[SciPy-User] Curve fitting questions
josef.pktd@gmai...
josef.pktd@gmai...
Tue Oct 19 15:17:43 CDT 2010
On Tue, Oct 19, 2010 at 3:20 PM, Gökhan Sever <gokhansever@gmail.com> wrote:
> On Tue, Oct 19, 2010 at 1:04 PM, <josef.pktd@gmail.com> wrote:
>> I think you could take logs and you would have a linear function in
>> param=log(x), and you could use linalg to solve for param, and then
>> transform back exp(param). Or this would give you a starting value if
>> you want the non-linear optimization.
>
> Using 3 and 5 data-points the curve_fit usually does a good job, even without
> the initial estimates provided. When it's necessary we usually constrain the
> initial parameters with max CCN concentration for C (param[0]) and a typical
> k (param[1]) values.
>
> This even works with 2 data-points:
>
> I[34]: ccn_ss1 = [0.27, 0.34]
>
> I[35]: ccn_conc1 = np.array([383.51237409766452, 424.82669523141652])
>
> I[36]: tfit2, pcov2 = curve_fit(my_ck, ccn_ss1, ccn_conc1, p0=(424,
> 0.5), ftol=1)
>
> provides me reasonable estimations. However, having another data-point
> would surely
> improve the quality of the fit and estimations.
>
>>> ccn_ss1 = 0.27
>>> ccn_conc1 = 383.51237409766452
>>> # One data point estimation fails with IndexError: index out of range for array
>>> tfit3, pcov3 = curve_fit(my_ck, ccn_ss1, ccn_conc1, p0=tfit1, ftol=1)
>>
>> If you have one parameter to estimate and only one observations, then
>> you should be able to solve it exactly with one of the solvers/
>> rootfinders in scipy. optimize.
>>
>> Josef
>
> I want to estimate two parameters using one observation (which is a
> data-pair for my case --one for ccn_ss1 and one for ccn_conc1.)
> Probably, in this current version fsolve can't do give me any roots.
I remembered curve_fit wrongly, I didn't remember it switched data and
parameters in the argument list compare to leastsq.
I need to reread your example (later today).
Taking logs and using linalg is still more efficient (unless you
insist on an additive error term).
Josef
>
> def my_ck(x, a, b):
> return a*x**b
>
> fsolve(my_ck, x0=tfit1, args=(ccn_ss1, ccn_conc1), xtol=1)
>
> rather gives a couple of overflow warnings:
> Warning: overflow encountered in power
>
> In one data-pair situation my function looks like:
>
> a*x**b = 383.5
>
> Now there are two unknowns providing the x as ccn_ss1 as a*0.27**b =
> 383.5. I should make one more assumption otherwise it is still
> unsolvable. Probably making an assumption for a, then I can hand solve
> this easily. OK, with a = 350 assumption in 350*0.27**b == 383.5, here
> solving for b results ~ -0.065
>
> With some modifications on the original fitfunc:
>
> def my_ck(x):
> return 350*0.27**x - 383
>
> fsolve nicely estimates what I want.
>
> fsolve(my_ck, x0=0.5)
> -0.068815047568104443
>
>
>
>
>
> --
> Gökhan
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