Convert cell arrays to 4-D array

I've got a 35*10 cell array, which contains multiple matrices with the same size(256*256). I want to convert it into a 4-D array(256*256*10*35).

[Stephen23]

Important note to future readers: the accepted answer mixes up the data!

Answer 1

[Matt J]

Where C is your cell array:

M = reshape( cat(3,C{:})  , [256,256,10,35])

EDIT:

Or M = reshape( cat(3,C{:}) , [256,256,35,10]) ?

[Stephen23]

Note that this will not put the cell array 1st dimension along the numeric array third dimension, or the 2nd along the fourth, it will simply follow linear-indexing order, which will result in those two dimensions being mixed up in the final array. A permute or a transpose is required.

[Matt J]

Note that this will not put the cell array 1st dimension along the numeric array third dimension, or the 2nd along the fourth

I think it will -- because of the way cat is applied.

[Stephen23]

"I think it will -- because of the way cat is applied."

Nope, that is incorrect (for the reasons that I already explained in my previous comment). You could concatenate along any dimension >2 and it makes absolutely no difference to the linear-indexing order that your answer uses: this order will always mix up the data. Your answer is flawed without a transpose or a permute (simplest would be a transpose beforehand).

It very easy to demonstrate with a simpler example, which should also make the reason clear. Here I used a 3x5 cell array of 1x1 numeric arrays:

>> C = {11,12,13,14,15;21,22,23,24,25;31,32,33,34,45} % digits= row col
C = 11 12 13 14 15
    21 22 23 24 25
    31 32 33 34 35
>> M = reshape(cat(3,C{:}),[1,1,5,3])
M =
ans(:,:,1,1) =  11
ans(:,:,2,1) =  21
ans(:,:,3,1) =  31
ans(:,:,4,1) =  12
ans(:,:,5,1) =  22
ans(:,:,1,2) =  32
ans(:,:,2,2) =  13
ans(:,:,3,2) =  23
ans(:,:,4,2) =  33
ans(:,:,5,2) =  14
ans(:,:,1,3) =  24
ans(:,:,2,3) =  34
ans(:,:,3,3) =  15
ans(:,:,4,3) =  25
ans(:,:,5,3) =  45

Here we can see for example, that along the 3rd dimesion M(1,1,:,1) (which has size 5) contains the values 11, 21, 31, 12, and 22. Your code has mixed up the elements of those two dimensions, exactly as I said it would: three elements from the first column, two elements from the second column! Lets look along the third dimension:

M(1,1,:,1) = 11,21,31,12,22
M(1,1,:,2) = 32,13,23,33,14
M(1,1,:,3) = 24,34,15,25,45

Try it on any non-square example and you will find that your code always mixes up the data.

Compare to my (correct) answer:

>> M = cell2mat(permute(C,[4,3,2,1]))
M =
ans(:,:,1,1) =  11
ans(:,:,2,1) =  12
ans(:,:,3,1) =  13
ans(:,:,4,1) =  14
ans(:,:,5,1) =  15
ans(:,:,1,2) =  21
ans(:,:,2,2) =  22
ans(:,:,3,2) =  23
ans(:,:,4,2) =  24
ans(:,:,5,2) =  25
ans(:,:,1,3) =  31
ans(:,:,2,3) =  32
ans(:,:,3,3) =  33
ans(:,:,4,3) =  34
ans(:,:,5,3) =  45

Along the 3rd dimension M(1,1,:,1) (which has size 5) are the elements 11, 12, 13, 14, 15, thus keeping exactly the sequence of the orginal cell array's 2nd dimension (which has size 5), even though its dimensions have been swapped. Lets look at all of the 3rd dimension:

M(1,1,:,1) = 11,12,13,14,15
M(1,1,:,2) = 21,22,23,24,25
M(1,1,:,3) = 31,32,33,34,35

The 1st dimension of C has been correctly converted to the 3rd dimension of M.

In case my simple example is too simple for you, here is the complete 35x10 cell array:

>> C = arrayfun(@(n)rand(256,256),nan(35,10),'uni',0);
>> M = reshape(cat(3,C{:}), [256,256,10,35]); % your buggy answer
>> isequal(M(:,:,10,1),C{1,10})               % your buggy answer
ans = 0
>> M = cell2mat(permute(C,[4,3,2,1]));        % my correct answer
>> isequal(M(:,:,10,1),C{1,10})               % my correct answer
ans = 1

The onus of proof lay with me: for that I have provided two examples of why your answer is buggy, I have also explained why it is buggy, and also explained how it can be fixed.

[Matt J]

OK, I think you should allow for the possibility, though, that the OP had a typo and really meant to say that the final dimensions should be [256,256,35,10], since that requires the least data reorganization. In that case, my solution is valid with the following small change.

M = reshape( cat(3,C{:})  , [256,256,35,10])

It is also possible that the OP meant a 3rd alternative,

C=C.';
M = reshape( cat(3,C{:})  , [256,256,10,35])

Answer 2

[Stephen23]

Where C is your cell array:

M = cell2mat(permute(C,[4,3,2,1]))

[Wenyi Xiao]

Thank you for your help!

[Stephen23]

@Wenyi Xiao: my pleasure.

You should know that the answer you accepted gives the data in an mixed up order.