In parfor , why a variable cannot be classified?

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sm 2015-7-2

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回答： Walter Roberson 2015-7-3

采纳的回答： Walter Roberson

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In the code below I get an error

Error: The variable Matrix in a parfor cannot be classified.See Parallel for Loops in MATLAB, "Overview".

    parfor i=1:n
    try
        query = 'SELECT * FROM TABLE WHERE Criterion = '' criteria(i) ''';
        curs = exec(conn, query);
        curs = fetch(curs);
        close(curs);
        results = curs.Data;
        A = f(results);
        B = g(results);
        C = h(results);
        D = k(results);
        Matrix(i,6:9) = horzcat(A,B,C,B);
    catch
        Matrix(i,6:9) = {0,0,0,0};
    end
end

Any ideas what I am missing?

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Walter Roberson 2015-7-2

Do A, B, C, D come out as cell arrays? At the moment it looks plausible that you have storing a numeric vector in one case and a vector of cells in the other case.

sm 2015-7-2

Yes, A,B,C and D are all cells. Also, note that the parfor loop runs fine as a simple for loop.

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Answer 1

Walter Roberson 2015-7-3

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write to

tMatrix(i,:) = horzcat(A,B,C,B);

or write {0,0,0,0} there. Then after the parfor, let

Matrix(:,6:9) = tMatrix;

The difficulty you are running into is that you are trying to update only part of one row of a matrix, which requires that the matrix be input to the parfor as well as output from the parfor. parfor can handle rows that are only input and rows that are only output, but it doesn't like input-update-output as your code is requiring. The adjusted version I show has the entire row of tMatrix be output, and that is fine with parfor.

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Answer 2

Brendan Hamm 2015-7-2

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Preallocate your matrix before the loop:

Matrix = nan(n,9)

Although lookingat your catch statement is Matrix a cell array? That is confusing.

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sm 2015-7-2

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I do pre-allocate the cell array before the parfor loop

 n = size(data,1); 
Matrix = cell(n,14);
Matrix(:,1:5) = data;
parfor i=1:n
 ....
end

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