Creating submatrix with variable indexing

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James Lee
James Lee 2020-2-21
评论： Rik 2020-2-26
I have a larger multidimensional array of data that I want to grab a section of, but the data I'm interested in extracting starts at a different index depending on the vector. I know the values the starting index and ending index of the data i want to extract, but I'm not sure how to get matlab to do create this new matrix (the data is of a fixed length). Due to the data size I do not want to use nested for loops, which would have made this easy.
Thanks!
(edit, one approach that hasn't worked is the creation of a 1s and 0s matrix multiplied by the original data set and then removing the zeros. The data of interest contains zeros, so using a nonzeros approach led to problems).
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James Lee
James Lee 2020-2-21
I have (for simplicity) an (a x b x c) array. For simplicity, the data I want to get is 1000 data points long, but it starts at a different place in each vector. For example, it might be (1,1,1:1000), (1,2,301:1300), (2,1,56:1055), etc. I want to create a new (a x b x c) array, where each c vector is taken from the original matrix but is only of length 1000 of the data I want. I have an (a x b) matrix that gives the initial index of the data I wish to select.
Turlough Hughes
Turlough Hughes 2020-2-22
You're potentially very close to getting this. If your logical matrix is an a x b x c array with true values at indices you wish to extract then instead of multiplying the logical matrix you should use the logical matrix as the index itself. Look up logical indexing in matlab. .

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回答（1 个）

Rik
Rik 2020-2-22
The code below isn't elegant, but it should do what you want. The example is for just 2x2xN. Despite the name, the slow method is faster for larger matrices, because the JIT can optimize a lot there.
data=rand(2,2,1500);
inds=randi(size(data,3)-1000,size(data,1),size(data,2));
[X,Y,Z]=ndgrid(1:size(data,1),1:size(data,2),1:1000);
Z=Z+inds;
ind=sub2ind(size(data),X,Y,Z);
new_fast=data(ind);
new_slow=zeros(size(X));
for x=1:size(data,1)
for y=1:size(data,2)
new_slow(x,y,:)=data(x,y,inds(x,y)+(1:1000));
end
end
%test if the results are the same, should return 0 (or close to it)
max(abs(new_fast(:)-new_slow(:)))
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Rik
Rik 2020-2-26
Did my answer solve your question? If not, feel free to comment with your remaining issues.

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