Re: Extracting a Function's Domain and Image
- To: mathgroup at smc.vnet.net
- Subject: [mg67991] Re: [mg67970] Extracting a Function's Domain and Image
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 20 Jul 2006 06:04:35 -0400 (EDT)
- References: <200607190921.FAA21378@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 19 Jul 2006, at 11:21, 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} > Because what happens when you evaluate Cases[h,f[x_]->x,Infinity] is completely different from what you obviously intended. First of all, because of your definition of f, f[x_] is evaluated to 0 and then Cases[h,0->x,Infinity] finds one 0 in the DownValues of f and performs the requested replacement by x. You can avoid this simply by using Cases[h,HoldPattern[f[x_]]->x,Infinity] {a,b,_} Andrzej Kozlowski Karakida,Tokyo,Japan
- References:
- Extracting a Function's Domain and Image
- From: Bruce Colletti <vze269bv@verizon.net>
- Extracting a Function's Domain and Image