How to slice these variables?
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Joerg Pfannmoeller
on 9 Nov 2021
Commented: Joerg Pfannmoeller
on 14 Mar 2022
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 Comments
Rik
on 9 Nov 2021
If it is merely the detection that is preventing execution, you could consider putting this in a separate function.
Also, shouldn't you be using Zcoor in your third call?
Accepted Answer
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
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)
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)
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