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