Question

I’ve created a codebook using k-means of size 4000×300 (4000 centroids, each with 300 features). Using the codebook, I then want to label an input vector (for purposes of binning later on). The input vector is of size Nx300, where N is the total number of input instances I receive.

To compute the labels, I calculate the closest centroid for each of the input vectors. To do so, I compare each input vector against all centroids and pick the centroid with the minimum distance. The label is then just the index of that centroid.

My current Matlab code looks like:

function labels = assign_labels(centroids, X)
labels = zeros(size(X, 1), 1);

% for each X, calculate the distance from each centroid
for i = 1:size(X, 1)
    % distance of X_i from all j centroids is: sum((X_i - centroid_j)^2)
    % note: we leave off the sqrt as an optimization
    distances = sum(bsxfun(@minus, centroids, X(i, :)) .^ 2, 2);
    [value, label] = min(distances);
    labels(i) = label;
end     

However, this code is still fairly slow (for my purposes), and I was hoping there might be a way to optimize the code further.

One obvious issue is that there is a for-loop, which is the bane of good performance on Matlab. I’ve been trying to come up with a way to get rid of it, but with no luck (I looked into using arrayfun in conjunction with bsxfun, but haven’t gotten that to work). Alternatively, if someone know of any other way to speed this up, I would be greatly appreciate it.

Update

After doing some searching, I couldn’t find a great solution using Matlab, so I decided to look at what is used in Python’s scikits.learn package for ‘euclidean_distance’ (shortened):

 XX = sum(X * X, axis=1)[:, newaxis]
 YY = Y.copy()
 YY **= 2
 YY = sum(YY, axis=1)[newaxis, :]
 distances = XX + YY
 distances -= 2 * dot(X, Y.T)
 distances = maximum(distances, 0)

which uses the binomial form of the euclidean distance ((x-y)^2 -> x^2 + y^2 – 2xy), which from what I’ve read usually runs faster. My completely untested Matlab translation is:

 XX = sum(data .* data, 2);
 YY = sum(center .^ 2, 2);
 [val, ~] = max(XX + YY - 2*data*center');

Answer 1

You can vectorize it by converting to cells and using cellfun:

[nRows,nCols]=size(X);
XCell=num2cell(X,2);
dist=reshape(cell2mat(cellfun(@(x)(sum(bsxfun(@minus,centroids,x).^2,2)),XCell,'UniformOutput',false)),nRows,nRows);
[~,labels]=min(dist);

Explanation:

• We assign each row of X to its own cell in the second line
• This piece @(x)(sum(bsxfun(@minus,centroids,x).^2,2)) is an anonymous function which is the same as your distances=... line, and using cell2mat, we apply it to each row of X.
• The labels are then the indices of the minimum row along each column.