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一种可能的解决方案如下。请注意，我自由地为您的变量分配有意义的名称，因为诸如A和Start Time A（甚至不是有效的 Matlab 标识符）之类的名称非常容易混淆。您还可以看到您的矩阵 C和Start Time C是多余的，因为所有信息都已经编码在A和 B中Start Time A。

% The values to put in the result matrix.
value = [5 6 7;
         7 5 6];
% Column index where each sequence starts in the result matrix.
start = [2 3 7;
         1 6 8];
% The length of each sequence, i.e. how often to put the value into the result.
count = [1 2 3;
         3 1 2];

% Determine the longest row. Note: At this place you could also check, if all 
% rows are of the same length. The current implementation pads shorter rows with
% zeros.
max_row_length = max(start(:, end) + count(:, end) - 1);

% Allocate an output matrix filled with zeros. This avoids inserting sequences
% of zeros afterwards.
result = zeros(size(start, 1), max_row_length);

% Finally fill the matrix using a double loop.
for row = 1 : size(start, 1)
    for column = 1 : size(start, 2)
        s = start(row, column);
        c = count(row, column);
        v = value(row, column);
        result(row, s : s + c - 1) = v;
    end
end

result是_

result =

     0     5     6     6     0     0     7     7     7
     7     7     7     0     0     5     0     6     6

按照要求。

如何修改上述代码以解决 3D 矩阵。

示例：第三维的大小为 2。矩阵值

value(:,:,1) = [5 6 7;
                7 5 6];
value(:,:,2) = [6 5 7;
                6 7 5];

start(:,:,1) = [2 3 7;
                1 6 8];
start(:,:,2) = [1 5 6;
                2 5 9];

count(:,:,1) = [1 2 3;
                3 1 2];
count(:,:,2) = [2 1 3;
                2 3 1];

我希望我的结果矩阵是

result(:,:,1) =[0 5 6 6 0 0 7 7 7;
                7 7 7 0 0 5 0 6 6]
result(:,:,2) =[6 6 0 0 5 7 7 7 0;
                0 6 6 0 7 7 7 0 5]

如何制作代码以产生结果。谢谢

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1 回答 1

0

如果问题没有完全回答，这应该让您了解如何解决问题。我没有测试我在这里输入的代码，只是据我所知简单地修改了你提供的内容。

result = zeros(size(start, 1), max_row_length, size(start,3)); 

% Finally fill the matrix using a double loop. 
for depth = 1:size(start,3)
    for row = 1 : size(start, 1) 
        for column = 1 : size(start, 2) 
            s = start(row, column, depth); 
            c = count(row, column, depth); 
            v = value(row, column, depth); 
            result(row, s : s + c - 1, depth) = v; 
        end 
    end 
end
于 2012-07-26T14:55:22.067 回答