1
$\begingroup$

There is a translation operation from a numeric variable to a string variable:
a[i_, j_] := Symbol["a" <> ToString[i] <> ToString[j]]

How could the reverse operation be done? Something like that:
Symbol["a" <> ToString[i_] <> ToString[j_]] := a[i, j]

$\endgroup$
8
  • 6
    $\begingroup$ Why on earth would you want to do that? $\endgroup$ Commented May 20, 2023 at 12:17
  • 2
    $\begingroup$ I am getting a headache just trying to understand what you are trying to do :) Are you trying to convert a12 to a[1,2] for example? can you give an explicit example of an input and what the output should be? $\endgroup$
    – Nasser
    Commented May 20, 2023 at 12:24
  • 1
    $\begingroup$ Anyways, ReleaseHold[With[{symbol = Symbol["a" <> ToString[i] <> ToString[j]]},HoldForm[symbol = a[i, j]]] ] might work. $\endgroup$ Commented May 20, 2023 at 12:25
  • $\begingroup$ @Nasser, I have many string variables, I want to somehow automate data entry for them. But how can I distinguish them from each other in order to apply a loop to them? $\endgroup$
    – Mam Mam
    Commented May 20, 2023 at 12:48
  • 6
    $\begingroup$ I've read all of your recent questions but I'm also confused. I have a strong feeling that you're asking XY problem repeatedly. What on earth are you trying to design? $\endgroup$
    – xzczd
    Commented May 20, 2023 at 13:17

2 Answers 2

3
$\begingroup$

As with all the others, I do not understand what are you after. However, in the past, I met a comparable problem myself, where I had a good reason to do something like this. Of course, without a clear understanding of your aim, I can give a useless solution. Anyway, let us try.

You defined the function which acts only on tensors with the head a transforming them into a string and then into a symbol:

a[i_, j_] := Symbol["a" <> ToString[i] <> ToString[j]]

Let us check

a[i, j] // FullForm
a[i, j] // Head

returns aij and Symbol correspondingly. That's OK.

Let us introduce the function

toTensor[expr_Symbol] := 
 Module[{x, y, z}, 
  ToExpression@Characters[ToString[expr]] /. {x_, y_, z_} :> x[y, z]]

Now let us apply it to aij:

Clear[a];
toTensor[aij]

(* a[i, j] *)

The key point here is to first apply Clear[a]. It is because you have already previously defined a[i,j] such that it immediately transforms into aij. Therefore, without clearing it will always return aij. The symbol with a different first letter you can transform without clearing

toTensor[bij]

(* b[i, j]  *)

Hope this helps.

Have fun!

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the explanation! $\endgroup$
    – Mam Mam
    Commented May 20, 2023 at 19:37
  • 1
    $\begingroup$ Simpler: ToExpression@Characters[ToString[expr]] /. {x__} :> (First@{x})[Sequence @@ Rest@{x}] $\endgroup$ Commented May 20, 2023 at 22:05
3
$\begingroup$

I feel like we're going in circles. Over the last 2 days you've asked several questions about associating atomic symbols with headed expressions in some fashion. I'm almost certain that you don't need to do any of this. But I can't help you figure out what to do instead without knowing what problem you're actually trying to solve. And I don't mean the ultimate problem, like whatever advanced math or physics problems you're dealing with. I just mean the computational problem you're struggling with. Can you give an example that shows a bit more of the problem than you've shown us so far?

In the meantime, I'll take a stab in the dark...

Maybe want you want is to use a data structure rather than individual atomic values. You keep mentioning loops, so instead of things like a[1,2] <--> a12, maybe you use Part instead: a[[1,2]]. So, you can do things like this:

a = {{1, 2}, {3, 4}};
a[[1, 2]]
(* 2 *)

Or loops:

Do[
  Do[
    Print[a[[i, j]]],
    {j, 2}],
  {i, 2}]

Or maps:

Map[f, a, {2}]
(* {{f[1], f[2]}, {f[3], f[4]}} *)

Or threading, or matrix operations, or etc etc.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.