how to fit a step function in python

Question

I have a question about fitting a step function using scipy routines like curve_fit. I have trouble making it vectorized, for example:

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

xobs=np.linspace(0,10,100)
yl=np.random.rand(50); yr=np.random.rand(50)+100
yobs=np.concatenate((yl,yr),axis=0)

def model(x,rf,T1,T2):
#1:   x=np.vectorize(x)
    if x<rf:
        ret= T1
    else:
        ret= T2
    return ret
#2: model=np.vectorize(model)
popt, pcov = curve_fit(model, xobs, yobs, [40.,0.,100.])

It says

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

If I add #1 or #2 it runs but doesn't really fit the data:

OptimizeWarning: Covariance of the parameters could not be estimated     category=OptimizeWarning)
[ 40.          50.51182064  50.51182064] [[ inf  inf  inf]
[ inf  inf  inf]
[ inf  inf  inf]]

Anybody know how to fix that? THX

Answer 1

Here's what I did. I retained xobs and yobs :

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

xobs=np.linspace(0,10,100)
yl=np.random.rand(50); yr=np.random.rand(50)+100
yobs=np.concatenate((yl,yr),axis=0)

Now, Heaviside function must be generated. To give you an overview of this function, consider the half-maximum convention of Heaviside function:

In Python, this is equivalent to: def f(x): return 0.5 * (np.sign(x) + 1)

A sample plot would be:

xval = sorted(np.concatenate([np.linspace(-5,5,100),[0]])) # includes x = 0
yval = f(xval)
plt.plot(xval,yval,'ko-')
plt.ylim(-0.1,1.1)
plt.xlabel('x',size=18)
plt.ylabel('H(x)',size=20)

Now, plotting xobs and yobs gives:

plt.plot(xobs,yobs,'ko-')
plt.ylim(-10,110)
plt.xlabel('xobs',size=18)
plt.ylabel('yobs',size=20)

Notice that comparing the two figures, the second plot is shifted by 5 units and the maximum increases from 1.0 to 100. I infer that the function for the second plot can be represented as follows:

or in Python: (0.5 * (np.sign(x-5) + 1) * 100 = 50 * (np.sign(x-5) + 1)

Combining the plots yields (where Fit represents the above fitting function)

The plot confirms that my guess is correct. Now, assuming that YOU DO NOT KNOW how did this correct fitting function come about, a generalized fitting function is created: def f(x,a,b,c): return a * (np.sign(xb) + c) , where theoretically, a = 50 , b = 5 , and c = 1 .

Proceed to estimation:

popt,pcov=curve_fit(f,xobs,yobs,bounds=([49,4.75,0],[50,5,2])) .

Now, bounds = ([lower bound of each parameter (a,b,c)],[upper bound of each parameter]) . Technically, this means that 49 < a < 50, 4.75 < b < 5, and 0 < c < 2.

Here are MY results for popt and pcov :

pcov represents the estimated covariance of popt. The diagonals provide the variance of the parameter estimate [Source] .

Results show that the parameter estimates pcov are near the theoretical values.

Basically, a generalized Heaviside function can be represented by: a * (np.sign(xb) + c)

Here is the code that will generate parameter estimates and the corresponding covariances:

import numpy as np
from scipy.optimize import curve_fit

xobs = np.linspace(0,10,100)
yl = np.random.rand(50); yr=np.random.rand(50)+100
yobs = np.concatenate((yl,yr),axis=0)

def f(x,a,b,c): return a * (np.sign(x-b) + c) # Heaviside fitting function

popt, pcov = curve_fit(f,xobs,yobs,bounds=([49,4.75,0],[50,5,2]))
print 'popt = %s' % popt
print 'pcov = \n %s' % pcov

Finally, note that the estimates of popt and pcov vary.

python scipy