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Re: Replace in an elegant way

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122877] Re: Replace in an elegant way
  • From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
  • Date: Mon, 14 Nov 2011 07:06:13 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <j9o3hc$o$1@smc.vnet.net>

On Sun, 13 Nov 2011 09:44:12 -0000, Mirko <dashiell at web.de> wrote:

> Hi all,
> I have following equation:
>
> ((1 + theta[m])^2)/(1 - m) - lambda*((theta[m] + ((lambda + m -  
> lambda*m)*(1 + theta[m])*(1 + theta[m] - lambda*theta[m]))/(1 - m) +
>      (m*Derivative[1][theta][m])/lambda))
>
> I want to replace all m that are not an argument of theta[m] or the  
> Derivative. However, if I use
> /., it replaces all m. If I use Replace[expr,m->qq] it doesn't replace  
> anything. (qq is the one I want to replace with). If I specify any  
> level, it doesn't change anything, unless I use -1, but then I replace  
> everything.
>
> Do you know how I solve this problem?
> Right now I use:
> //. Derivative[y_][theta][m] -> Derivative[y][theta][dd] //.
>     theta[m] -> theta[dd] /. m -> qq //.
>   Derivative[y_][theta][dd] -> Derivative[y][theta][m] //.
>  theta[dd] -> theta[m]
>
> but this is not elegant at all (and takes slightly longer).
>

While still inelegant (since it contains replacements of items with  
themselves), the following is something of an improvement over your  
existing method:

expr /. {
   deriv : Derivative[__][theta][m] :> deriv,
   arg : theta[m] :> arg,
   m -> qq
  }



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