Speed optimization of partial inner product (norm)

For a row vector, the norm can be written as

sqrt(sum(P.^2))

or

sqrt(P*P')

The latter is about twice as fast. Now I have a 4D matrix with dimensions [100,100,100,70], and would like to take the norm of the last dimension to yield a matrix of dimension [100,100,100]. This works:

sqrt(sum(P.^2,4))

but is too slow. Does anyone know a way to speed this up (perhaps in a similar way as the 1D case?)

Answer 1

[Matt J]

This may help

http://www.mathworks.com/matlabcentral/fileexchange/29035-dnorm2

[lvn]

Thanks a lot! I just tested this function and it works, but unfortunately my data is in single format and converting it to double first makes it again very slow.

[Matt J]

I'm sure it wouldn't be hard to modify the code to accept singles.

[Jan]

So you could ask the author, Ivn. Do you want the output to be in single format also?

[Matt J]

Jan, if you're going to take that modification on, I would just request that the summations/accumulations in the norm calculation still be done in double precision, regardless of the class of the input/output (or that there be an option to do so).

I also vote that the output class should match the input class.

[lvn]

Jan, I am sorry I didn't see your comment until now (after I tagged the question answered, I didn't open it anymore).

This norm is still a bottleneck in my program and I would therefore be very grateful if you could make a version with both in and output having single format.

Answer 2

[Ernst Jan]

My results show that the first is actually faster:

n = 10000;
P1 = rand(1,n);

tic
A1 = sqrt(sum(P1.^2));
toc

tic
A2 = sqrt(P1*P1');
toc

tic
A3 = sqrt(sum(P1.*P1));
toc

P2 = rand([100,100,100,70]);
tic
A4 = sqrt(sum(P2.*P2,4));
toc

tic
A5 = sqrt(sum(P2.^2,4));
toc

Elapsed time is 0.000044 seconds.
Elapsed time is 0.000141 seconds.
Elapsed time is 0.000031 seconds.
Elapsed time is 0.307783 seconds.
Elapsed time is 0.309741 seconds.

Please provide a code example?

[lvn]

Here are my results:

    >> P=rand(100,1);
    >> tic; for k=1:1000000,  sqrt(P'*P); end; toc;
    Elapsed time is 0.918256 seconds.
    >> tic; for k=1:1000000,  sqrt(sum(P.^2)); end; toc;
    Elapsed time is 1.533144 seconds.

But to be clear, the question is related more specifically to the 4D case.

[Matt J]

@Ernst

You're using way too small a value of n to see a meaningful comparison. Here's what I get with n=1e7

Elapsed time is 0.031045 seconds.
Elapsed time is 0.008693 seconds.
Elapsed time is 0.030998 seconds.