将单元格数组的单元格数组转换为矩阵矩阵 [英] Convert cell array of cell arrays to matrix of matrices

问题描述

我可以将矩阵的单元格数组转换为矩阵:

I can convert a cell array of matrices to matrix:

>> C={[1,1]; [2,2]; [3,3]};
>> cell2mat(C)
ans =
     1     1
     2     2
     3     3

这可以.但是，我想将包含其他单元格数组的单元格数组转换为矩阵:

This is OK. But, I want to convert a cell array including other cell arrays to a matrix:

>> C={{1,1}; {2,2}; {3,3}};    
>> cell2mat(C)
Error using cell2mat (line 53)
Cannot support cell arrays containing cell arrays or objects.

因此，所需的输出是:

>> mycell2mat({{1,1}; {2,2}; {3,3}})
ans =
     1     1
     2     2
     3     3

该怎么做?

我也想对多维对象做同样的事情:

I want to do same thing for multidimensional ones also:

>> mycell2mat({{1,1;1,1}; {2,2;2,2}; {3,3;3,3}})
ans(:,:,1) =

     1     1
     1     1

ans(:,:,2) =

     2     2
     2     2

ans(:,:,3) =

     3     3
     3     3

推荐答案

老实说，我从不喜欢cell2mat速度慢，所以我提出了使用

To be honest, I never liked cell2mat for being slow, so I've come up an alternative solution using comma-separated lists instead!

这非常简单，只需使用冒号运算符和垂直合并所有矢量:

This is fairly simple, just use the colon operator and concatenate all vectors vertically:

C = {[1,1]; [2,2]; [3,3]};
A = vertcat(C{:})

所以我们得到:

A =
    1   1
    2   2
    3   3

转换单元格数组的单元格

这有点棘手.由于它是单元格数组的单元格数组，因此我们必须通过两次使用冒号和 reshape 放入所需的矩阵.

Transform a cell array of cell arrays

This is a bit trickier. Since it's a cell array of cell arrays, we'll have to obtain a vector of all elements by a double use of the colon and horzcat, and then reshape it into the desired matrix.

C = {{1,1}; {2,2}; {3,3}};
V = [size(C{1}), 1]; V(find(V == 1, 1)) = numel(C);
A = reshape([horzcat(C{:}){:}], V)

所以我们得到:

A =
    1   1
    2   2
    3   3

V的操作可确保正确调整A的形状，而不必显式指定输出尺寸(不幸的是，我没有为此找到一个衬板).这也适用于多维单元阵列:

The manipulation of V makes sure that A is reshaped correctly without having to specify the output dimensions explicitly (unfortunately, I didn't find a one liner for this). This also works for multi-dimensional cell arrays as well:

C = {{1, 1; 1, 1}; {2, 2; 2, 2}; {3, 3; 3, 3}};
V = [size(C{1}), 1]; V(find(V == 1, 1)) = numel(C);
A = reshape([horzcat(C{:}){:}], V)

A(:,:,1) = 
    1   1
    1   1

A(:,:,2) =   
    2   2
    2   2

A(:,:,3) =    
    3   3
    3   3

PS

我认为最后一个示例的正确结果应该是6×2矩阵，而不是2×2×3.但是，这不是您所要求的，因此是不合时宜的.

P.S

I think the correct result for the last example should be a 6-by-2 matrix instead of a 2-by-2-by-3. However, that is not what you requested, so it's off-topic.

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