Scipy's curve_fit / leastsq become slower when given the Jacobian?

Question

So I wad reading the documentation about curve_fit here. It contains the following example:

import numpy as np
import scipy.optimize as so

def func(x, a,b,c ):
    return a * np.exp(-b * x) + c

a,b,c = 2.5, 1.3, 0.5
nx = 500
noiseAlpha = 0.5
#
xdata = np.linspace(0, 4, nx)
y = func(xdata, a,b,c)
ydata = y + noiseAlpha * np.random.normal(size=len(xdata))

If I call curve_fit now, it will approximate the derivatives since I didn't provide anything. Let's time it:

%%timeit
popt, pcov = so.curve_fit(func, xdata, ydata )
1000 loops, best of 3: 787 µs per loop

In fact curve_fit calls leastsq (doc here), which accepts a Dfun argument for computing the Jacobian. So I did this:

def myDfun( abc, xdata, ydata, f ) :
    a,b,c = abc
    ebx = np.exp(-b * xdata)
    res = np.vstack( ( ebx, a * -xdata * ebx, np.ones(len(xdata)) ) ).T
    return res

And I timed it again:

%%timeit
popt, pcov = so.curve_fit(func, xdata, ydata, Dfun=myDfun )
1000 loops, best of 3: 858 µs per loop

Uh? Using the Jacobian is slower than approximating it? Did i do something wrong?

Answer 1

Not really an answer, but my feeling is it dependents on the size of the problem. For small size (n=500), the extra time spend in evaluating jacobian (with the supplied jac) probably don't pay off in the end.

n=500, with jab:

100 loops, best of 3: 1.50 ms per loop

Without:

100 loops, best of 3: 1.57 ms per loop

n=5000, with jab:

100 loops, best of 3: 5.07 ms per loop

Without:

100 loops, best of 3: 6.46 ms per loop

n=50000, with jab:

100 loops, best of 3: 49.1 ms per loop

Without:

100 loops, best of 3: 59.2 ms per loop

Also you may want to consider rewrite the jacobian function, for example getting rid of the expensive .T() can bring about ~15% speed up:

def myDfun2( abc, xdata, ydata, f ) :
    a,b,c = abc
    ebx = np.exp(-b * xdata)
    res = np.hstack( ( ebx.reshape(-1,1), (a * -xdata * ebx).reshape(-1,1), np.ones((len(xdata),1)) ) )
    return res