Is there any better alternative of four nested loops?

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Sheet 2023-2-7

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https://ww2.mathworks.cn/matlabcentral/answers/1908160-is-there-any-better-alternative-of-four-nested-loops

编辑： DGM 2023-2-8

Hi,

I have certain matrices e1, e2, e3 & e4 corresponding to u1, u2, u3 & u4 of sizes h1, h2, h3, & h4 respectively. I have to search for

(a row of e1)+(a row of e2)+ (a row of e3) = -(a row of e4)

and report corresponding rows of u1, u2, u3 & u4 with name A, B, C & D. Here is my code:

for i1=1:h1(1,1)

for j1=1:h2(1,1)

for k1=1:h3(1,1)

R=[e1(i1,:)+e2(j1,:)+e3(k1,:), i1, j1, k1];

for z=1:h4(1,1)

if e4(z,:)==-R(1:(length(R)-3))

A=u1(R(1,length(R)-2),:)

B=u2(R(1,length(R)-1),:)

C=u3(R(1,length(R)),:)

D=u4(z,:)

end

Is there any short cut? Please help me.

Thank you in advance.

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Sheet 2023-2-8

Thank you.

Can we implement genetic algorithm or any other search method into it? How?

DGM 2023-2-8

编辑：DGM 2023-2-8

I don't know. I wouldn't be the one to ask about GA stuff.

If the values in the arrays have some sort of order to them, there might be simplifications.

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Answer 1

Constantino Carlos Reyes-Aldasoro 2023-2-7

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Hi

In Matlab, if you think in matrices you can avoid many many many loops. For example, if you want to find if a column of a certain matrix (call that a) is equal to a different matrix (call that b), you can do that directly in a single command:

a = [1 2 3 4; 2 3 4 5; 3 4 5 6; 1 2 3 9;0 1 0 1]
a = 5×4
     1     2     3     4
     2     3     4     5
     3     4     5     6
     1     2     3     9
     0     1     0     1
b = [2 3 4 2 1]';
a==repmat(b,[1 4])
ans = 5×4 logical array
   0   1   0   0
   0   1   0   0
   0   1   0   0
   0   1   0   0
   0   1   0   1

So, I compared matrix b with all the columns of a, to do that I repeated b (repmat) 4 times. You will see that all the elements of the second column are the same as those elements of b and thus all are 1. In the fourth column, there is just one element that is the same. So, what I want is all the elements to be the same, so I will use all, and eventually, what I want is to know which column is the one that matches, so I will use find:

all(a==repmat(b,[1 4]))
ans = 1×4 logical array
   0   1   0   0
find(all(a==repmat(b,[1 4])))
ans = 2

So without any loops, I have found which column of a matches b. Use these ideas and you may still need one loop but in general using matrices and matrix operations is much faster and efficient than using loops.