H
Harold Fellermann
Hi,
I need to perform leastSquaresFit of a model that is given by a
differential equation for which there seems to be no analytic
solution. So, I am trying to solve the ODE numerically (using
scipy.integrate.odeint) within the function I provide to
leastSquaresFit as a model:
def func(L, t, a, k) :
return -k * L * (1 - ( 1 - a*L**(-1./3) )**3.)
def model((k, L0, a), t) :
solution = odeint( func, array(L0[0]), array([0,t]), args=(a,k) )
return L0 - solution[1][0]
params, chisq = leastSquaresFit(model, params, data)
Unfortunately, this approach runs into an error (ValueError: shape
mismatch: objects cannot be broadcast to a single shape) that seems to
stem from the fact that leastSquaresFit is based on automatic
derivation (DerivVar), and according to the manual "the function [that
defines the model] may only use the mathematical functions known to
the module FirstDerivatives".
What is a good solution or workaround to this problem which appears to
be quite a standard situation to me?
Thanks for any help, harold.
I need to perform leastSquaresFit of a model that is given by a
differential equation for which there seems to be no analytic
solution. So, I am trying to solve the ODE numerically (using
scipy.integrate.odeint) within the function I provide to
leastSquaresFit as a model:
def func(L, t, a, k) :
return -k * L * (1 - ( 1 - a*L**(-1./3) )**3.)
def model((k, L0, a), t) :
solution = odeint( func, array(L0[0]), array([0,t]), args=(a,k) )
return L0 - solution[1][0]
params, chisq = leastSquaresFit(model, params, data)
Unfortunately, this approach runs into an error (ValueError: shape
mismatch: objects cannot be broadcast to a single shape) that seems to
stem from the fact that leastSquaresFit is based on automatic
derivation (DerivVar), and according to the manual "the function [that
defines the model] may only use the mathematical functions known to
the module FirstDerivatives".
What is a good solution or workaround to this problem which appears to
be quite a standard situation to me?
Thanks for any help, harold.