Suppose I want to calculate a vector or an array (say, for simplicity a 4-by-1 vector), but each element contains independent numerically evaluated functions $f_1,f_2,f_3,f_4$ and each takes long time. Is there a way to parallelize computation of $f_i$?

The documentation and Stackexchange I searched so far (maybe I missed some) seems to be catering for a situation where we have a "master function" $F[i]$ whose entries are parallelized by using ParallelTable. Does that mean the only way (perhaps valid anyway) to do this is to simply define $F[i\_]:= f_i$? What is the "best practice" for this?

Update: the functions $f_i$ are a priori unrelated, but they depend on the same set of variable, so I was trying to Parallelize this inside a Module. I am also trying to avoid defining too many functions which is why if I can help it I want to avoid defining $F$ (because this will occur many times in the notebook).

  • 3
    $\begingroup$ For a start, you could try Parallelize[{f1, f2, f3, f4}, Method -> "CoarsestGrained"]. ParallelSubmit may also be helpful. For more relevant answers, please provide an example of your functions, or even some made up sample code that represents your problem. $\endgroup$
    – MarcoB
    Jan 14, 2022 at 22:20

2 Answers 2


Here is an example using ParallelSubmit:

f1[x_] = x;
f2[x_] = x^2;
f3[x_] = x^3;
f4[x_] = x^4;

(* {2, 4, 8, 16} *)
ParallelMap[Construct[#, x] &, {f1, f2, f3, f4}]
(*    {f1[x], f2[x], f3[x], f4[x]}    *)
  • 1
    $\begingroup$ You need to add a Method -> "FinestGrained" to ensure one calc per kernel at a time. $\endgroup$
    – Edmund
    Jan 15, 2022 at 17:19
  • $\begingroup$ @Edmund that's not necessary. Try f[x_] := Module[{}, Pause[1]; x^2] and then ParallelMap[Construct[#, x] &, {f, f, f, f}] // AbsoluteTiming, which on my computer takes about 1 second, meaning that the execution was one task per kernel. It looks like the default method can deal with this situation just fine. $\endgroup$
    – Roman
    Jan 15, 2022 at 17:29
  • 1
    $\begingroup$ Add the option will guarantee it to be the case. Automatic behavior can change. $\endgroup$
    – Edmund
    Jan 15, 2022 at 17:37

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