Re: ReplaceAll doesn't replace

*To*: mathgroup at smc.vnet.net*Subject*: [mg32572] Re: ReplaceAll doesn't replace*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Sun, 27 Jan 2002 03:29:04 -0500 (EST)*References*: <a2tsuk$ea9$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Ken The reason for the difference is bracketing. Here are the main steps in the evaluations #^2 & /@ (x /. x -> {a, b, c}) Map[#^2 &, x /. x -> {a, b, c})] Map[#^2 &, {a, b, c}] {#^2 & [a], {#^2 & [a],{#^2 & [a]} {a^2, b^2, c^2} (#^2 & /@ x) /. x -> {a, b, c} Map[#^2 & , x]/. x -> {a, b, c} x/. x -> {a, b, c} {a,b,c} -- 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 "Ken Morgan" <kmorga51 at calvin.edu> wrote in message news:a2tsuk$ea9$1 at smc.vnet.net... > I understand how the following will replace x with the list > > #^2 & /@ (x /. x -> {a, b, c}) > > to generate > > {a^2, b^2, c^2} > > But, why isn't x replaced at the beginning of the evaluation in the > following > > (#^2 & /@ x) /. x -> {a, b, c} > > since it generates > > {a, b, c} > > What I really want to know is: What is it about the Function function that > doesn't allow ReplaceAll to "replace all" at the beginning of an evaluation? > > Thanks, > Ken Morgan > kmorga51 at calvin.edu > >