In parfor , why a variable cannot be classified?
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sm
le 2 Juil 2015
Réponse apportée : Walter Roberson
le 3 Juil 2015
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 commentaires
Walter Roberson
le 2 Juil 2015
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.
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Walter Roberson
le 3 Juil 2015
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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Brendan Hamm
le 2 Juil 2015
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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