Re: Evaluate, Replace, and Hold
- To: mathgroup at smc.vnet.net
- Subject: [mg77207] Re: Evaluate, Replace, and Hold
- From: dh <dh at metrohm.ch>
- Date: Tue, 5 Jun 2007 06:59:13 -0400 (EDT)
- References: <f40gib$65a$1@smc.vnet.net>
Hi,
you want to evaluate a function of 2 parameters over a range of every
parameter. "Outer" will do this. Here is a simple example for the syntax
of the parameter you indicated:
Remove[fun];
fun[pts_,params_]:=
Outer[(Print["p1=",#1];model/.#1/.x->#2)&,params,pts,1]
model=a+b x;
params={{a\[Rule]1,b\[Rule]2},{a\[Rule]2,b\[Rule]1}};
pts={1,2,3};
fun[pts,params]
hope this helps, Daniel
grenander at gmail.com wrote:
> This is a newbie question. I have a parametric form of a model and I
> would like to evaluate it with different parameters on a list of
> values.
>
> model = a x + b;
>
> Different sets of parameters:
>
> p1 = {a->1, b->0};
> p2 = {a->1, b->1};
>
> x for which I'd like to evaluate the model:
>
> pts = Range[10];
>
> I would like to define a function to evaluate the model at pts using
> any of the sets of paramters:
>
> f[pts, p1]
> f[pts, p2]
>
> I would also like it to work for single points:
>
> f[1, p1]
> f[2, p1]
>
> I've tried various forms for a function definition that dont work. I
> can do the following for a particular set of parameters:
>
> Function[{x}, Evaluate[model /. p1]] /@ pts
>
> but how can I turn it into a function of also the parameters. The
> below doesn't work, for example:
>
> f[pts_, p_]:=Function[{x}, Evaluate[model /. p] ] /@ pts
>
> What I know is from descending the documentation tree (v6.0) starting
> from the examples in FindFit. I have a fundamental misunderstanding of
> replacement rules and delayed evaluation.
>
> Thanks.
>
>