在MATLAB

Question

查找與矩陣相同的值最接近的元素考慮下面的矩陣：

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

當我指定一個元素的位置，我想獲得具有相同值最接近的元素的位置。例如，如果我選擇第3行第3列中的元素，即'6'，我希望獲得最接近'6'的行和列值，在這種情況下，它位於第2行第4列。同樣，對於第4行第4列中的'1'，最近的是在第4行，第5行和第4行第6列，其中任何一個都很好。我查了'bwdist'和'找到'的功能，但他們沒有給出正確的結果。任何人都可以幫助我解決這個問題？

編輯：

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) 

「最低」功能將不會在這裏工作爲其中所需的元素存在將被轉換到零，並且每個位置「分鐘」掃描矩陣行方向，並給出第一zero.The以下的位置是這樣的情況下：

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 

即使有一個「5」旁邊的「5」排在4，第5欄，「5」在第10行，選擇第2列。

Answer 1

假設A爲輸入2D矩陣，這可能是一種方法 -

%// 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

這種方法不需要任何工具箱。如果不存在具有相同值的其他條目，則返回[]。

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);