Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: ReplaceAll (/.)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69213] RE: [mg69197] ReplaceAll (/.)
  • From: "David Park" <djmp at earthlink.net>
  • Date: Fri, 1 Sep 2006 18:41:27 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Bruce,

Try

(a b) /. x_Times -> List @@ x // Trace

If you want delayed evaluation of the rhs of the rule you have to use
RuleDelayed.

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/

From: Bruce Colletti [mailto:vze269bv at verizon.net]
To: mathgroup at smc.vnet.net

Re Mathematica 5.2.0.0.

Why doesn't:

               (a b) /. x_Times -> List @@ x

return {a,b}?  My (flawed) reasoning is that a b has form Times[a,b], and so
x matches to Times[a,b].  List@@x replaces Times with List to yield
List[a,b], or {a,b}...or so it would seem.

The explanation is somehow linked to the fact that:

               (a b) /. x_Times :> List @@ x

returns {a,b}, but I don't see why this works but not the above.

Although List@@(a,b) returns {a,b}, I want to use the above ReplaceAll
function (this example here is taken from a larger problem).

Thankx.

Bruce



  • Prev by Date: Please help: How to use Mathematica to get Parametric solution for a transcendental equation?
  • Next by Date: Re: ReplaceAll (/.)
  • Previous by thread: Re: Re: ReplaceAll (/.)
  • Next by thread: Re: ReplaceAll (/.)