Question

取一组随机坐标（x，y，z），它将成为我的3x3x3矩阵的中心（也被视为局部最小值）。我有一个函数J，它接受那些坐标，进行计算并返回一个数字。如果这26个点中的任何一个更小，那将是我下一个矩阵的中心。如果我找不到更小的值，矩阵的半径增加1，我们再次运行循环。我的问题是：如何仅通过＆＃34; shell＆＃34;多维数据集，而不是为之前测试的值调用函数？

我试图在下面说明它（这里是2d，但你明白了）..点是经过测试的值，＆＃34;？＆＃34;是需要计算并与当地最小值进行比较的那些。

这是代码

minim=100;

%%the initial size of the search matrix 2*level +1
level=1;
x=input('Enter the starting coordinate for X : ');
y=input('Enter the starting coordinate for Y : ');
z=input('Enter the starting coordinate for Z : ');

%%The loop
if(level<=10)
for m=x-level:x+level
    for n=y-level:y+level
        for p=z-level:z+level
            A(m,n,p)=J(m,n,p);
            if A(m,n,p)<minim
               minim=A(m,n,p);
               x=m;y=n;z=p;
               level=1;
            else
                level=level+1;

                %<<----shell loop here ---->>

            end

        end
    end
    end
else

%Display global min
display(minim, 'Minim');
%Coordinates of the global min
[r,c,d] = ind2sub(size(A),find(A ==minim));

display(r,'X');
display(c,'Y');
display(d,'Z');
end

Answer 1

您可以使用逻辑索引，但我不确定您是否会通过这样做获得速度。重建min的位置有点尴尬，但是像这样你可以摆脱所有的for循环。

A = rand(7,7,7);
%"shell" mask for extraction
B = ones(5,5,5);
B = padarray(B,[1,1,1]);
B = logical(abs(B-1));

[val, ind] = min(A(B))

%reconstruct location
tmp = zeros(1,sum(B(:)));
tmp(ind) = 1;
C = zeros(size(A));
C(B) = tmp;
[~, ind] = max(C(:));
[r,c,d] = ind2sub(size(A),ind);

Answer 2

这是一种可以在单个循环中访问所有“shell”元素的方法：

clear;
%// a cube matrix to play with
A=nan(5,5,5);

n=length( A(:,1,1) ); %// Assuming cube matrix
%// lets change all ot the "shell" elements to 1
for i=1:n
        % 1st and nth level
        A(1,i,1)=1;
        A(i,1,1)=1;
        A(n,i,1)=1;
        A(i,n,1)=1;
        A(1,i,n)=1;
        A(i,1,n)=1;
        A(n,i,n)=1;
        A(i,n,n)=1;

        % 2nd to (n-1)th level
        A(1,1,i)=1;
        A(1,n,i)=1;
        A(n,1,i)=1;
        A(n,n,i)=1;
end

请注意，角落元素不止一次到达。结果矩阵：

>> A
A(:,:,1) =
     1     1     1     1     1
     1   NaN   NaN   NaN     1
     1   NaN   NaN   NaN     1
     1   NaN   NaN   NaN     1
     1     1     1     1     1

A(:,:,2) =   
     1   NaN   NaN   NaN     1
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
     1   NaN   NaN   NaN     1

A(:,:,3) =
     1   NaN   NaN   NaN     1
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
     1   NaN   NaN   NaN     1

A(:,:,4) =   
     1   NaN   NaN   NaN     1
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN
     1   NaN   NaN   NaN     1

A(:,:,5) =   
     1     1     1     1     1
     1   NaN   NaN   NaN     1
     1   NaN   NaN   NaN     1
     1   NaN   NaN   NaN     1
     1     1     1     1     1

Answer 3

这是一个简单的C ++解决方案

这是一个代表2D的5x5x5立方体：

a[i][j][0]
1   1   1   1   1   
1   1   1   1   1   
1   1   1   1   1
1   1   1   1   1
1   1   1   1   1

a[i][j][1]
1   1   1   1   1
1   0   0   0   1   
1   0   0   0   1
1   0   0   0   1
1   1   1   1   1

a[i][j][2]
1   1   1   1   1
1   0   0   0   1   
1   0   0   0   1
1   0   0   0   1
1   1   1   1   1

a[i][j][3]
1   1   1   1   1
1   0   0   0   1   
1   0   0   0   1
1   0   0   0   1
1   1   1   1   1

a[i][j][4]
1   1   1   1   1   
1   1   1   1   1
1   1   1   1   1
1   1   1   1   1
1   1   1   1   1

这是立方体解析的代码：

int a[5][5][5]
int matrix_size = 2*level+1;

for(int z=0;z<matrix_size;z++)

if(z==0 || z= matrix_size-1)
{
    for(int i=0;i<matrix_size-1;i++)
        for(int j=0;j<matrix_size-1;j++)
            {
                //compare minim with  a[i][j][z];
            }
}
else
    for(int i=0;i<matrix_size-1;i++)
        {
            if(i==1 || i==matrix_size-1)
                {   
                    for(int j=0;j<matrix_size-1;j++)
                        //compare minim with  a[i][j][z];
                }
            else
                {
                    //compare minim with  a[i][1][z] and  a[i][matrix_size-1][z];
                }
        }

Matlab For循环仅适用于＆＃34; shell＆＃34;矩阵

3 个答案: