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In a multistep code, I am trying to use module to define a function such that (working example)

CR[i_] := {{i*5, 3, 5, i, 2 i, 5 i}, {1, 2, 3, 4, 5, 6}}

A[EF_] := Module[{cr = CR[EF]}, c[t_] = ParallelTable[cr[[t, m]], {m, 1, 6}]; {c[1], c[2]}]

A[1]

by using this expression, I am getting my o/p in about 14 secs. However, along with the o/p, there also comes an error message 'Part::pkspec1: The expression t$ cannot be used as a part specification.'

But, when I replace equal to (=) by set delayed (:=) in the definition of c[t], error message is gone, but I get my o/p in about 4 minutes instead of 14 secs.

So, my question is how to go about it..so that I get my o/p in reasonable time, but without any error message. Will appreciate any suggestion.

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There are few problems with your code.

First, I guess that A[r] and B[r] are some nested constructs. In the simplest case, lists. Is it right?

If yes, let us for the sake of example take the following:

A[r_] := {r, r^2, r^3, r^4};
B[r_] := {r^2, r^3, r^4, r^5};

Second, you cannot define c[t] as a function of t in your code, since t is a dummy variable. Therefore, c is not a function of t but a constant. I would define it as follows:

c = Table[a[[t]] + b[[t]], {t, 1, 4}];

Third, the use of the ParallelTableonly pays off if the table is very long or the evaluation of its components is very time-consuming, or both. Otherwise, its use makes the evaluation considerably longer. For this reason, I used Table instead. If you have a good reason to use the ParallelTable in your code, you may replace one with another.

Fourth, you need to use a semicolon (;) after each operator entering the Module except for the last one.

Fifth, as soon as c cannot be defined as the function of t the c[t] statement staying at the end of the Modulehas no sense.

With all these notes one finds this:

 AA[r_] = Module[{a = A[r], b = B[r]}, 
      c = Table[a[[t]] + b[[t]], {t, 1, 4}]; c]

(*    {r + r^2, r^2 + r^3, r^3 + r^4, r^4 + r^5}   *)

and, say,

AA[2]

(*  {6, 12, 24, 48}  *)

Think also, if it is better to define AA[r] using Setor SetDelayed?

Have fun!

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  • $\begingroup$ Now, I have added a working example in the original post. It explains my problem better. Thanks $\endgroup$
    – user49535
    Commented Sep 7, 2021 at 11:52
  • $\begingroup$ In this case, your problem is solved by a more economical code: c[z_] := Take[CR[i], {z}] // Flatten Check c[1] and c[2]. $\endgroup$ Commented Sep 7, 2021 at 12:59

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