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MathGroup Archive 2000

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Re: Making a function dynamically define another conditional function...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21745] Re: Making a function dynamically define another conditional function...
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Thu, 27 Jan 2000 22:56:28 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <86md60$2cr@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Paul,


with

TestFn[data_List] := Module[{args, lst}, ClearAll[f];
    lst = {x + First[#] , x == Last[#]} & /@ data;
    ( f[x_] = #) & /@ (ReleaseHold[Condition[#1, Hold[#2]] & @@@ lst])
    ]

and

TestFn[{{1, 2}, {3, 4}, {5, 6}}]

f[x_] = 1 + x /; x == 2
f[x_] = 3 + x /; x == 4
f[x_] = 5 + x /; x == 6

Since the Condition[] has the HoldAll attribute your version can't work.
Using my construction with out the Hold[] in Condition[#1,Hold[#2]] will 
also not work because the condition is evaluatet and fails. When the
result us enclosed in Condition[] one can release the Hold[] bcause
Condition[] will not evaluate. Finaly the function definition can be
created.

Hope that helps
  Jens

Paul Howland wrote:
> 
> 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

ZZ


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