restate in logical indexing

Question

Amy 2017-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/322836-restate-in-logical-indexing

评论： Jan 2017-2-1

采纳的回答： Stephen23

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Dear all,

How could I restate the following code using logical indexing?

for kk=1:rxnumber %columns
    for ll=1:jj %rows
        if A(ll,kk)>0
            C(ll,A(ll,kk))=B(ll,kk); 
        end
    end
end

What does this code do:

If I have matrices

A= [ 0 0 7; 0 5 0; 0 0 0; 0 0 0]; 
B= [ 0 0 0.2; 0 0.3 0; 0 0 0; 0 0 0];

Then I would like to produce the following matrix:

C= [0 0 0 0 0 0 0.2; 0 0 0 0 0.3 0 0; 0 0 0 0 0 0 0; 0 0 0 0 0 0 0];

Or:

C=zeros(4,7);
B(1,3)--> C(1,7).
B(2,2)--> C(2,5).

Why do I find it hard to figure out a logical indexing code myself: if the value in A is zero, then nothing should happen. I don't know how to write a code where the 0 values are ignored.

A solution that is not elegant would be:

C=zeros(4,max(max(A))+1).
A(A==0)=max(max(A))+1.

Followed by:

C(:,A(:,:))=B(:,:).
C(:,8)=[];

Which realizes:

C= zeros(4,8).
A= [ 8 8 7; 8 5 8; 8 8 8; 8 8 8]; 
C= [0 0 0 0 0 0 0.2; 0 0 0 0 0.3 0 0; 0 0 0 0 0 0 0; 0 0 0 0 0 0 0];

How can I make an elegant solution for this?

Thank you in advance!

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Answer 1

Stephen23 2017-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/322836-restate-in-logical-indexing#answer_252893

编辑：Stephen23 2017-2-1

在 MATLAB Online 中打开

One way would be to use sub2ind, which generates linear indices (not logical indices):

>> A = [0,0,7; 0,5,0; 0,0,0; 0,0,0];
>> B = [0,0,0.2; 0,0.3,0; 0,0,0; 0,0,0];
>> X = A>0;
>> [R,~] = find(X);
>> C = zeros(size(B,1),max(A(X)));
>> C(sub2ind(size(C),R,A(X))) = B(X)
C =
 0.00000   0.00000   0.00000   0.00000   0.00000   0.00000   0.20000
 0.00000   0.00000   0.00000   0.00000   0.30000   0.00000   0.00000
 0.00000   0.00000   0.00000   0.00000   0.00000   0.00000   0.00000
 0.00000   0.00000   0.00000   0.00000   0.00000   0.00000   0.00000

If you really want to use logical indices, then you can use the same method to create a logical array:

 L = false(size(B,1),max(A(X)));
 L(sub2ind(size(L),R,A(X))) = true;

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Amy 2017-2-1

Thank you very much!

Usually 'find' is not recommended for computational efficiency, if I recall correctly.

Because of your answer I thought of the possibility of using 'reshape' to compute the linear indices.

I am not sure which method would be most efficient computationally.

Jan 2017-2-1

@Amy: If you can solve a problem using logical indexing, find wastes time usually. But here find is used to obtain the row indices. You can try which version is faster for your data.

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