Re: Replace in an elegant way

*To*: mathgroup at smc.vnet.net*Subject*: [mg122887] Re: Replace in an elegant way*From*: Heike Gramberg <heike.gramberg at gmail.com>*Date*: Mon, 14 Nov 2011 07:08:01 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201111130943.EAA29993@smc.vnet.net>

You can actually do something like this: ((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)) /. {theta'[m] -> theta'[m], theta[m] -> theta[m], m -> dd} This works because ReplaceAll compares each part of the original expression with the elements in the replacement list in turn until the first match is found. As soon as a match is found, that part nor any of its subparts are checked further. So for example, if a match for theta[m] is found in the example above, it's replaced with theta[m] and none of its subparts (i.e. m inside the brackets) are checked against the replacement list. Heike. On 13 Nov 2011, at 10:43, Mirko 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). >

**References**:**Replace in an elegant way***From:*Mirko <dashiell@web.de>