How to slice these variables?

Question

Joerg Pfannmoeller 2021-11-9

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582089-how-to-slice-these-variables

评论： Joerg Pfannmoeller 2022-3-14

采纳的回答： Edric Ellis

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

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Rik 2021-11-9

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?

Joerg Pfannmoeller 2022-3-14

Thank you. There is a type-o in indices_3, as you found.

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Answer 1

Edric Ellis 2021-11-9

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582089-how-to-slice-these-variables#answer_827084

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