Question statement: Consider a Table with any number of indices {i,1,imax},{j,1,jmax},{k,1,kmax},.... Now I want to apply the same independent operation to each one of the elements while parallelizing the computation. What is the most concise way to do so?


So far, I have been using ParallelTable. However, this will separately take the lowest-order sub-array and paralellize over it separately. This may end up using fewer kernels at a time than available and slow down the process.

Consider the following example of a Table with 3 rows and 3 columns that I want to simplify in parallel:

ExTab = Table[..., {i,1,3},{j,1,3}];
SimplExTab = ParallelTable[FullSimplify[ExTab[[i,j]]],{i,1,3},{j,1,3}];

Now I have 10 kernels available, but ParallelTable will only use 3 kernels at a time, since it will only parallelize over each row at a time, even though it could use 9 and cut the time to roughly a third.

Of course, I could use Flatten on ExTab, apply ParallelTable, and reconstruct the structure of SimplExTab in the next step. This seems like an inelegant exercise, however. There should be a more concise way to do this. What is it?

More concrete example:

Let me give a little bit more precise example of my use case. Consider an example tensor

ExTens[n_,k_]:= Module[ {IndexStringList,TableString},
 IndexStringList = Table[ToString[{ToExpression["i:<>ToString[l]],1,k}],{l,1,n}];
 TableString = "Table[ExpandAll[Product[(x - RandomInteger[100])^RandomInteger[10],{m,1,18}]]";
 Do[TableString = TableString<>","<>IndexStringList[[l]],{l,1,n}];
 TableString = TableString<>"]";

For a given $n,k$, this function will randomly generate an $n$-index table (or tensor) in dimension $k$. Each component is a polynomial with integer coefficients, and variable powers of $x$ appearing. The thing that is not entirely obvious from the generated expressions is that every component can be factored using at most 18 positive integer roots of various multiplicity.

Now I want to recover the fact that the polynomials are obviously quite simple. Mathematica is able to do this by threading over the Table sequentially:

RandomTens = ExTens[5,5]

This does give the simple result of each component factorized by using the 18 integer roots. However, you start to notice the non-zero computation time even in this reasonably simple case ($\sim$ minutes on my machine).

Attempt at fully parallelizing concrete example:

I attempted to accelerate the example by using ParallelMap[] as suggested by Daniel Huber in the comments. The idea is that one can use ParallelMap to apply a function to each component of a simple Table:

-> {{f[a],f[b]},{f[c],f[d]}

This takes all parts of the table {{a,b},{c,d}} of depth one (which in this case are the table components), and applies $f$ to them. However, this stops working for RandomTens, since its components are expressions that have a longer depth due to being polynomials. So I instead use:


This does parallelize the computation partially, but surprisingly enough, it is only as good as the ParallelTable[] method given above! In particular, it never uses the full 10 cores, it launches only $k$ cores, and goes again "row by row" in some sense!

  • 1
    $\begingroup$ Try e.g.: ExTab = Array[Subscript[a, #1, #2] &, {3, 3}]; SimplExTab = ParallelMap[f, ExTab, {-2}] $\endgroup$ Apr 20, 2023 at 12:13
  • $\begingroup$ @DanielHuber Interesting approach, thanks. I am just curious whether this is error prone to the structure of the components of the table. For example, if I do ExTab = {{a,b,c},{d,e,f},{g,h,i}}, I have to use ParallelMap[f, ExTab, {-1}] instead of having {-2} in the level specification to ge the desired result. (This is since the desired level has changed because of the lack of subscripts.) That is, can one perhaps detect the lowest level of a Table automatically? $\endgroup$
    – Void
    Apr 20, 2023 at 13:03
  • 1
    $\begingroup$ Yes there is the function "Depth" that gives the max. numbers of indices+1 $\endgroup$ Apr 20, 2023 at 14:12
  • $\begingroup$ @DanielHuber Nice, thanks. I ended up using ArrayDepth[] instead, it gives me precisely on which level are the components of the multi-index table. However, ParallelMap[] still does not seem to parallelize efficiently in that case (see edit of post). Any ideas? $\endgroup$
    – Void
    Apr 21, 2023 at 11:52
  • $\begingroup$ Did you allow MMA to use all 10 cores? I observe that one core is always less loaded than the others. $\endgroup$ Apr 21, 2023 at 12:57

1 Answer 1


After exploring a few options in Mathematica and elsewhere, I came to the conclusion that there is probably no in-built function or concise expression that would do what I am looking for (speaking of v13.2 MMa here).

I believe that the reason for this is that a general array or list can have a rather complex and varied structure, and brute-forcing the parallelization over all components may or may not give a better result than the default, conservative Mathematica treatment.

I have written my own function that does what I mention in the OP:

ParTabSimp[f_,T_] :=   Block[{FlatT,Tind,Tlength,SubT,Tcomps,FlatSimpT,k,IndexList,IndexSum},
  FlatT = Flatten[T];   (*Flattened version of T*)
  Tind = ArrayDepth[T]; (*Number of indices of tensor*)
  SubT = T;             
  Do[Tlength[i] = Length[SubT]; SubT = SubT[[1]],{i,1,Tind}]; 
  (*The assumption is that T has a tensor-like structure, that is, it is a table of i1*i2*... numbers, no variable-length structures etc. are assumed*)
  Tcomps = Product[Tlength[i],{i,1,Tind}]; (*Tlength[i] tells you the range of the i-th index*)
  FlatSimpT = ParallelTable[f[FlatT[[i]]],{i,1,Tcomps}];
  IndexSum = " ";IndexList = " ";
    IndexSum = IndexSum<>"+i"<>ToString[j]; 
    IndexList = IndexList<>",{i"<>ToString[j]<>",1,"<>ToString[Tlength[j]]<>"}"

This function breaks up a multi-dimensional table T into its single entries, and applies the function f[] on each component while using all kernels available, then reconstructs the original tensor structure.

The use of the function is for example:ParTabMap[FullSimplify[#]&, RandomTens], using the random tensor RandomTens from the OP, and this can provide a significant edge over ParallelMap[FullSimplify[#]&,RandomTens,{ArrayDepth[RandomTens]}] in some use cases.


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.