Converting a cell type data to a 4d format

Hi,
I struck into a phase where my data came into one format and I need to reshape it to other format. I have tried reshape and permute, yet, I did not achieve the ultimate result. Would you please help me to get the result. And I would really appreciate if you would please explain your solution.
Input:
A={5 x 1} cell type data where every cell is in this format {300 x 18 single}
Expected Output:
I expect to have an output where each element of A will be formatted as 18 x 300 x 1 x 1 structure.
I have tried to implement the following code to format A but I did not achieve the expected result.
z = cellfun(@(X) permute(X,[3 2 1]),A,'UniformOutput',false);
I am looking for your advice in this regard.
Thanks

3 comentarios

madhan ravi
madhan ravi el 26 de Jun. de 2019
Bose I would expect a 300 X 18 X 1 X 5 matrix , please illustrate.
Saugata Bose
Saugata Bose el 26 de Jun. de 2019
@madhan: Hi. I would expect each cell from A(of size 300 x 18) will turn into a 18 x 300 x 1 x 1, 18 x 300 x 1 x 2, 18 x 300 x 1 x 3, 18 x 300 x 1 x 4 and 18 x 300 x 1 x 5 format. I am not sure though, whether it is possible or not. what you think?
per isakson
per isakson el 26 de Jun. de 2019
What's the meaning of the colons in A(1,:1,:1,:1) ?

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 Respuesta aceptada

madhan ravi
madhan ravi el 26 de Jun. de 2019
Try this:
AA=cellfun(@(x) x.',A,'un',0);
Wanted = cat(4,AA{:})

3 comentarios

@Madhan. thank you very much for your solution. But I expect the output would be like this:
A(1,:1,:1,:1)=0.5,
A(1,:2,:1,:1)=0.7
....
A(1,:300,:1,:1)=0.8
...
A(18,:300,:1,:1)=0.99.
...
...
...
A(18,:300,:1,:5)=0.89, (here, the values are considered at random
But the solutions does not appear like this. May be it was my mistake not elaborating the idea properly. What kind of change does this code require then?(For you and others I edit the question with the expected outputs)
thanks,
madhan ravi
madhan ravi el 26 de Jun. de 2019
Editada: madhan ravi el 26 de Jun. de 2019
Bose keep the facts straight the latter comment you made seems to contradict the original question before you edited.
:1 is not a valid MATLAB syntax
Saugata Bose
Saugata Bose el 26 de Jun. de 2019
@Madhan. I absolutely agree. this is my wrong understanding. So to access each of the 4 d element seperately I need to acces by A(:,:,:,1). Thanks for such a solution. I'll edit my post to undo my wrong output expectation.

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Más respuestas (1)

KSSV
KSSV el 26 de Jun. de 2019
% CReate random input for demo
A = cell(5,1) ;
for i = 1:5
A{i} = rand(300,18) ;
end
B = zeros(18,300,5) ;
for i = 1:5
B(:,:,i) = A{i}' ;
end

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el 25 de Jun. de 2019

Editada:

el 26 de Jun. de 2019

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