The way you've written the code, the number of outputs in yBinned depends on how many of the bins of X were populated.
edges = -180:1:180;
X = rand(100,1)*360-180;
Y = rand(100,1);
bins = discretize(X,edges);
yBinned = arrayfun(@(i)Y(bins==i),unique(bins),'UniformOutput', false);
size(yBinned)
numel(unique(bins))
If you want yBinned to correspond to the bin number, you'll need empty placeholders for cases where the bin is empty.
Method 1: somewhat wasteful, but keeps the one-liner aspect. Instead of applying the function to the populated bins (unique(bins)) apply it to the potential bins (numel(edges)-1)
yBinned = arrayfun(@(i)Y(bins==i),1:numel(edges)-1,'UniformOutput', false);
size(yBinned)
Method 2: Uses the same approach as you took to calculate the filled bins, but then distributes them into a vector. I think this is harder to read...but I suppose more efficient (I'd probably pull the unique(bins) part out if performance was really a concern).
yBinnedFull = arrayfun(@(i)Y(bins==i),unique(bins),'UniformOutput', false);
yBinned = cell(numel(edges)-1,1); % initialize in case trailing bins weren't populated
yBinned(unique(bins)) = yBinnedFull;
Note that these produce very slightly different results: method 1's empties are a 0x1 double, and method 2's empties are a 0x0 double.
