Use replace in rule

Question

I would like to define two rules: one for the material and thickness the second one for volume and mass

rule1={t->1/2,row->1000};
rule2={v->t^3,m->v*row};

now if I used the following:

rule2/.rule1={v->1/8,m->v*1000}

how to replace v in rule2 to get the following:

{v->1/8,m->125}

Why not just use Set and allow MMA to do the substitutions for you: t = 1/2; row = 1000; vv = t^3; mm = vv row; Thread[{v, m} -> {vv, mm}] — MelaGo, Commented May 19, 2021 at 2:48
Somewhat convoluted: ReplacePart[rule2, 2 -> (rule2[[2]] /. rule2[[1]])] /. rule1 — Bob Hanlon, Commented May 19, 2021 at 2:49
I dont want to set the parameters to values because I am not sure what is the material that I'll be using (design problem). But after conducting the full study I will replace all parameters by the appropriate values. @MelaGo — Love Eva, Commented May 19, 2021 at 2:54
(rule2 /. rule1) /. (lhs_ -> rhs_) :> (lhs -> (rhs /. (rule2 /. rule1))) — Bob Hanlon, Commented May 19, 2021 at 3:24
I would turn your rules into equations (e.g. Equal@@@rule1) then use Solve. — mikado, Commented May 19, 2021 at 9:33

kglr · Accepted Answer · 2021-05-19 04:04:14Z

3

ReplaceAll[rule1] @ MapAt[ReplaceAll[rule2], rule2, {All, 2}]

{v -> 1/8, m -> 125}

answered May 19, 2021 at 4:04

kglr

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$\begingroup$ What changes do I need yo do to the above code to work with Mathematica 7 ? $\endgroup$
– Love Eva
Commented May 19, 2021 at 16:21
$\begingroup$ @LoveEva, try ReplaceAll[MapAt[ReplaceAll[#, rule2] &, rule2, {All, 2}], rule1] or MapAt[# /. rule2 &, rule2, {All, 2}] /. rule1. $\endgroup$
– kglr
Commented May 19, 2021 at 18:55

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