Question

如何有效地将单元格数组中的值赋给数值数组？ A和B总是任何大小的方形矩阵。为简单起见，我只分配了小矩阵。如果例如A {1} = [2 3 8];那么我想在B矩阵的第二行和第三列中分配值8。如，

 Input 

  A=[1x3 double]    [1x3 double]    [1x3 double]   [1x3 double]
    [1x3 double]    [1x3 double]    [1x3 double]   [1x3 double]
    [1x3 double]    [1x3 double]    [1x3 double]   [1x3 double]
    [1x3 double]    [1x3 double]    [1x3 double]   [1x3 double]


 B=zeros(4,4);

 A{1}=[2 3 8];
 A{2}=[3 4 7];
 and so on...

 Output
 B(2,3)=8;
 B(3,4)=7;
 and so on...

Answer 1

使用spconvert：

B = full(spconvert(cat(1, A{:})));

Answer 2

你可以不用循环来做到这一点：

B=zeros(4,4);

A{1}=[2 3 8];
A{2}=[3 4 7];

C = cell2mat(A);
C = reshape(C,3,size(C,2)/3)';
indices =  sub2ind(size(B), C(:,1), C(:,2));
B(indices) = C(:,3);

对于您的示例，结果为：

B =

 0     0     0     0
 0     0     8     0
 0     0     0     7
 0     0     0     0

Answer 3

使用sub2ind和indexing

%// creating a sample cell array. Replace this with your actual cell array
A = {[1 2 8] [1 1 4]; [2 1 5] [2 2 1]};

%// Reshaping the cell array to kx3 matrix
Amat = cat(1,A{:});          %// from luis' answer

%// taking first column as row sub, 2nd col as col sub
%// The subs are converted into linear indices to the size of A or B
ind = sub2ind(size(A),Amat(:,1),Amat(:,2));

%// You have done this (helps in fixing the dimensions of B)
B = zeros(size(A));

%// Taking last col of 'Amat' as values and assigning to corresponding indices
B(ind) = Amat(:,3);

<强>结果：

Answer 4

我建议在A单元格上循环一次以找到最大索引值，比方说max_i,max_j。

然后初始化B，如：

B=zeros(max_i,max_j)

然后在A单元格上再次循环，并将值分配给相应的B元素。

编辑：添加了示例代码：

max_i=0
max_j=0
for k=1:size(A,1)
    for l=1:size(A,2)
         max_i=max([A{k,l}(1),max_i])
         max_j=max([A{k,l}(2),max_j])
end

B=zeros(max_i,max_j)

for k=1:size(A,1)
    for l=1:size(A,2)
         B(A{k,l}(1),A{k,l}(2))=A{k,l}(3)
end

将单元格数组中的值分配给数字arrray

4 个答案: