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matlabnormalizenorm
dragonmnl
dragonmnl

Reputation: 15558

How should I normalize a vector in Matlab where the sum is 1?

I need to normalize a vector of N integers so that:

For instance:

If I have a vector

V = [2,2,1,0]

the normalized vector should should be:

V_norm = [0.4,0.4,0.2,0]  % 0.4+0.4+0.2 = 1

I tried with many solutions found in this community and on the web and finally I did it with this code:

part = norm(V);
if part > 0
  V_norm = V/part;
else % part = 0 --> avoid "divide by 0" 
  V_norm = part;
end

The problem this works if:

but if I have a different case,although the result is proportional,the sum is not 0. For instance:

   V = [1,0,1]
   V_norm = [0.74,0,0.74]

   V = [1,1,1]
   V_norm = [0.54,0.54,0.54]

(I'm not sure if the number are correct because I can't use Matlab right now but I'm sure the sum is > 1 )

Ahy hint?

Thank you in advance

Upvotes: 10

Views: 24569

Answers (4)

drumath
drumath

Reputation: 170

I will leave here my solution for "normalizing" the rows of a matrix with positive entries, where normalization means that the row-sums are 1 after the procedure.

Summary

normmat = normr(sqrt(mat)).^2

Minimal Example

mat is your matrix with positive entries

mat = [1 2 3; 2 3 4]

mat =

     1     2     3
     2     3     4

Now we do the one-liner:

normmat = normr(sqrt(mat)).^2

normmat =

    0.1667    0.3333    0.5000
    0.2222    0.3333    0.4444

The sum of each row is 1.

sum(normmat, 2)

ans =

    1.0000
    1.0000

Proportions within one row are constant

normmat ./ mat

ans =

    0.1667    0.1667    0.1667
    0.1111    0.1111    0.1111

Not necessarily the most efficient but it's one line.

Upvotes: 0

O. Th. B.
O. Th. B.

Reputation: 1353

What you need to do is, I believe, normalize using the 1-norm (taxicab norm):

v = [2, 2, 1, 0];
v_normed = v / norm(v, 1); % using the 1-norm

Variable v_normed should now be [0.4, 0.4, 0.2, 0.0]. The 1-norm of v_normed will equal 1. You can also sum the vector (similar to the 1-norm, but without applying the absolute function to each value), but the range of that sum will be between -1 to 1 in the general case (if any values in v are below 0). You could use abs on the resulting sum, but mathematically it will no longer qualify as a norm.

Upvotes: 10

Deve
Deve

Reputation: 4658

If there are no furhter conditions to your normalization than you gave at the beginning of your question, a possible solution would be

V = [3 4 -2];
S = sum(V);
if (S == 0)
    % no solution
else
    V_norm = V ./ S;
end
sum(V_norm)

Upvotes: 6

Eitan T
Eitan T

Reputation: 32930

... the normalized vector should should be:

v_norm = [0.4, 0.4, 0.2, 0]; % 0.4+0.4+0.2 = 1

That depends. What is your norm function?

norm(x) in MATLAB returns the standard norm, meaning the sum of the squares of the elements of a normalized vector x is 1.

In your example:

v = [1, 1, 1];         %# norm(v) = sqrt(1^2+1^2+1^2) = ~1.7321
v_norm = v / norm(v);  %# v_norm = [0.5574, 0.5574, 0.5574]

sum(v_norm .^ 2) indeed yields 1, but sum(v_norm) does not, as expected.


I need to normalize a vector of N integers so that each value is proportional to its original value (the value will be between 0 and 1) and the sum of all values is 1.

What do you mean by "normalize"? Does that mean dividing by a value which is a valid mathematical norm function, according to the norm definition?

What do you mean by "proportional"? Does that imply that all elements are multiplied by that same number? If it does, and it's a valid mathematical norm, you cannot guarantee that the sum of the elements will always be 1.
For instance, consider v = [1, -2]. Then sum(v) = -1.

Or maybe sum is the function you're looking for, but it doesn't mathematically qualify as a norm, because a norm is a function that assigns a strictly positive length or size to all vectors in a vector space.
In the example above, sum(v) is negative.


Ahy hint?

You can choose either:

  1. sum(x), which fulfills both requirements but doesn't qualify as a norm function since it can yield negative values.
  2. norm(x, 1), as OleThomsenBuus suggested, which actually calculates sum(abs(x(:))).
    It won't fulfill both your requirements unless you restrict your vector space to non-negative vectors.

Upvotes: 9

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