first post here.

I'm currently completely stuck and frustrated with mathematica. In retrospective, it might had been a mistake to use it for this, I'm actually considering to somehow export the results I have so far into python. Disregarding that: What I have right now are two nested lists/matrices, one of which is empty and is supposed to be filled by the nested list. The nested list is organized in such a way that each line has a value, and 2 positional values, to insert into the matrix. I've been trying to use while loops, for loops etc. and everytime it just returns an unchanged empty matrix to me.

Some "pseudo-code" I've been actually using:

Test1 = {{1, 1, 1}, {2, 3, 2}, {3, 1, 1}}
FinTest1 = ConstantArray[0, {3, 3}]
i = 1;
While[i < 4, Print[i];
  j = 1;
  While[j < 4,
   n = 1;
   While[n < 4,
    If[i === Test1[[n, 3]] And j === Test1[[n, 2]],
     FinTest1[[i, j]] = FinTest1[[i, j]] + Test1[[n, 1]]];

This one is a heavily simplified version of what I am currently doing (Mine has around a million of data and the matrix is around 50 times larger). As you might see, at least that's something I expected to get, is a matrix in the form of:


Instead I get the empty matrix back, as if nothing happened with it. I've also noticed that the iterables stay the same within each loop, meaning i is always 1 inside the inner-most While loop, even though it should iterate.

Mind you, these are just my calculations, that's all I've been using Mathematica for so far. It's the first time I use some kind of "proper code" in this. But it turns out that somehow it treats values inside the loops differently from normally. Is there a way to circumvent that?

  • $\begingroup$ To start with, your syntax for And is wrong. $\endgroup$
    – lericr
    Commented Mar 30, 2022 at 16:41
  • $\begingroup$ I'm making some assumptions, but maybe you're trying to do something like this: SparseArray[Rule[{#1, #2}, #3] & @@@ Test1, {3, 3}] $\endgroup$
    – lericr
    Commented Mar 30, 2022 at 16:46
  • $\begingroup$ @lericr It's mostly experimental. Though it didn't seem to have an issue with it. Replace seems to be a wonderful function. Though now I'm having a problem with drawing a matrixplot/arrayplot. All I get is "Argument {Matrix} at position 1 is not a list of lists". Even though it is... Does Wolfram have a maximum matrix size which it is willing to plot? (360x180) $\endgroup$
    – Hewe
    Commented Mar 30, 2022 at 17:13
  • $\begingroup$ that sounds like a new question. but you should trust the error message and try to fix that first. $\endgroup$
    – lericr
    Commented Mar 30, 2022 at 17:23
  • $\begingroup$ @lericr Managed to find the mistake. Replace seems to cause the problem by turning the matrix into a ragged list (somehow a few elements simply seem to disappear). SpraseArray on the other hand seems to have a problem with decimal values (around 10^-4) $\endgroup$
    – Hewe
    Commented Mar 30, 2022 at 17:30

1 Answer 1


Instead of using While, it is simpler to process the Test1 variable as follows:

Test1 = {{1, 1, 1}, {2, 3, 2}, {3, 1, 1}};
FinTest1 = ConstantArray[0, {3, 3}];

Replace[Test1, {i_, j_, k_} :> AddTo[FinTest1[[j,k]], i], {1}];

{{4, 0, 0}, {0, 0, 0}, {0, 2, 0}}

It is also possible to modify SparseArray options to do this as well, e.g.:

old = SystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries"];
SetSystemOptions["SparseArrayOptions" -> "TreatRepeatedEntries" -> Total];

Then, using SparseArray:

SparseArray[Replace[Test1, {i_, j_, k_} :> {j, k} -> i, {1}]] //Normal

{{4, 0}, {0, 0}, {0, 2}}

Finally, reset the SparseArray options:

  • $\begingroup$ Thanks. It actually calculated it extremely fast, but there seems to be a problem with it removing some elements from the matrix, or at least making wolfram think that it's a ragged list which cannot be converted into an array nor matrix. SparseArray causes problems with the following error message: The left-hand side of {89,0}->0.000109064 in {{89, -29} (shortened) 0.0000853162, {90, 7} -> 0.0000215934} is not a position or a pattern \ that will match the position of an element in an array with depth $\endgroup$
    – Hewe
    Commented Mar 30, 2022 at 17:33
  • $\begingroup$ Right, this is for the 360x180 matrix I'm using. I've checked the dimensions of the inputs, and they're correct, 360 180. $\endgroup$
    – Hewe
    Commented Mar 30, 2022 at 17:41
  • $\begingroup$ Mathematica starts indexing matrices with 1, not 0. j and k must be positive integers. $\endgroup$
    – Carl Woll
    Commented Mar 30, 2022 at 17:54
  • $\begingroup$ Oh right. Thanks $\endgroup$
    – Hewe
    Commented Mar 30, 2022 at 18:21

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