RE: ReplaceAll doesn't replace
- To: mathgroup at smc.vnet.net
- Subject: [mg32566] RE: [mg32550] ReplaceAll doesn't replace
- From: "David Park" <djmp at earthlink.net>
- Date: Sun, 27 Jan 2002 03:28:55 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Ken,
Map[f, expr] or f /@ expr applies f to each element on the first level in
expr.
But there is nothing at the first level of the expression x, so x is
returned unchanged.
#^2 & /@ x
x
and, of course,
x /. x -> {a, b, c}
{a, b, c}
David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/
> From: Ken Morgan [mailto:kmorga51 at calvin.edu]
To: mathgroup 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
>
>