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

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

Answer 1

[Edric Ellis]

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]

Thanks a lot! It works!! I still have another question, if I want to assign values to part of a variable in each parfor loop like this:

B = zeros(4,3);
in = [0 4 8 12];
A = zeros(12,1);
B(:,1) = [1 1 1 1];
B(:,2) = [2 2 2 2];
B(:,3) = [3 3 3 3];

parfor i = 1:size(B,2)
    tmpColumn = A(in(i)+1:in(i+1));
    tmpColumn = B(:,i);
    A(in(i)+1:in(i+1)) = tmpColumn;
end
disp(A)

It shows the same problem as before. How to deal with kind of problem? In my case, the size of the B(:,i) in each loop can be different.

[Edric Ellis]

In general, parfor requires outputs to be either "sliced" variables or "reduction" variables. In this case, what you're trying to do is a bit like a sliced output, but (as you've discovered) the way you're using A is not valid. Basically, to be valid, the indexing has to be similar to A(:,i) or A(i,:) etc. In this case, you're attempting to write to a possibly arbitrary range of values in the vector A. (In this specific case, the range of values you're writing happens to be such that you could get around the problem by treating A as a matrix).

You could fix this by using concatenation, and making A a "reduction" variable:

B = zeros(4,3);
in = [0 4 8 12];
B(:,1) = [1 1 1 1];
B(:,2) = [2 2 2 2];
B(:,3) = [3 3 3 3];
A = zeros(0, 1);
parfor i = 1:size(B,2)
    A = [A; B(:,i)];
end

Starting parallel pool (parpool) using the 'Processes' profile ...
Parallel pool using the 'Processes' profile is shutting down.

disp(A)

     1
     1
     1
     1
     2
     2
     2
     2
     3
     3
     3
     3

[Yanda CHEN]

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

Looking forward to your replay.

[Edric Ellis]

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