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Re: Length of a held expression
*To*: mathgroup at smc.vnet.net
*Subject*: [mg96599] Re: Length of a held expression
*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
*Date*: Tue, 17 Feb 2009 06:23:46 -0500 (EST)
*Organization*: Uni Leipzig
*References*: <gn8jdj$7qa$1@smc.vnet.net> <gncmhe$flg$1@smc.vnet.net>
*Reply-to*: kuska at informatik.uni-leipzig.de
Hi,
and Rules[] will not help ??
{a->10,b->20} intead of assigments ??
Regards
Jens
Nikolaus Rath wrote:
> Hi,
>
> Nikolaus Rath <Nikolaus at rath.org> writes:
>> Hello,
>>
>> How can I get the length of a list in a Hold[] expression?
>
> It seems that in my attempt to produce a short example, I also
> simplified away the actual problem. So here it comes in slightly
> longer form.
>
> I have written a function that propagates uncertainties in the
> parameters of a function into the uncertainty of the result. It is
> used as follows:
>
> error[a*b, {a, b}, {aErr, bErr}]
>
> --> {a b, Sqrt[Abs[aErr b]^2 + Abs[a bErr]^2]}
>
> so it yields the formula again, together with its error.
>
> However, usually the parameter already have values assigned to them
> when I call the function. As was correctly pointed out, I can easily
> circumvent this problem by packing the call into a Block[]:
>
> a = 10;
> b = 20;
> aErr = 1;
> bErr = 2;
> Block[{a,b},
> error[a*b, {a, b}, {aErr, bErr}]]
>
>
> However, I was wondering if it might not be possible to handle this
> case entirely in the error function, so that I can call
>
> a = 10;
> b = 20;
> aErr = 1;
> bErr = 2;
> error[a*b, {a, b}, {aErr, bErr}]
>
> and get the correct result. *So I am not really solving an actual
> problem here*, I'm just playing with Mathematica's programming model
> to see if it would allows me to do such a thing.
>
>
> I figured out that, in order to be really able to work with the
> unevaluated expression I cannot to wrap it in Hold[], but I have to
> "escape" all the parameters that might otherwise be evaluated. So what
> I'm doing is this:
>
>
> SetAttributes[error, HoldAll];
> error::varno = "There must be the same number of variables and errors";
> error[expr_, vars_, errs_] :=
> Module[{safeExpr, safeVars, res, escapeRule, restoreRule, varno},
> varno = Length[vars];
>
> If[varno != Length[errs],
> Message[error::varno];
> Throw[$Failed];
> ];
>
> (* This replaces the given variables (which may have global definition=
> s) by local ones *)
> escapeRule = Table[
> Extract[Hold[vars], {1, i}, HoldPattern] -> safeVars[i],
> {i, varno}];
>
> (* This restores the original values of the variables *)
> restoreRule = Table[safeVars[i] -> vars[[i]], {i, varno}];
>
> (* Escape variables, so that we can differentiate the expression *)
> safeExpr = ReleaseHold[Hold[expr] /. escapeRule];
>
> (* Calculate result with error *)
> {expr,
> Sqrt[Plus @@ Table[
> Abs[D[safeExpr, safeVars[i]] errs[[i]] /. restoreRule ]^2,
> {i, varno}]]}
> ]
>
> This indeed works nicely, except for one thing: in order to replace
> all the global variables in the expression by local ones, I have to
> iterate over the list of parameters, which in turn requires me to know
> it's length.
>
> However, if I'm calculating the length with Length[vars] as above,
> vars is evaluated first which may, in theory, change its length. (I
> cannot really come up with an example where this may actually happen,
> but as I said I'm not trying to solve a real problem anyway).
>
>
> Now that I've written this all up so nicely, I also realize that the
> solution with Length[Unevaluated[vars]] will also work for this case.
> I'm mailing this anyway now, since I've already spend so much time
> writing it. Maybe someone else can come up with a more elegant version
> to do the above, so this mail wasn't entirely pointless :-).
>
>
>
> Best,
>
> -Nikolaus
>
> --
> =C2=BBTime flies like an arrow, fruit flies like a Banana.=C2=AB
>
> PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C
>
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