Store mxn matrix within a for loop

Question

Jake 2022-7-19

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1763435-store-mxn-matrix-within-a-for-loop

编辑： Rik 2022-7-19

How can I store results of a for loop when one iteration gives m x n matrix and not a 1 x n matrix?

a = rand(20,12);
for k=length(a)
    store_this = reshape((a(k,:)), 3, [])';
end

**EDIT

The sample code I wanted to share is as follows, and the one above is wrong as pointed out :)

a = rand(20,12);
for k=1:size(a,1)
    store_this = reshape((a(k,:)), 3, [])';
end

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Rik 2022-7-19

Your loop is confusing.

Did you intend to have a single iteration?

Did you intend to use max(size(a))? Judging from your indexing, you may have wanted size(a,1) instead.

If you would have multiple iterations, that would mean your array gets overwritten every time. Is that intentional?

Jake 2022-7-19

You're right! The sample code I've shared was wrong. I've corrected it in the question too.

As you see, each iteration creates 4 x 3 matrix under store_this. What I'm trying is to store those 4 x 3 matrices in each iteration one after the other. So for instance, store_this would look like a 8 x 3 matrix after its second iteration. I'm not sure if that's feasible, or detailed enough (?).

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Answer 1

Rik 2022-7-19

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1763435-store-mxn-matrix-within-a-for-loop#answer_1010690

编辑：Rik 2022-7-19

在 MATLAB Online 中打开

I'm not sure this is the optimal way to do it, but this does what you explain by using a cell array as an intermediary step.

a = rand(20,12);
store_this=cell(1,size(a,1));
for k=1:size(a,1)
    store_this{k} = reshape((a(k,:)), 3, [])';
end
store_this=vertcat(store_this{:})
store_this = 80×3
    0.2288    0.3095    0.1840
    0.9412    0.6852    0.0284
    0.5210    0.5932    0.5208
    0.1174    0.2460    0.1937
    0.3876    0.2077    0.4256
    0.7620    0.6441    0.3932
    0.8775    0.5043    0.9992
    0.8888    0.4054    0.3534
    0.4653    0.9296    0.3296
    0.1698    0.7805    0.6041

In this case your entire loop can be replaced by a single line:

x=reshape(a.',3,[]).';
isequal(x,store_this),store_this
ans = logical
   1
store_this = 80×3
    0.2288    0.3095    0.1840
    0.9412    0.6852    0.0284
    0.5210    0.5932    0.5208
    0.1174    0.2460    0.1937
    0.3876    0.2077    0.4256
    0.7620    0.6441    0.3932
    0.8775    0.5043    0.9992
    0.8888    0.4054    0.3534
    0.4653    0.9296    0.3296
    0.1698    0.7805    0.6041