Is there a simple way to apply a function to the RHS of a substitution?

Question

Suppose I have solved a system of equations, resulting in a solution of the form:

sol = {{f[a1] -> SomeComplicatedFunction[x, a1]}, {f[a2] -> SomeComplicatedFunction[x, a2]}, {f[a3] -> SomeComplicatedFunction[x, a3]}}

I want to use the list of substitutions sol later on, but first I want to apply some method myFunction to the rhs of the substitution. I.e., the desired output is

sol = {{f[a1] -> myFunction[SomeComplicatedFunction[x, a1]]}, {f[a2] -> myFunction[SomeComplicatedFunction[x, a2]]}, {f[a3] -> myFunction[SomeComplicatedFunction[x, a3]]}}

I have tried extracting all the RHSs from sol, applying myFunction to the list, then setting this back to the original f[]s, but this seems to be very inefficient. Is there some canonical Mathematica way of applying rules to solutions while keeping the solutions intact?

score 11 · Accepted Answer · answered Oct 01 '19 at 00:19

sol = {
   {f[a1] -> SomeComplicatedFunction[x, a1]},
   {f[a2] -> SomeComplicatedFunction[x, a2]},
   {f[a3] -> SomeComplicatedFunction[x, a3]}};

sol2 = sol /. Rule[lhs_, rhs_] :> Rule[lhs, myFunction[rhs]]

(* {{f[a1] -> myFunction[SomeComplicatedFunction[x, a1]]}, {f[a2] -> 
   myFunction[SomeComplicatedFunction[x, a2]]}, {f[a3] -> 
   myFunction[SomeComplicatedFunction[x, a3]]}} *)

kglr · Answer 2 · 2023-05-04T10:50:15.507

Additional alternatives:

You can use MapAt to map myFunction at positions {All, -1, -1}:

sol3 = MapAt[myFunction, {All, -1, -1}] @ sol

{{f[a1] -> myFunction[SomeComplicatedFunction[x, a1]]},
{f[a2] -> myFunction[SomeComplicatedFunction[x, a2]]},
{f[a3] -> myFunction[SomeComplicatedFunction[x, a3]]}}

You can assign new values at desired positions using Part assignment:

sol4 = sol;
sol4[[All, -1, -1]] = myFunction /@ sol4[[All, -1, -1]];
sol4

same result

score 1 · Answer 3 · answered May 04 '23 at 10:44

You can also replace SomeComplicatedFunction with the composition of myFunction and SomeComplicatedFunction:

sol /. SomeComplicatedFunction -> myFunction @* SomeComplicatedFunction

{{f[a1] -> myFunction[SomeComplicatedFunction[x, a1]]}, 
 {f[a2] -> myFunction[SomeComplicatedFunction[x, a2]]},
 {f[a3] -> myFunction[SomeComplicatedFunction[x, a3]]}}

score 1 · Answer 4 · answered May 04 '23 at 23:25

Or:

sol /. Rule -> (Rule[#, myFunction @ #2] &)

(*
{{f[a1] -> myFunction[SomeComplicatedFunction[x, a1]]}, 
 {f[a2] -> myFunction[SomeComplicatedFunction[x, a2]]},
 {f[a3] -> myFunction[SomeComplicatedFunction[x, a3]]}}
*)

Safer:

Apply[Rule[#, myFunction@#2] &, sol, {2}]

Or:

# -> myFunction @ #2 & @@@ # & /@ sol

Or:

Dataset[Association /@ sol][All, Map@myFunction] // Normal // Normal

Is there a simple way to apply a function to the RHS of a substitution?

4 Answers4