Converting a cell type data to a 4d format

2019 6 25
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更新时间：2019 6 26
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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

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madhan ravi
madhan ravi 2019-6-26
Bose I would expect a 300 X 18 X 1 X 5 matrix , please illustrate.
Saugata Bose
Saugata Bose 2019-6-26
@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 2019-6-26
What's the meaning of the colons in A(1,:1,:1,:1) ?

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madhan ravi
madhan ravi 2019-6-26

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Try this:
AA=cellfun(@(x) x.',A,'un',0);
Wanted = cat(4,AA{:})

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Saugata Bose
Saugata Bose 2019-6-26
@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 2019-6-26
编辑：madhan ravi 2019-6-26
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 2019-6-26
@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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KSSV
KSSV 2019-6-26

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% 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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