Function scale (PR#2209)

Ben Bolker bolker@zoo.ufl.edu
Thu, 24 Oct 2002 10:21:22 -0400 (EDT)

```  The function is behaving as documented ...

"If `scale' is `TRUE' then scaling is done by dividing the (centered)
columns of `x' by their root-mean-square, and if `scale' is `FALSE',
no scaling is done.

"The root-mean-square for a column is obtained by computing the
square-root of the sum-of-squares of the non-missing values in the
column divided by the number of non-missing values minus one."

It says "root-mean-square", not "standard error" or "standard
deviation".  If you set center=TRUE, then the two are equivalent. If you
*want* it to scale by standard deviation just set scale=sd(l) [or in the
matrix case, scale=apply(l,2,sd)].

Ben Bolker

On Thu, 24 Oct 2002 lucas@toulouse.inra.fr wrote:

> I found a problem with scale function,
> while using center=FALSE and scale=TRUE:
>
> all column are not divided by standard error,
> but divided by sqrt (1/(n-1) sum Xi^2 )
>
> Example:
> > l<- c(1,2,3)
> > scale(l,F,T)
>           [,1]
> [1,] 0.3779645
> [2,] 0.7559289
> [3,] 1.1338934
> attr(,"scaled:scale")
> [1] 2.645751
>
> 2.645751 = sqrt( 1/2 * (1+4+9) )
>
>
> Antoine Lucas & Aymeric Labourdette
>
> --
> Antoine  Lucas
> INRA, Unité de biométrie et     |  Tel 05 61 28 53 34
> intelligence artificielle       |  Fax 05 61 28 53 35
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>
>
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