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