Question

考虑以下矩阵：

0 3 0 1 1 4
1 3 5 6 7 0
2 5 6 2 6 1
4 4 2 1 5 1

当我指定元素的位置时，我想获得具有相同值的最近元素的位置。例如，如果我选择第3行第3列中的元素，即＆＃39; 6＆＃ 39;，我想获得最近的＆＃39; 6＆＃39;的行和列值，在这种情况下，它是在第2行第4列。同样，对于＆＃39; 1＆＃ 39;在第4行，第4列，最近的是在第4行，第5行和第4行，第6行，其中任何一个都很好。我已经查找了＆＃39; bwdist＆＃39;和＆＃39;找到＆＃39;功能，但他们没有给出正确的结果。任何人都可以帮助我吗？

修改

a1 = randi(10,10,5); 
disp(a1);
%// For an array of search numbers
search_array = a1(4,5);
disp(search_array);
%%// Find the linear index of the location
[~,ind] = min(abs(bsxfun(@minus,a1(:),search_array')));%//'

%%// Convert the linear index into row and column numbers
[x,y] = ind2sub(size(a1),ind)

＆＃39; min＆＃39;函数在这里不起作用，因为存在所需元素的每个位置将被转换为零并且“＃min;＆＃39; min逐行扫描矩阵并给出第一个零的位置。以下是这种情况：

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

即使有一个＆＃39; 5＆＃39;就在＆＃39; 5＆＃39;旁边在第4行，第5栏，＆＃39; 5＆＃39;在第10行，第2列被选中。

Answer 1

假设A是输入的二维矩阵，这可能是一种方法 -

%// Row and column indices of the "pivot"
row_id = 4;
col_id = 5;

%// Get the linear index from row and column indices
lin_idx = sub2ind(size(A),row_id,col_id)

%// Logical array with ones at places with same values
search_matches = false(size(A));
search_matches(A==A(lin_idx)) = 1;

%// Create a logical array with just a single 1 at the "pivot"
A_pivot = false(size(A));
A_pivot(lin_idx) = 1;

%// Use BWDIST to find out the distances from the pivot to all the places
%// in the 2D matrix. Set the pivot place and places with non-similar
%// values as Inf, so that later on MIN could be used to find the nearest
%// same values location
distmat = bwdist(A_pivot)
distmat(lin_idx) = Inf
distmat(~search_matches)=Inf

[~,min_lin_idx] = min(distmat(:))
[closest_row_idx,closest_col_idx] = ind2sub(size(A),min_lin_idx)

Answer 2

此方法不需要任何工具箱。如果不存在具有相同值的其他条目，则返回[]。

A = [0 3 0 1 1 4
     1 3 5 6 7 0
     2 5 6 2 6 1
     4 4 2 1 5 1];                     %// data matrix
pos_row = 3;                           %// row of reference element
pos_col = 3;                           %// col of reference element

ref = A(pos_row,pos_col);              %// take note of value
A(pos_row,pos_col) = NaN;              %// remove it, to avoid finding it as closest
[ii, jj] = find(A==ref);               %// find all entries with the same value
A(pos_row,pos_col) = ref;              %// restore value
d = (ii-pos_row).^2+ (jj-pos_col).^2;  %// compute distances
[~, ind] = min(d);                     %// find arg min of distances
result_row = ii(ind);                  %// index with that to obtain result
result_col = jj(ind);

在matlab中找到矩阵中具有相同值的最接近元素

2 个答案: