Re: Extracting a Function's Domain and Image
- To: mathgroup at smc.vnet.net
- Subject: [mg68011] Re: Extracting a Function's Domain and Image
- From: "sashap" <pavlyk at gmail.com>
- Date: Thu, 20 Jul 2006 06:04:52 -0400 (EDT)
- References: <e9ku8v$l1v$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
It is because, your definition implies f[x_] evaluates to zero. This
Cases sees
Cases[h,0->x,Infinity]. Thus zero is picked up from h and replaced with
x.
You should use
In[33]:= Cases[h, HoldPattern[f][x_] -> x, {0, Infinity}]
Out[33]= {"a", "b", _}
Oleksandr Pavlyk
Wolfram Research
Bruce Colletti wrote:
> Re Mathematica 5.2.2.
>
> Why does the last line of code return {x} and not {a,b,_}, while if the f[_]=0 is struck, the answer is {a,b}?
>
> I want to extract the domain and image of a user-defined function, and tech support recommended using DownValues (pretty handy). If you have an easier approach, please share it. Thanks.
>
> Bruce
>
> Remove@f;
> f[_]=0;
> f["a"]=1;
> f["b"]=2;
> Clear@x;
> h=DownValues@f
> Cases[h,f[x_]->x,Infinity]
>
> Out[52]=
> {HoldPattern[f[a]]:>1,HoldPattern[f[
> b]]:>2,HoldPattern[f[_]]:>0}
>
> Out[53]=
> {x}