Re: ReplaceAll doesn't replace

*To*: mathgroup at smc.vnet.net*Subject*: [mg32571] Re: ReplaceAll doesn't replace*From*: atelesforos at hotmail.com (Orestis Vantzos)*Date*: Sun, 27 Jan 2002 03:29:02 -0500 (EST)*References*: <a2tsuk$ea9$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

This is a Trace[..] of your second expression: In[7]:= 2 (#1 & ) /@ x /. x -> {a, b, c} Out[7]= 2 {(#1 & ) /@ x, x} {x -> {a, b, c}, x -> {a, b, c}} x /. x -> {a, b, c} {a, b, c} As you can see, ReplaceAll never gets to see the square Function at all. The reason for this is that f/@a returns a no matter what f is...the reason for this is that the operator /@ applies f to the first level of a, which just isn't there! #^2& /@ {x} /. x->{a,b,c} will do the job just fine though. Orestis