Question

dt1 dt2 dt3 dt
1   3   6   10
2   4   1   5 
3   6   5   3
4   7   4   1
5   1   2   4
6   2   8
7   8
8
9
10

我有上面的数据，我想在一个矩阵（10 x 4）中组合。最大行数为10.我创建了一个zeros矩阵。但是，我有一个问题，因为数据没有相同的尺寸。如何获得如下输出？我应该对数据进行排序并将缺少的值替换为0？

1   1   1   1  
2   2   2   0  
3   3   0   3   
4   4   4   4   
5   0   5   5  
6   6   6   0   
7   7   0   0
8   8   8   0
9   0   0   0
10  0   0   10

Answer 1

以下是@gnovice对之前类似问题

的回答的修改版本

%# group the variables. If you would be generating them in a loop, you could use the loop
%# to group them, i.e. have something like
%# for i=1:n
%#    dtCell{i} = "function that generates dt_i"
%# end

dtCell = {dt1,dt2,dt3,dt};
nCells = length(dtCell);
maxVal = max(cellfun(@max,dtCell)); %# this way, I don't have to know vector orientation

%# you could replace the loop with calls to cellfun. 
%# While this may make you feel more Matlab-ish, it wouldn't be
%# faster, or more readable
out = zeros(maxVal,nCells);
for iCell = 1:nCells
   idx = dtCell{iCell}; %# this assignment is just for clarity 
   out(idx,iCell) = idx;
end

Answer 2

让我们定义示例数据：

dt1 = randi(10,10,1)-1;
dt2 = randi(10,7,1)-1;
dt3 = randi(10,6,1)-1;
dt4 = randi(10,5,1)-1; %// example data. Column vectors.

一种方法是使用bsxfun创建一个遮罩，然后在遮罩指示的位置填充值：

dt = {dt1, dt2, dt3, dt4}; %// collect data into a cell array
n = cellfun(@numel, dt); %// length of each vector
mask = bsxfun(@le, (1:max(n)).', n); %// create mask
result = zeros(size(mask)); %// initiallize result with zeros
result(mask) = vertcat(dt{:}); %// fill in values.

示例结果（随机数据）：

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

如何将不均匀矩阵组合成单个矩阵？

2 个答案: