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Re: Making a function dynamically define another conditional function...
*To*: mathgroup at smc.vnet.net
*Subject*: [mg21755] Re: Making a function dynamically define another conditional function...
*From*: "Allan Hayes" <hay at haystack.demon.co.uk>
*Date*: Thu, 27 Jan 2000 22:56:36 -0500 (EST)
*References*: <86md60$2cr@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Paul:
We need to get the values inside the assignments before evaluating them:
TestFn[data_List] := Module[{args},
ClearAll[f];
Do[
args = data[[i]];
(f[x_] = x + #1 /; x == #2) & @@ args,
{i, Length[data]}
]]
TestFn[{{a, A}, {b, B}}]
?f
"Global`f"
f[x_] = a + x /; x == A
f[x_] = b + x /; x == B
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Paul Howland" <paul.howland at nc3a.nato.int> wrote in message
news:86md60$2cr at smc.vnet.net...
>
> How can I make a function dynamically define a conditional function?
>
> Given a list of arguments {{a,A}, {b,B}, ...} I want to write a function
> that will take these arguments, and generate a new function, f say,
> which is defined as (for example):
> f[x_] := x+a /; x==A
> f[x_] := x+b /; x==B
> etc.
>
> So, the obvious solution is to define a function as follows:
>
> In[1] := TestFn[data_List] := Module[{args},
> ClearAll[f];
> Do[
> args = data[[i]];
> f[x_] = x + args[[1]] /; x==args[[2]],
> {i, Length[data]}
> ]]
>
> and call it using something like TestFn[{{1,2},{3,4},{5,6}}].
>
> But this doesn't work (see attached notebook) as it appears that
> Mathematica does not evaluate any part of the condition at the time of
> definition, so args[[2]] remains unevaluated. As a consequence, the
> resulting function definition is not properly defined.
>
> So, the obvious solution to this is to wrap Evaluate[] around the
> condition (i.e. define the function as f[x_] = x + args[[1]] /;
> Evaluate[x == args[[2]]]. And this appears to work. If you do ?f, then
> you see a function comprising a number of conditional definitions.
> However, if you come to use the function, then it appears that
> Mathematica does not perform the condition test that appears in the
> definition! It simply uses the first definition it finds.
>
> What is going on?! How can I make this work?
>
> I attach a notebook with example code. [Contact Paul to
> get this notebook - Moderator]
>
> Many thanks for any help.
>
> Paul
>
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