Re: ReplaceAll (/.)

• To: mathgroup at smc.vnet.net
• Subject: [mg69203] Re: [mg69197] ReplaceAll (/.)
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 1 Sep 2006 18:41:01 -0400 (EDT)
• References: <200609011041.GAA25660@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```On 1 Sep 2006, at 11:41, Bruce Colletti wrote:

> 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
>

You need to use RuleDelayed (:>) rather than Rule (->) because, if
you use just Rule, List@@x is first evaluated and gives:

List@@x

x

So all you are actually doing is

(a b) /. x_Times -> x

Andrzej Kozlowski

```

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