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 >