Question

见https://stackoverflow.com/questions/41810306/appointment-scheduling ....

Answer 1

你的例子不起作用。您正在为第二个元素编制索引两次（顺便说一句，floor (rand (n) * x)的一个不错的替代方法是使用randi()）：

octave> M = randi(10, 3)
M =

   9   2   5
   9   3   1
   2   8   7

octave> v = [2;2];
octave> M(v)
ans =

   9
   9

octave> M([2;2])
ans =

   9
   9

执行所需操作的正确方法是使用适用于任意维度的sub2ind()。

octave> M(sub2ind (size (M), 2, 2))
ans =  3
octave> M = randi (10, 3, 3, 3)
M =

ans(:,:,1) =

    6    3   10
    1    7    9
    7    6    8

ans(:,:,2) =

    7    9   10
    9    4    5
    8    5    5

ans(:,:,3) =

    3    5   10
    8    3   10
    4    9    4

octave> M(sub2ind (size (M), 1, 2, 3))
ans =  5

Answer 2

我编辑了sub2ind函数，因此可以使用向量。

的工作原理如下：

M(sub2ind2(dims, V));

我可能会在接下来的几天发送修改后的sub2ind2函数。

[编辑]

function ind = sub2ind2 (dims, varargin)    
  if (nargin > 1)
    if (isvector (dims) && all (round (dims) == dims))
      nd = length (dims);
      v = varargin{1};
      vlen = length (v)
      dims(vlen) = prod (dims(vlen:nd));
      dims(vlen+1:nd) = [];
      scale = cumprod (dims(:));
      for i = 1:vlen
    arg = v(i);
    if (isnumeric (arg) && isequal (round (arg), arg))
      if (i == 1)
        if (all (arg(:) > 0 & arg(:) <= dims(i)))
          ind = first_arg = arg;
        else
          error ("sub2ind: index out of range");
        endif
      else
        if (size_equal (first_arg, arg))
          if ((i > nd && arg == 1) || all (arg(:) > 0 & arg(:) <= dims(i)))
        ind += scale(i-1) * (arg - 1);
          else
        error ("sub2ind: index out of range");
          endif
        else
          error ("sub2ind: all index arguments must be the same size");
        endif
      endif
    else
      error ("sub2ind: expecting integer-valued index arguments");
    endif
      endfor
    else
      error ("sub2ind: expecting dims to be an integer vector");
    endif
  else
    print_usage ();
  endif


endfunction

矢量作为矩阵坐标

2 个答案: