Re: What is a good way of returning a function from a Module[]?

*To*: mathgroup at smc.vnet.net*Subject*: [mg83392] Re: What is a good way of returning a function from a Module[]?*From*: Norbert Marxer <marxer at mec.li>*Date*: Mon, 19 Nov 2007 06:15:32 -0500 (EST)*References*: <fhmesn$8pg$1@smc.vnet.net> <fhp36k$276$1@smc.vnet.net>

On 18 Nov., 11:12, Peter Pein <pet... at dordos.net> wrote: > Szabolcs Horv=E1t schrieb: > > > > > > > What is an elegant way of returning a function from a Module[]? > > > Module[{a, b}, a = 1; b = a+1; (a+b+#)&] does not work because > > Function[] holds its arguments. > > > The best way I could find was > > > Module[{p, q}, p = 1; q = p+1; With[{a = p, b = q}, (a+b+#)&]] > > > Is there a nicer/more concise way of doing this? This is a simplified > > example, but the important points are: > > > 1. The returned function may depend on more than one parameter ('a' and > > 'b'; let's forget that in this case their sum could have been computed > > inside the Module[]). > > > 2. 'a' and 'b' are not calculated independently. The value of 'a' is > > needed to find 'b' > > > Szabolcs > > In Version 5.2: > > map Evaluate onto the function: > > Module[{a, b}, a = 1; b = a + 1; > Evaluate /@ (a + b + #1 & )] > > 3 + #1 & > > This works even for functions with arguments which got names already > used in the module: > > Module[{a, b, x = Pi, y = E}, > a = 1; b = a + 1; Evaluate /@ > Function[{x, y}, (a + b + x)/ > (y - b + a)]] > > Function[{x$, y$}, (3 + x$)/(-1 + y$)] > > Peter- Zitierten Text ausblenden - > > - Zitierten Text anzeigen - Hello You do not even need Map. You can just wrap your Function body into an Evaluate: Module[{a, b}, a = 1; b = a + 1; Evaluate[a + b + #1] & ] Best Regards Norbert Marxer