab as plt

def func(kd,p0,l0):
    return 0.5*(-1-((p0+l0)/kd) + np.sqrt(4*(l0/kd)+(((l0-p0)/kd)-1)**2))

The sum of the squares of the residuals is the function of kd we're trying to minimize:

def residuals(kd,p0,l0,PLP):
    return PLP - func(kd,p0,l0)

Here I generate some random data. You'd want to load your real data here instead.

N=1000
kd_guess=3.5  # <-- You have to supply a guess for kd
p0 = np.linspace(0,10,N)
l0 = np.linspace(0,10,N)
PLP = func(kd_guess,p0,l0)+(np.random.random(N)-0.5)*0.1

kd,cov,infodict,mesg,ier = optimize.leastsq(
    residuals,kd_guess,args=(p0,l0,PLP),full_output=True,warning=True)

print(kd)

yields something like

3.49914274899

This is the best fit value for kd found by optimize.leastsq.

Here we generate the value of PLP using the value for kd we just found:

PLP_fit=func(kd,p0,l0)

Below is a plot of PLP versus p0. The blue line is from data, the red line is the best fit curve.

plt.plot(p0,PLP,'-b',p0,PLP_fit,'-r')
plt.show()