CODEWITHSUNDEEP
arraysmatlabmatrixvectorizationnormalization
mcExchange
mcExchange

Reputation: 6494

L2 Normalize a 3 dimensional matrix in Matlab

Is there a quick way of normalizing each row of a 3 dimensional Matrix without resorting to slow for loops in Matlab?

Say my input data looks like this: 

d(:,:,1) =
 1     2     3
 4     5     6

d(:,:,2) =
 7     8     9
10    11    12

I know that I can get the norm of each row by using 

norms = sqrt(sum(d.^2,2))
norms(:,:,1) =
3.7417
8.7750

norms(:,:,2) =
13.9284
19.1050

But how to divide now the second dimension with these norm values? I know that in 2 dims I can use ./ however this seems not to work for 3 dimensional data.

Upvotes: 2

Views: 1439

Answers (1)

rayryeng
rayryeng

Reputation: 104474

bsxfun is your friend:

out = bsxfun(@rdivide, d, norms);

What this does is that it temporarily creates a 3D matrix that replicates each row of norms for as many columns as there are in d and it divides each element in an element-wise manner with d and norms.

We get:

>> d = cat(3, [1 2 3; 4 5 6], [7 8 9; 10 11 12]);
>> norms = sqrt(sum(d.^2,2));
>> out = bsxfun(@rdivide, d, norms)

out(:,:,1) =

    0.2673    0.5345    0.8018
    0.4558    0.5698    0.6838


out(:,:,2) =

    0.5026    0.5744    0.6462
    0.5234    0.5758    0.6281

We can also verify that each row is L2-normalized by determining the sum of squares along each row independently and ensuring that each result sums to 1:

>> sum(out.^2, 2)

ans(:,:,1) =

    1.0000
    1.0000


ans(:,:,2) =

    1.0000
    1.0000

If the approach with bsxfun doesn't quite make sense, an alternative you could use is to create a matrix that respects the same dimensions as d by using repmat... then you can perform the element-wise division you desire:

>> out = d ./ repmat(norms, [1 size(d,2) 1])

out(:,:,1) =

    0.2673    0.5345    0.8018
    0.4558    0.5698    0.6838


out(:,:,2) =

    0.5026    0.5744    0.6462
    0.5234    0.5758    0.6281

With repmat you specify how many times you want the matrix to be copied in each dimension. We only want the matrix to be replicated over the columns while the number of rows and slices are the same... hence the vector [1 size(d,2) 1] that specifies how many times you want the matrix copied in each dimension.

Actually, this is what bsxfun does under the hood without you having to deal with the headaches of creating this temporary matrix. This replication is done for you without having you think about it.

Upvotes: 3

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