# How to assign values to arrays inside the PARFOR loop in Parallel Computing Toolbox?

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Yanda CHEN on 20 Sep 2023
Commented: Edric Ellis on 21 Sep 2023
I tried to assign valued to a matrix at specified locations at each loop, but it did not work. It showed that the variable 'A' cannot be classified in parfor-loop. Maybe the problem came from the sliced variable 'b(:,i)'. But I do not know how to modify it.
A = rand(10, 10);
%b = (2:4);
b = zeros(3,4)
b(:,1) = [1 2 3];
b(:,2) = [2 3 4];
b(:,3) = [3 4 5];
b(:,4) = [4 5 6];
parfor i = 1:4
A(b(:,i), i) = ones(3, 1);
end

Edric Ellis on 20 Sep 2023
To assign into A here, you need to follow the rules of sliced variables in parfor. In this case, you need to modify your code to assign to a whole row of column of A each time round the loop, like this:
A = rand(10, 10);
b = zeros(3,4)
b = 3×4
0 0 0 0 0 0 0 0 0 0 0 0
b(:,1) = [1 2 3];
b(:,2) = [2 3 4];
b(:,3) = [3 4 5];
b(:,4) = [4 5 6];
parfor i = 1:4
tmpColumn = A(:, i);
tmpColumn(b(:,i)) = ones(3, 1);
A(:, i) = tmpColumn;
end
Starting parallel pool (parpool) using the 'Processes' profile ... Parallel pool using the 'Processes' profile is shutting down.
disp(A)
1.0000 0.9683 0.1756 0.8298 0.3531 0.3313 0.3907 0.9286 0.0291 0.0008 1.0000 1.0000 0.3131 0.2572 0.3250 0.9650 0.7758 0.3500 0.0066 0.4571 1.0000 1.0000 1.0000 0.3183 0.4465 0.7105 0.6725 0.6469 0.6300 0.8838 0.8736 1.0000 1.0000 1.0000 0.8056 0.5376 0.0758 0.4828 0.6726 0.4277 0.7046 0.2091 1.0000 1.0000 0.4532 0.8524 0.7978 0.7761 0.6336 0.8924 0.7084 0.9754 0.0910 1.0000 0.3777 0.5662 0.7766 0.8363 0.1298 0.4706 0.8724 0.1200 0.6708 0.3640 0.6861 0.1135 0.0157 0.7365 0.7295 0.9218 0.9134 0.2101 0.6980 0.4086 0.1757 0.7772 0.1264 0.9909 0.0295 0.4074 0.2208 0.4582 0.4297 0.0378 0.1022 0.1332 0.5524 0.5354 0.8923 0.1826 0.9337 0.4130 0.9842 0.6144 0.4617 0.9431 0.6865 0.5676 0.2264 0.4760
Yanda CHEN on 21 Sep 2023
Thank you so much for your reply. What if the values of B changes in each loop, so that A can not be recognized as a matrix:
B0 = rand(12,12);
in = [0 9 13 29 38];
ind = cell(4,1);
ind{1} = [1 2 3];
ind{2} = [4 5];
ind{3} = [6 7 8 9];
ind{4} = [10 11 12];
A = zeros(38,1);
parfor i = 1:size(ind,1)
B = B0(ind{i},ind{i});
A(in(i)+1:in(i+1)) = B(:);
end
Edric Ellis on 21 Sep 2023
In this case, you need to work a bit harder to make a vector fragment of the right (variable) size to append:
B0 = rand(12,12);
in = [0 9 13 29 38];
ind = cell(4,1);
ind{1} = [1 2 3];
ind{2} = [4 5];
ind{3} = [6 7 8 9];
ind{4} = [10 11 12];
A = zeros(0,1);
parfor i = 1:size(ind,1)
B = B0(ind{i},ind{i});
numToAppend = in(i+1) - in(i);
valsToAppend = zeros(numToAppend, 1);
valsToAppend(1:numel(B)) = B(:);
A = [A; valsToAppend];
end