Reputation: 2454
I need to interpolate data coming from an instrument using a gaussian fit. To this end I thought about using the curve_fit
function from scipy
.
Since I'd like to test this functionality on fake data before trying it on the instrument I wrote the following code to generate noisy gaussian data and to fit it:
from scipy.optimize import curve_fit
import numpy
import pylab
# Create a gaussian function
def gaussian(x, a, b, c):
val = a * numpy.exp(-(x - b)**2 / (2*c**2))
return val
# Generate fake data.
zMinEntry = 80.0*1E-06
zMaxEntry = 180.0*1E-06
zStepEntry = 0.2*1E-06
x = numpy.arange(zMinEntry,
zMaxEntry,
zStepEntry,
dtype = numpy.float64)
n = len(x)
meanY = zMinEntry + (zMaxEntry - zMinEntry)/2
sigmaY = 10.0E-06
a = 1.0/(sigmaY*numpy.sqrt(2*numpy.pi))
y = gaussian(x, a, meanY, sigmaY) + a*0.1*numpy.random.normal(0, 1, size=len(x))
# Fit
popt, pcov = curve_fit(gaussian, x, y)
# Print results
print("Scale = %.3f +/- %.3f" % (popt[0], numpy.sqrt(pcov[0, 0])))
print("Offset = %.3f +/- %.3f" % (popt[1], numpy.sqrt(pcov[1, 1])))
print("Sigma = %.3f +/- %.3f" % (popt[2], numpy.sqrt(pcov[2, 2])))
pylab.plot(x, y, 'ro')
pylab.plot(x, gaussian(x, popt[0], popt[1], popt[2]))
pylab.grid(True)
pylab.show()
Unfortunately this does not work properly, the output of the code is the following:
Scale = 6174.816 +/- 7114424813.672
Offset = 429.319 +/- 3919751917.830
Sigma = 1602.869 +/- 17923909301.176
And the plotted result is (blue is the fit function, red dots is the noisy input data):
I also tried to look at this answer, but couldn't figure out where my problem is.
Am I missing something here? Or am I using the curve_fit
function in the wrong way? Thanks in advance!
Upvotes: 0
Views: 2944
Reputation: 2823
As I said in a comment, if you provide a reasonable initial guess, the fit succeeds, i.e. call curve_fit
like that:
popt, pcov = curve_fit(gaussian, x, y, [50000,0.00012,0.00002])
Upvotes: 1
Reputation: 1813
I agree with Olaf in so far as it is a question of scale. The optimal parameters differ by many orders of magnitude. However, scaling the parameters with which you generated your toy data does not seem to solve the problem for your actual application. curve_fit
uses lestsq
, which numerically approximates the Jacobian, where numerical problems arise because of the differences in scale (try to use the full_output
keyword in curve_fit
).
In my experience it is often best to use fmin
which does not rely on approximated derivatives but uses only function values. You now have to write your own least-squares function that is to be optimized.
Starting values are still important. In your case you can make sufficiently good guesses by taking the maximum amplitude for a
and the corresponding x-values for b
and c
.
In code, it looks like this:
from scipy.optimize import curve_fit,fmin
import numpy
import pylab
# Create a gaussian function
def gaussian(x, a, b, c):
val = a * numpy.exp(-(x - b)**2 / (2*c**2))
return val
# Generate fake data.
zMinEntry = 80.0*1E-06
zMaxEntry = 180.0*1E-06
zStepEntry = 0.2*1E-06
x = numpy.arange(zMinEntry,
zMaxEntry,
zStepEntry,
dtype = numpy.float64)
n = len(x)
meanY = zMinEntry + (zMaxEntry - zMinEntry)/2
sigmaY = 10.0E-06
a = 1.0/(sigmaY*numpy.sqrt(2*numpy.pi))
y = gaussian(x, a, meanY, sigmaY) + a*0.1*numpy.random.normal(0, 1, size=len(x))
print a, meanY, sigmaY
# estimate starting values from the data
a = y.max()
b = x[numpy.argmax(a)]
c = b
# define a least squares function to optimize
def minfunc(params):
return sum((y-gaussian(x,params[0],params[1],params[2]))**2)
# fit
popt = fmin(minfunc,[a,b,c])
# Print results
print("Scale = %.3f" % (popt[0]))
print("Offset = %.3f" % (popt[1]))
print("Sigma = %.3f" % (popt[2]))
pylab.plot(x, y, 'ro')
pylab.plot(x, gaussian(x, popt[0], popt[1], popt[2]),lw = 2)
pylab.xlim(x.min(),x.max())
pylab.grid(True)
pylab.show()
Upvotes: 2
Reputation: 425
Looks like some numerical instabilities are creeping into the optimizer. Try scaling the data. With the following data:
zMinEntry = 80.0*1E-06 * 1000
zMaxEntry = 180.0*1E-06 * 1000
zStepEntry = 0.2*1E-06 * 1000
sigmaY = 10.0E-06 * 1000
I get estimates of
Scale = 39.697 +/- 0.526
Offset = 0.130 +/- 0.000
Sigma = -0.010 +/- 0.000
Compare that to the true values:
Scale = 39.894228
Offset = 0.13
Sigma = 0.01
The minus sign of sigma can of course be ignored.
This gives the following plot
Upvotes: 1