如何通过考虑speciall另一个矩阵输入值到矩阵

Question

一种可能的解决方案如下。请注意，我自由地为变量指定了有意义的名称，因为名称如A和Start Time A（它甚至不是有效的Matlab标识符）很容易混淆。你也可以看到你的矩阵 C和Start Time C是多余的，因为所有的信息已经编码在A, B和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 

的result是

result = 

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

的要求。

如何修改上述解决3D矩阵的代码。

例如：第三个维度的尺寸为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] 

如何使代码以使结果。谢谢

Answer 1

这应该给你一个如何解决这个问题的想法，如果它没有完全回答它。我还没有测试过我在这里提供的代码，只是简单地修改你所提供的内容，尽我所知。

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