# How to slice these variables?

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Joerg Pfannmoeller on 9 Nov 2021
Hi,
can somebody help me to slice these variables in a parfor loop?
indices(:,1) = max(min(ceil(vector(:,1)-Xcoor),nx),1);
indices(:,2) = max(min(ceil(vector(:,2)-Ycoor),ny),1);
indices(:,3) = max(min(ceil(vector(:,3)-Ycoor(idx_parfor)),nz),1);
I currently use the construction:
indices_1 = max(min(ceil(vector(:,1)-Xcoor),nx),1);
indices_2 = max(min(ceil(vector(:,2)-Ycoor),ny),1);
indices_3 = max(min(ceil(vector(:,3)-Ycoor(idx_parfor)),nz),1);
Is there any way to use a single variable indices again? In order to allow for indexing of the variable indices? The parfor variable is of course idx_parfor. An indexed solution would be unavoidable if there are more than 3 indices.
Best Joerg
##### 2 CommentsShowHide 1 older comment
Joerg Pfannmoeller on 14 Mar 2022
Thank you. There is a type-o in indices_3, as you found.

Edric Ellis on 9 Nov 2021
I think what you're trying to do is something like this, which doesn't work:
N = 4;
out = zeros(N, 3);
try
parfor idx = 1:N
out(idx,1) = idx;
out(idx,2) = -idx;
out(idx,3) = 2*idx;
end
catch E
disp(E.message)
end
Error: Unable to classify the variable 'out' in the body of the parfor-loop. For more information, see Parallel for Loops in MATLAB, "Solve Variable Classification Issues in parfor-Loops".
The way to fix this is to assign a whole column of out all in one indexing expression:
N = 4;
out = zeros(N, 3);
parfor idx = 1:N
o1 = idx;
o2 = -idx;
o3 = 2*idx;
% Single sliced indexing assignment into `out`
out(idx,:) = [o1, o2, o3];
end
disp(out)
1 -1 2 2 -2 4 3 -3 6 4 -4 8
In some situations, you can use for loops inside parfor.
N = 4;
M = 3;
out = zeros(N, M);
parfor idx = 1:N
% inner for-loop over the columns of `out`
for jdx = 1:M
out(idx,jdx) = 1000 * idx + jdx;
end
end
disp(out)
1001 1002 1003 2001 2002 2003 3001 3002 3003 4001 4002 4003
Joerg Pfannmoeller on 14 Mar 2022
Thanks. It works now.

R2021b

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