How to normalize a vector in MATLAB efficiently? Any related built-in function? [closed]

Answer 1

The only problem you would run into is if the norm of V is zero (or very close to it). This could give you Inf or NaN when you divide, along with a divide-by-zero warning. If you don't care about getting an Inf or NaN, you can just turn the warning on and off using WARNING:

oldState = warning('off','MATLAB:divideByZero');  % Return previous state then
                                                  %   turn off DBZ warning
uV = V/norm(V);
warning(oldState);  % Restore previous state

If you don't want any Inf or NaN values, you have to check the size of the norm first:

normV = norm(V);
if normV > 0,  % Or some other threshold, like EPS
  uV = V/normV;
else,
  uV = V;  % Do nothing since it's basically 0
end

If I need it in a program, I usually put the above code in my own function, usually called unit (since it basically turns a vector into a unit vector pointing in the same direction).

Answer 2

I took Mr. Fooz's code and also added Arlen's solution too and here are the timings that I've gotten for Octave:

clc; clear all;
V = rand(1024*1024*32,1);
N = 10;
tic; for i=1:N, V1 = V/norm(V);         end; toc % 7.0 s
tic; for i=1:N, V2 = V/sqrt(sum(V.*V)); end; toc % 6.4 s
tic; for i=1:N, V3 = V/sqrt(V'*V);      end; toc % 5.5 s
tic; for i=1:N, V4 = V/sqrt(sum(V.^2)); end; toc % 6.6 s
tic; for i=1:N, V1 = V/norm(V);         end; toc % 7.1 s
tic; for i=1:N, d = 1/norm(V); V1 = V*d;end; toc % 4.7 s

Then, because of something I'm currently looking at, I tested out this code for ensuring that each row sums to 1:

clc; clear all;
m = 2048;
V = rand(m);
N = 100;
tic; for i=1:N, V1 = V ./ (sum(V,2)*ones(1,m));                end; toc % 8.2 s
tic; for i=1:N, V2 = bsxfun(@rdivide, V, sum(V,2));            end; toc % 5.8 s
tic; for i=1:N, V3 = bsxfun(@rdivide, V, V*ones(m,1));         end; toc % 5.7 s
tic; for i=1:N, V4 = V ./ (V*ones(m,m));                       end; toc % 77.5 s
tic; for i=1:N, d = 1./sum(V,2);V5 = bsxfun(@times, V, d);     end; toc % 2.83 s
tic; for i=1:N, d = 1./(V*ones(m,1));V6 = bsxfun(@times, V, d);end; toc % 2.75 s
tic; for i=1:N, V1 = V ./ (sum(V,2)*ones(1,m));                end; toc % 8.2 s

Answer 3

I don't know any MATLAB and I've never used it, but it seems to me you are dividing. Why? Something like this will be much faster:

d = 1/norm(V)
V1 = V * d

Answer 4

By the rational of making everything multiplication I add the entry at the end of the list

    clc; clear all;
    V = rand(1024*1024*32,1);
    N = 10;
    tic; for i=1:N, V1 = V/norm(V);         end; toc % 4.5 s
    tic; for i=1:N, V2 = V/sqrt(sum(V.*V)); end; toc % 7.5 s
    tic; for i=1:N, V3 = V/sqrt(V'*V);      end; toc % 4.9 s
    tic; for i=1:N, V4 = V/sqrt(sum(V.^2)); end; toc % 6.8 s
    tic; for i=1:N, V1 = V/norm(V);         end; toc % 4.7 s
    tic; for i=1:N, d = 1/norm(V); V1 = V*d;end; toc % 4.9 s
    tic; for i=1:N, d = norm(V)^-1; V1 = V*d;end;toc % 4.4 s

Answer 5

Fastest by far (time is in comparison to Jacobs):

clc; clear all;
V = rand(1024*1024*32,1);
N = 10;
tic; 
for i=1:N, 
    d = 1/sqrt(V(1)*V(1)+V(2)*V(2)+V(3)*V(3)); 
    V1 = V*d;
end; 
toc % 1.5s

Answer 6

The original code you suggest is the best way.

Matlab is extremely good at vectorized operations such as this, at least for large vectors.

The built-in norm function is very fast. Here are some timing results:

V = rand(10000000,1);
% Run once
tic; V1=V/norm(V); toc           % result:  0.228273s
tic; V2=V/sqrt(sum(V.*V)); toc   % result:  0.325161s
tic; V1=V/norm(V); toc           % result:  0.218892s

V1 is calculated a second time here just to make sure there are no important cache penalties on the first call.

Timing information here was produced with R2008a x64 on Windows.

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EDIT:

Revised answer based on gnovice's suggestions (see comments). Matrix math (barely) wins:

clc; clear all;
V = rand(1024*1024*32,1);
N = 10;
tic; for i=1:N, V1 = V/norm(V);         end; toc % 6.3 s
tic; for i=1:N, V2 = V/sqrt(sum(V.*V)); end; toc % 9.3 s
tic; for i=1:N, V3 = V/sqrt(V'*V);      end; toc % 6.2 s ***
tic; for i=1:N, V4 = V/sqrt(sum(V.^2)); end; toc % 9.2 s
tic; for i=1:N, V1=V/norm(V);           end; toc % 6.4 s

IMHO, the difference between "norm(V)" and "sqrt(V'*V)" is small enough that for most programs, it's best to go with the one that's more clear. To me, "norm(V)" is clearer and easier to read, but "sqrt(V'*V)" is still idiomatic in Matlab.