Re: defining a function whose parameter must be a function

*To*: mathgroup at smc.vnet.net*Subject*: [mg131027] Re: defining a function whose parameter must be a function*From*: Roman <rschmied at gmail.com>*Date*: Tue, 4 Jun 2013 02:01:44 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130601102829.0E29D6A2D@smc.vnet.net> <koehga$2t2$1@smc.vnet.net>

Thanks a lot to Bob and David for answers! Bob's first part of test-calling FreeQ[f[Unique[], Unique[]], f] looks like a good idea, but I believe his second part can be improved since it would fail on a function like Function[{x, y}, x] which happens to not depend on one of its arguments. Something like this: FunctionWithTwoParametersQ[f_] := Quiet[ Check[FreeQ[f[Unique[], Unique[]], f], False, {Function::slotn, Function::fpct}], {Function::slotn, Function::fpct}] David's comment on performance is spot-on: test-calling f[Unique[], Unique[]] may incur an unbearable cost if f is slow to execute when called with nonnumeric arguments but fast with numeric arguments. This makes the idea of test-calling the function f[Unique[], Unique[]] a bit unclean since in principle I would only like to find out whether or not I *can* call the function, and not to actually call it and execute it all the way through. I would hope that the Mathematica internals would allow me to do this somehow, i.e. to test-call f[Unique[], Unique[]] but stop after the pattern-matching step, and not go into the actual function execution. I don't know what I'd have to Hold[] in order to achieve this, though. In practice I will probably be using something closer to this, which combines Bob's idea and David's warning: NumericFunctionWithTwoNumericArgumentsQ[f_] := Quiet[NumericQ[f[Random[], Random[]]], {Function::slotn, Function::fpct}] In a concrete case one might even replace the Random[] calls by zero values if this is guaranteed not to cause problems. Any thoughts on this would be appreciated. Cheers! Roman Am Sonntag, 2. Juni 2013 06:25:46 UTC+2 schrieb Bob Hanlon: > f1[x_] = x; > > f1[x_, y_] = x^2 - y^2; > > f1[x_, y_, z_] = x^2 - y^2 - 3 z; > > > > > > f2 = #1^2 - #2^2 &; > > f3 = Function[{x, y}, x^2 - y^2]; > > f4 = #1 &; > > f5 = Function[{x}, x]; > > f6[a_, b_, c_] = a^2 - b^2 - 3 c; > > > > > > F[f_] := f[2, 3] /; > > FreeQ[f[Unique[], Unique[]], f] && > > Length[Union[ > > Cases[f[Unique[], Unique[]], > > _Symbol?(! NumericQ[#] &), > > Infinity]]] == 2 > > > > > > F /@ {f1, f2, f3, f4, f5, f6} > > > > > > {-5, -5, -5, F[#1 &], F[Function[{x}, x]], F[f6]} > > > > > > > > Bob Hanlon > > > > > > > > > > > Dear all, > > > I am trying to define a function F which will only execute if its > > > parameter is a function with two parameters. Let's say I define it thus, > > > without any checks on the parameter pattern f: > > > > > > In[1] := F[f_] := f[2,3] > > > > > > There are several ways of calling F: > > > 1) pass it a function of two parameters: > > > In[2] := F[Function[{a,b},a^2-b^2]] > > > Out[2] = -5 > > > > > > 2) pass it an anonymous function of two parameters: > > > In[3] := F[#1^2-#2^2 &] > > > Out[3] = -5 > > > > > > 3) pass it a pre-defined function: > > > In[4] := g[a_,b_] = a^2-b^2; > > > In[5] := F[g] > > > Out[5] = -5 > > > > > > My question is: how can I define a pattern in the definition of F[f_] such > > > that this function F will execute these three cases while not executing if > > > called with any other kind of parameter? The following calls should fail, > > > for example: > > > > > > In[6] := F[Function[{a,b,c},a^2-b^2-3c]] > > > Out[6] = F[Function[{a,b,c},a^2-b^2-3c]] > > > > > > In[7] := F[#1^2-#2^2-3#3 &] > > > Out[7] = F[#1^2-#2^2-3#3 &] > > > > > > In[8] := h[a_,b_,c_] = a^2-b^2-3c; > > > In[9] := F[h] > > > Out[9] = F[h] > > > > > > Further, for bonus points, if there are multiple definitions of a > > > function, I'd like to pick the one with two parameters: > > > In[10] := k[a_,b_] = a^2-b^2; > > > In[11] := k[a_,b_,c_] = a^2-b^2-3c; > > > In[12] := F[k] > > > Out[12] = -5 > > > > > > Thanks for any help! > > > Roman > > > > > >

**References**:**defining a function whose parameter must be a function with two parameters***From:*Roman <rschmied@gmail.com>