Re: Block vs. ReplaceAll
- To: mathgroup at smc.vnet.net
- Subject: [mg79968] Re: Block vs. ReplaceAll
- From: ben <benjamin.friedrich at gmail.com>
- Date: Thu, 9 Aug 2007 06:27:45 -0400 (EDT)
- References: <f9c1cs$6jg$1@smc.vnet.net>
Dear Neil, I frequently encounter situations where I need an extra symbol to pass the value of a variable ( classical example: f = x; g[x_] := f; g[1]==x f=x g[xval_]:=f/.{x->xval} g[1]==1) Why not use something similiar here? sol2={aval->a,bval->b}/.Last[solution] Block[ {a=aval/.sol2,b=bval/.sol2} , ... ] Bye On 8 Aug., 11:11, Neil Stewart <neil.stew... at warwick.ac.uk> wrote: > Take a function f of some global variables and a function g that depends on > f. > > f := a^2 + b^2 (* f is a function of a and b, a and b are global variables *) > g := Count[Table[f, {10}], 0] (* g in turn depends on f *) > > Solutions can be found for a and b that minimize f. > > solution = NMinimize[{f, a > 0 && b > 0}, {a, b}] > {0., {a -> 0., b -> 0.}} > > But how is it best to use the solution? > > Block[{a = 0, b = 0}, f] (* Fast *) > 0 > > or > > f /. Last[solution] (* Slow *) > 0 > > For this trivial example obviously both are fast, but when f is non-trivial, > Block[] is much faster. > > Further for g, only Block[] works as I intend (counting the number of times > ten calls to f evaluate 0): > > Block[{a = 0, b = 0}, g] (* Fast and works as I intend *) > 10 > > g /. Last[solution] (* Slow, and doesn't count the number of times f is zero as I intended *) > > I understand why this happens: Count[] counts the fs in the table in symbolic > form, before a and b are replaced and f is evaluated at 0. > > My question is, how is it best to take "solution" - which is a list of > replace rules - and use it in a Block[] statement - which requires a list of > assignments? I need the speed that Block[] gives by making replacements before > f is evaluated. I'm keen to keep the simplicity of having model parametes as > global variables to avoid having to pass model parameters explicitly to > model functions as this is messy with a non-trivial model.