# 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
*>* http://genopole.toulouse.inra.fr/~lucas
*>*
*>*
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