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

*To*: mathgroup at smc.vnet.net*Subject*: [mg21777] Re: [mg21733] Making a function dynamically define another conditional function...*From*: BobHanlon at aol.com*Date*: Thu, 27 Jan 2000 22:56:58 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Here is one approach: Clear[f]; ?f "Global`f" testList = {{a, A}, {b, B}, {1, 2}, {3, 4}, {5, 6}}; TestFn[data_List] := ToExpression[ "f[x_] := x+" <> ToString[#[[1]]] <> "/;x==" <> ToString[#[[2]]] & /@ data] TestFn[testList]; ?f "Global`f" f[x_] := x + a /; x == A f[x_] := x + b /; x == B f[x_] := x + 1 /; x == 2 f[x_] := x + 3 /; x == 4 f[x_] := x + 5 /; x == 6 f[#[[2]]] & /@ testList {a + A, b + B, 3, 7, 11} Bob Hanlon In a message dated 1/26/2000 5:20:51 AM, paul.howland at nc3a.nato.int writes: >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] >