C-order reshape of multi-dimensional array in matlab

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Yijun Li 2019-3-10

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评论： Stephen23 2022-9-7

Hello,

I have a question regarding reshape in matlab, it seems that matlab reshapes N-dimensional array according to Fortran-order, however, I would like a C-order reshape, that is a line-wise reshape.

Suppose now there is a matrix M with 4 rows and 4 columns as follows:

M = [1   2   3   4
 6   7   8
10  11  12
14  15  16]

Then M is going to be reshaped to a 3D array with dimension (2,2,4) shown in the picture below. Does anyone know how to do this in a generalized way (suitable for even higher dimension)?

Thanks!

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Answer 1

Jan 2019-3-10

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编辑：Jan 2019-3-10

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Reshaping in line-wise order is not meaningful, because reshaping means to keep the order of elements in the memory and the order is defined columnwise in Matlab. But you can simply change the order by permute:

x = rand(2,3,4);
y = permute(x, [2, 1, 3:ndims(x)]);
result = reshape(y, [whateverYouWant])

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Yijun Li 2019-4-22

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@Jan

Hello Jan,

Unfortunately your solution fails the mission, maybe I have not explained my problem well, but anyway I have solved the problem in a generalized way with your insight.

Let me explain my solution according to the example specified in my problem:

Suppose again matrix M, which is going to be reshaped to a 3D array with dimension (2,2,4) shown in the picture above.

M = [1   2   3   4
 6   7   8
10  11  12
14  15  16]

As matrix M is arranged in C-order (line-wise), first transpose data to Fortran-order (column-wise).

data = M.'

Then specify the dimension along each axis from low to high (3rd to 1st).

dimension_low2high = [4,2,2]

Next reshape the data in Fortran-order (column-wise) from low-dimension to high dimension.

ndarray = reshape(data,dimension_low2high)

Finally swap the position of low dimension and high dimension to meet C-order.

ndarray = permute(ndarray,[ndims(ndarray):-1:1])

Then have a check whether the new indexing is correct, according to this example, the following code block will show integers from 1 to 16.

for i1=1:2
    for i2=1:2
        for i3=1:4
            ndarray(i1,i2,i3)
        end
    end
end

Steven Lord 2022-9-7

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That's the mapping between the dimensions of your original array and the dimensions of the result.

A = reshape(1:24, [4 3 2]);
szA = size(A)
szA = 1×3
     4     3     2
p = [2 1 3];
B = permute(A, p);
szB = size(B)
szB = 1×3
     3     4     2

A is size 4 in its first dimension. Because the first element of p is 2, B is size 4 in its 2nd dimension.

A is size 3 in its second dimension. Because the second element of p is 1, B is size 3 in its 1st dimension.

I was very careful to distinguish first and second from 1st and 2nd in those descriptions.

Stephen23 2022-9-7

"Therefore I do not understand, why my code "fails the mission"."

I accepted your answer, Jan, because it is clear that permuting the first two dimensions is the general solution.

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