How to use vectors instead of loops?

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Sherwin 2020-8-23

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评论： Stephen23 2020-8-28

Hi, I am new to coding and I seem to be using too many loops/cells which significantly slows down my codes.

I want to define a 200x1 matrix every cell of which is a 8X17 matrix and the cells of this one are 1x3 vectors. The values in the 1x3 vectors are generated randomly (from different sets of data: Hmin, Hmax, and Wmax are 8x17 matrices of data which have been imported as cell arrays. They are simple numbers which provide a range for the randomly generated values in 1x3 vectors.).

I want to run a genetic algorithm and this would define 200 members of my initial populations. I have tried this in MATLAB using cell arrays; but they significantly slow down my code. This is what I have tried in MATLAB.

I would be grateful if someone would help me to avoid this.

P = cell (200, 1);
for i = 1:200
    P{i,1} = cell(8, 17);
    for j = 1:8
        for k = 1:17
            P{i,1}{j,k} = zeros(1,3);
            P{i,1}{j,k}(1,1) = randi(Hmax{j}(k));
            P{i,1}{j,k}(1,2) = randi([Hmin{j}(k),Hmax{j}(k)]);
            P{i,1}{j,k}(1,3) = randi(Wmax{j}(k));
        end
    end
end

Thank you so much.

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David Goodmanson 2020-8-24

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Hi Sherwin,

You have four indices all together: two for the 8x17 matrix, one for the 3 components of the vector, and one for the 200 members. You can store all of this stuff in a four-dimensional array. There are several ways to do that, but suppose the 8x17 matrix is seen as basic. Then a 4d array can be initialized as

P = zeros(8,17,3,200).

After the array is filled, then x = P(:,:,2,123) is an 8x17 matrix containing the second vector component, for member 123. x = P(:,:,:,123) is an 8x17x3 array that contains all the information about member 123. If you display it, it can be seen as three 8x17 matrices stacked in layers, each layer containing one of the components of v. P(5,9,1,:) is a vector that contains the (5,9) element of the matrix for vector component 1, listed out for all 200 members. And so on.

A slight complication is that operations like this give an object that is still considered to be a 4d array, but some of those four dimensions often have only one component. For example,

size(P(:,:,2,123))
ans = 8  17  1   1

You can use the squeeze operator to reduce the dimensionality.

m = squeeze(P(:,:,2,123))

results in a 2d matrix, size 8x17.

Sherwin 2020-8-28

Thank you so much.

Stephen23 2020-8-28

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