| Peter Pein 2006-05-30, 4:15 am |
| David Park schrieb:
> Dear MathGroup,
>
> Here is a routine that mimics a more complicated routine I was working on. I think it has strange behavior. The routine uses a conditional test on the pattern variables that as a side effect computes a parameter used in the replacement. When used on a l
ist this doesn't work because Mathematica does all of the tests first, leaving only the parameter from the last test, and then does all of the replacements. Not what I want.
>
> The sample routine looks at expressions of the form f[a,b]c. If c ===a it returns f[a^2,b]. If c === b it returns f[a,b^2] and otherwise no replacement is done. I've added two Print statements to clarify what is happening.
>
> transformf[expr_] :=
> Module[{testQ, replacea},
>
> testQ[a_, b_, c_] :=
> Module[{},
> Print["Testing ", f[a, b]c];
> Which[
> c === a, replacea = True; True,
> c === b, replacea = False; True,
> True, False]];
>
> expr /. (f[a_, b_]c_) /; testQ[a, b, c] :>
> Module[{},
> Print["Processing " , f[a, b]c];
> If[replacea, f[a^2, b], f[a, b^2]]]
> ]
>
> If I Map this onto the following list it works as expected.
>
> transformf /@ {f[a, b]a, f[a, b]b}
> Testing a f[a, b]
> Processing a f[a, b]
> Testing b f[a, b]
> Processing b f[a, b]
> {f[a^2, b], f[a, b^2]}
>
> However, if I use the routine on the entire list I obtain:
>
> {f[a, b]a, f[a, b]b} // transformf
> Testing a f[a, b]
> Testing b f[a, b]
> Processing a f[a, b]
> Processing b f[a, b]
> {f[a, b^2], f[a, b^2]}
>
> which I regard as incorrect. The Help for ReplaceAll says...
>
> "ReplaceAll looks at each part of expr, tries all the rules on it, and then goes on to the next part of expr."
>
> That seems to me to be a vague and inaccurate statement. One might think that if the rule matched then the replacement would be done - or at least calculated. Then ReplaceAll would go on to look at the next part of the expression. Instead all the tests
are done first, and all the matching replacements are done afterwards. So in the above routine the replacea parameter is set to the last test result and then used in all of the replacements.
>
> That hardly seems fair. Is there a way around this behavior.
>
> David Park
> djmp@earthlink.net
> http://home.earthlink.net/~djmp/
>
> Everyone should study mathematics - just so they will know what it is to be wrong.
>
>
>
>
>
Hi David,
1.) Why don't you like mapping?
2.) SetAttributes[transformf,Listable];
{f[a,b]a,f[a,b]b}//transformf
--> {f[a^2, b], f[a, b^2]}
does not work?
HTH,
Peter
|