In parfor , why a variable cannot be classified?
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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?
2 件のコメント
Walter Roberson
2015 年 7 月 2 日
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.
採用された回答
Walter Roberson
2015 年 7 月 3 日
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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その他の回答 (1 件)
Brendan Hamm
2015 年 7 月 2 日
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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