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Combine cell array with contant matrices

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Ming
Ming 2016 年 7 月 17 日
コメント済み: Matt J 2016 年 7 月 18 日
Hi everyone,
Suppose I have a cell array A{1} A{2} A{3} ... A{n}, can I create a new cell array A_{1} A_{2} A_{3} ... A_{n}, such that A_{1}=[0 0 0;0 A{1} 0;0 0 0], A_{2}=[0 0 0;0 A{2} 0;0 0 0], A_{n}=[0 0 0;0 A{n} 0;0 0 0], without loop?
Is there any efficient way to perform this operation?
P.S. A{1:n} are all same size matrices, 0 represents a zero matrix.
thanks

回答 (2 件)

Matt J
Matt J 2016 年 7 月 17 日
編集済み: Matt J 2016 年 7 月 18 日
No, there is no way to avoid loops when working with cell arrays (including those hidden in cellfun). However, if you are storing things in cell arrays, it probably means your data isn't large enough to make efficiency a very big concern.
If A{1:n} are all same size matrices, it would have served you better to store them as a 3D array,
A3d=cat(3,A{:});
You could then create a similar, zero-padded 3d array by doing,
A3d_=zeros(...);
A3d_(i:j,m:n,:)=A3d;

Andrei Bobrov
Andrei Bobrov 2016 年 7 月 18 日
A1 = cat,3(A{:});
A2 = zeros(size(A1)+[2,2,0]);
A2(2:end-1,2:end-1,:) = A1;
A_ = reshape(num2cell(A2,[1,2]),[],1);
  1 件のコメント
Matt J
Matt J 2016 年 7 月 18 日
num2cell has hidden for loops in it...

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