Subtracting multiple vectors from each row of an a

问题:

I have a data set, X that is m x 2, and three vectors stored in a matrix C = [c1'; c2'; c3'] that is 3 x 2. I am trying to vectorize my code that finds, for each data point in X, which vector in C is closest (squared distance). I would like to subtract each vector (row) in C from each vector (row) in X, resulting in an m x 6 or 3m x 2 matrix of differences between the elements of X and the elements of C. My current implementation does this one row in X at a time:

for i = 1:size(X, 1)
    diffs = bsxfun(@minus, X(i,:), C);    % gives a 3 x 2 matrix result
    [~, idx(i)] = min(sumsq(diffs), 2);   % returns the index of the closest vector
                                          % in C to the ith vector in X
end

I want to get rid of this for loop and just vectorize the whole thing, but bsxfun(@minus, X, C) gives me a an error in Octave:

error: bsxfun: nonconformant dimensions: 300x2 and 3x2

Any ideas how I can "super-broadcast" my subtraction operation between these two matrices?

回答1:

The core of this problem is to compute a distance matrix D of size m x 3 that contains the pairwise distances between all data points in X and all data points in C. The Euclidean distance between the i-th vector x_i in X and the j-th vector c_j in C can be rewritten as:

|x_i-c_j|^2 = |x_i|^2 - 2<x_i, c_j> + |c_j|^2

where <,> refers to inner product. The right-hand side of this equation can be easily vectorized, because the inner product of all pairs is just X * C' which is BLAS3 operation. This way of computing the distance matrix is known as dist2 function in the book Pattern Recognition and Machine Learning by Christopher Bishop. I copy the function below with a little modification.

function D = dist2(X, C)        
    tempx = full(sum(X.^2, 2));
    tempc = full(sum(C.^2, 2).');
    D = -2*(X * C.'）;
    D = bsxfun(@plus, D, tempx);
    D = bsxfun(@plus, D, tempc);

The full here is used in case X or C is a sparse matrix.

Note: The distance matrix D computed this way might have tiny negative entries due to numerical rounding error. To guard against this case, use

D = max(D, 0);

The indices of the closest vector in C can be retrieved from D:

[~, idx] = min(D, [], 2);

回答2:

If you have the statistics toolbox, you can use pdist2:

PDIST2 Pairwise distance between two sets of observations. D = PDIST2(X,Y) returns a matrix D containing the Euclidean distances between each pair of observations in the MX-by-N data matrix X and MY-by-N data matrix Y.

So in your case,

[~, which_C] = min(pdist2(X,C), [], 2);

is what you're looking for.

Alternatively, you could use this beauty:

[~, which_c] = min(sum(bsxfun(@minus, X, permute(C, [3 2 1])).^2, 2), [], 3);

which wouldn't win any prizes for readability, robustness or manageability, but you will gain some speed (and the need for a toolbox, mind you :)