3D arrays manipulation and use of if else statments

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Mantas Vaitonis 2018-6-21

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重新打开： Mantas Vaitonis 2018-6-25

Hello Everyone,

Here is the situation. There are three 3D arrays. S (1x5x1000) that stores elements from 1:5, P (1x5x1000) that stores pairs for S, and idx (1x5x1000) to invoke pair. For example S is always the same and (1x5x1)=1 2 3 4 5, p (1x5x1)=3 1 5 2 3 and idx (1x5x1)=1 0 1 1 1. We have the situation where are two same (5-3 and 3-5) pairs with invoke signal, we need to change invoke signal from 1 to 0 that would look like 1 0 0 1 1, and we need to do this for the whole 3D array. Is there a way to achieve this without for loop, because this part is time sensitive? Simplified code

if idx(i)==1
   if p(i)=s(p(i))
      idx(i)=0
   end
end

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Mantas Vaitonis 2018-6-21

编辑：dpb 2018-6-21

在 MATLAB Online 中打开

Here is a small sample:

S(:,:,1) =1 2 3 4 5  
P(:,:,1) =3 1 1 3 2
idx(:,:,1) =1 0 1 1 0

in this case S(1,1,1) with P(1,1,1) is same pair as S(1,3,1) with P(1,3,1), thus idx should change its invode index from 1 to 0

idx[0 0 1 1 0]

Ex 2

S(:,:,2) =1 2 3 4 5
P(:,:,2) =5 4 3 1 1
idx(:,:,2) =1 0 1 1 0

in this case S(1,3,2) with P(1,3,2) pair can not be pair with it self, thus idx should be change to

idx [1 0 0 1 0]

Ex 3

S(:,:,3) =1 2 3 4 5
P(:,:,3) =4 5 2 1 5
idx(:,:,3) =1 0 0 1 1

in this case S(1,1,3) with P(1,1,3) is same pair as S(1,4,1) with P(1,4,1) and S(1,5,3) with P(1,5,3) can not be paired to it self, thus idx should be

idx[0 0 0 1 0]

Maybe now it is more clear?

Mantas Vaitonis 2018-6-22

编辑：Mantas Vaitonis 2018-6-22

在 MATLAB Online 中打开

If i convert 3D arrays to 2D I can achieve what I want, but it is very slow.

 cc=P(P);
[nn1,nn2]=size(P);
for j=1:nn1
    idx1(j)={find(idx(j,:)>0)};
        tempIdx=ones(size(idx1{j}));
        for i=1:nn2  
            if(i==cc(j,i))
        if any(cc(j,i)==idx1{j})
            if tempIdx(cc(j,i)==idx1{j})==1
            idx1{j}(cc(j,i)==idx1{j})=NaN;
            tempIdx(pairs2(j,i))=0;
            end
        end
            end
        end
end