[SciPy-user] parameter bounds using leastsq
Stephen Walton
stephen.walton at csun.edu
Tue Nov 8 10:53:12 CST 2005
Eric Zollars wrote:
>Robert-
> Could you flesh out this answer some more? I've had to do this in the
>past and was sure I was missing something. In the simplest case a set of:
>y1 = (b0 + b1*x1 + b2*x2)1
>.
>.
>yn = (b0 + b1*x1 + b2*x2)n
>
>In fortran I would pass x as a 2d matrix to the function misfit. What
>do you do in scipy if x is a vector at each point y?
>
>
You would write your function f accordingly. Because Scipy is vector
oriented, a simple
y=b0+b1*x[0]+b2*x[1]
would work, assuming x is a Scipy/Numeric array of shape (2,n).
By the way, I humbly suggest there was a typo in Robert's original post:
>>def f(beta, x):
>> # compute values y given parameters beta at points x
>>
>>def misfit(beta, x, y):
>> diff = y - f(beta, x)
>> return scipy.sum(diff*diff)
>>
>>beta_opt = scipy.optimize.fmin_cobyla(f, beta0, constraints, (x, y))
>>
>>
I think the "f" argument in the call to fmin_cobyla needs to be "misfit"
else you're not fitting to the data.
Steve Walton
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