如何在MATLAB中将单元阵列中不同大小的矩阵组合成矩阵

5 回答

• 0

  如果有可能在一行或几行中我会感到惊讶 . 你可能不得不自己做一些循环 . 以下是在不兼容的第一个尺寸长度的特定情况下实现的目标：

  A={[1 2 3; 4 5 6; 7 8 9]  [1 2 3; 4 5 6; 7 8 9; 9 9 9]  [1 2 3; 4 5 6; 7 8 9; 4 4 4]}
  
  maxsize = max(cellfun(@(x) size(x, 1), A));
  B = A;
  for k = 1:numel(B)
      if size(B{k}, 1) < maxsize
          tmp = B{k};
          B{k} = zeros(maxsize, size(tmp,1));
          B{k}(1:size(tmp,1),1:size(tmp,2)) = tmp;
      end
  end
  
  B = cat(2, B{:});
  

  现在B是：

  B =
  
       1     2     3     1     2     3     1     2     3
       4     5     6     4     5     6     4     5     6
       7     8     9     7     8     9     7     8     9
       0     0     0     9     9     9     4     4     4
  

  回复于 2024-04-19T21:34:14+08:00
• 0

  即使其他答案都很好，我也想提交我的，使用 cellfun .

  l = max(cellfun(@(x) length(x),A))
  
  B = cell2mat(cellfun(@(x) [x;zeros(l-length(x),3)], A, 'UniformOutput', 0));
  

  回复于 2024-04-19T21:34:14+08:00
• 2

  使用bsxfun的屏蔽功能 -

  %// Convert A to 1D array
  A1d = cellfun(@(x) x(:).',A,'Uni',0) %//'
  
  %// Get dimensions of A cells
  nrows = cellfun('size', A, 1)
  ncols = cellfun('size', A, 2)
  
  %// Create a mask of valid positions in output numeric array, where each of
  %// those numeric values from A would be put
  max_nrows = max(nrows)
  mask = bsxfun(@le,[1:max_nrows]',repelem(nrows,ncols))  %//'
  
  %// Setup output array and put A values into its masked positions
  B = zeros(max_nrows,sum(ncols))
  B(mask) = [A1d{:}]
  

  Sample run

  输入 -

  A={[1 2 3 5 6; 7 8 9 3 8]  [1 2 3; 4 5 6; 7 8 9; 9 9 9]  [1 2 3; 4 5 6; 7 8 9; 4 4 4]}
  

  输出 -

  B =
       1     2     3     5     6     1     2     3     1     2     3
       7     8     9     3     8     4     5     6     4     5     6
       0     0     0     0     0     7     8     9     7     8     9
       0     0     0     0     0     9     9     9     4     4     4
  

  回复于 2024-04-19T21:34:14+08:00
• 0

  我会使用一个很好的for循环来实现它，我认为这非常直观 .

  这是评论的代码：

  clc;clear var
  
  
  A={[1 2 3; 4 5 6; 7 8 9]  [1 2 3; 4 5 6; 7 8 9; 9 9 9]  [1 2 3; 4 5 6; 7 8 9; 4 4 4]};
  
  %// Find the maximum rows and column # to initialize the output array.
  MaxRow = max(cell2mat(cellfun(@(x) size(x,1),A,'Uni',0)));
  SumCol = sum(cell2mat(cellfun(@(x) size(x,2),A,'Uni',0)));
  
  B = zeros(MaxRow,SumCol);
  
  %// Create a counter to keep track of the current columns to fill
  ColumnCounter = 1;
  for k = 1:numel(A)    
      %// Get the # of rows and columns for each cell from A
      NumRows = size(A{k},1);
      NumCols = size(A{k},2);
  
      %// Fill the array
      B(1:NumRows,ColumnCounter:ColumnCounter+NumCols-1) = A{k};
  
      %// Update the counter
      ColumnCounter = ColumnCounter+NumCols;
  end
  disp(B)
  

  输出：

  B =
  
       1     2     3     1     2     3     1     2     3
       4     5     6     4     5     6     4     5     6
       7     8     9     7     8     9     7     8     9
       0     0     0     9     9     9     4     4     4
  

  回复于 2024-04-19T21:34:14+08:00
• 5

  [max_row , max_col] = max( size(A{1}) , size(A{2}) , size(A{3}) );
  A{1}(end:max_row , end:max_col)=0;
  A{2}(end:max_row , end:max_col)=0;
  A{3}(end:max_row , end:max_col)=0;
  B=[A{1} A{2} A{3}];
  

  回复于 2024-04-19T21:34:14+08:00