How to achieve zero mean and unit variance

I have a matrix 50x3000. How do I achieve zero mean and unit variance

Also How would i normalize the same after applying zero mean and unit variance?

[Image Analyst]

After you normalize to 0 mean and 1 var, why are you normalizing again? Why not go to your target values all in one operation? Why "normalize" twice? What's the mean and variance the second time?

[syd]

Ok sorry for asking in the wrong way. What I meant was once I get zero mean and assign to variable X, then do I dot standard deviation on this variable X, or on the original matrix?

[Image Analyst]

You do it on the shifted version of X. Subtracting the mean merely shifts X without narrowing its distribution. So then you divide to narrow or widen the distribution (X) without shifting it (because it's already shifted).

Answer 1

[Nils]

You can use the zscore function to zero mean / unit variance any array of data.

For a given matrix A,

A = reshape(zscore(A(:)),size(A,1),size(A,2));

will return the matrix A where all elements now follow a zero mean / unit variance distribution.

It is important to linearize A as the input to zscore, then reshape the output, because zscore will operate differently if you feed it an N-D array rather than a vector.

[sahana kp]

@Nils when i'm using this code(A = reshape(zscore(A(:)),size(A,1),size(A,2));) got an answer like this >> mean(cA3)

ans =

   1.0e-15 *

   -0.2722   -0.2722   -0.2722   -0.2722   -0.2722   -0.2722

>> var(cA3)

ans =

    1.0278    1.0278    1.0278    1.0278    1.0278    1.0278

is this correct?

[Walter Roberson]

1.0e-15 times those values is probably zero to within numeric round-off, depending on the range of the original values.

If you were to use

format long g

then you would see a less confusing output.

[sahana kp]

@Walter Roberson Now getting the answer like this >> mean(cA3)

ans =

Columns 1 through 5

     -2.72183709262942e-16     -2.72183709262942e-16     -2.72183709262942e-16     -2.72183709262942e-16     -2.72183709262942e-16

Column 6

     -2.72183709262942e-16

>> var(cA3)

ans =

Columns 1 through 5

          1.02777777777778          1.02777777777778          1.02777777777778          1.02777777777778          1.02777777777778

Column 6

          1.02777777777778
is this nomalized ?

[Walter Roberson]

Might be normalized. The 1e-16 means are round off error that are impossible to avoid in floating point

Answer 2

[Walter Roberson]

http://en.wikipedia.org/wiki/Feature_scaling#Standardization

[syd]

I found this http://stackoverflow.com/questions/8717139/zero-mean-and-unit-variance-of-a-signal

However when i tried this, some elements turned out negative

X=X-mean(X(:)); and unit variance by dividing by the standard deviation:

X=X/std(X(:));

[Image Analyst]

Of course some X will be negative. I'd say about half of them will be . Why are you expecting something else ?

[John D'Errico]

How would you have a mean of zero (and a non-zero variance) if some of the elements were not negative?