Fastest way to find the values and indices of the entries of a vector X that are closest to each entry of a matrix A.

Basically wondering if there is a faster way to do something like this:

X = [0:.05:1]; % the vector
A = rand(100); % the matrix
result_val = zeros(100);
result_idx = zeros(100);
for i = 1:100
    for j = 1:100
        [result_val(i,j), result_idx(i,j)] = min( abs(A(i,j) - X) );
    end
end

Answer 1

[Githin George]

You can vectorize the operation as shown below:

X = 0:0.05:1; % the vector
A = rand(5000); % the matrix
%% Vectorized Approach
tic
% Reshape X to create 1x1xsize(X) array
X = reshape(X, 1, 1, []);

% Calculate the absolute differences NxNxsize(X)
differences = abs(A - X);

% Find the minimum differences and their indices along dim=3
[result_val, result_idx] = min(differences, [], 3);
toc

Elapsed time is 0.701162 seconds.

%% Non Vectorized Approach
tic
result_val1 = zeros(5000);
result_idx1 = zeros(5000);
for i = 1:5000
    for j = 1:5000
        [result_val1(i,j), result_idx1(i,j)] = min( abs(A(i,j) - X) );
    end
end
toc

Elapsed time is 17.437640 seconds.

%%
disp("isequal(result_val,result_val1) output: "+ isequal(result_val1,result_val))

isequal(result_val,result_val1) output: true

Reshape: https://www.mathworks.com/help/matlab/ref/double.reshape.html#bud6j0x-1-sz1szN

[Kevin]

Nice, thanks!

[Image Analyst]

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Answer 2

[Matt J]

result_idx = reshape( interp1(X,1:numel(X),A(:),'nearest','extrap')  ,size(A));

result_val=abs(X(result_idx)-A);

[Matt J]

Speed comparison:

X = linspace(0,1,500); % the vector
A = rand(1000); % the matrix

%%Using min
tic
% Calculate the absolute differences NxNxsize(X)
differences = abs(A - reshape(X, 1, 1, []));

% Find the minimum differences and their indices along dim=3
[result_val, result_idx] = min(differences, [], 3);
toc

Elapsed time is 0.566279 seconds.

%%Using interp1
tic;
result_idx = reshape( interp1(X,1:numel(X),A(:),'nearest','extrap')  ,size(A));

result_val=abs(X(result_idx)-A);
toc

Elapsed time is 0.017314 seconds.

[Kevin]

Thank you -- also a very interesting solution!