[Rd] function "apply" and 3D arrays (PR#7221)

[Liaw, Andy]

The `problem', I think, is your expectation that the output of apply(a, 2,
var) to be of the same dimension as apply(a, 2, sd) if a has dimensions > 2.
Note that:

> sd(matrix(1:9, 3, 3))
[1] 1 1 1
> var(matrix(1:9, 3, 3))
     [,1] [,2] [,3]
[1,]    1    1    1
[2,]    1    1    1
[3,]    1    1    1

because var(), when given a matrix, returns the variance-covariance matrix
of the columns.

The output of sd() can be a bit surprising:

> sd(array(1:27, rep(3, 3)))
[1] 7.937254

This is because sd() looks like:

> sd
function (x, na.rm = FALSE) 
{
    if (is.matrix(x)) 
        apply(x, 2, sd, na.rm = na.rm)
    else if (is.vector(x)) 
        sqrt(var(x, na.rm = na.rm))
    else if (is.data.frame(x)) 
        sapply(x, sd, na.rm = na.rm)
    else sqrt(var(as.vector(x), na.rm = na.rm))
}

So for matrices and data frames, sd() returns the column standard
deviations.  Otherwise it treats the input as a vector and compute the SD.

Andy


> From: jaroslaw.w.tuszynski at saic.com
> 
> Full_Name: jarek tuszynski
> Version: 1.8.1
> OS: windows 2000
> Submission from: (NULL) (198.151.13.10)
> 
> 
> Example code:
> > a=array(1:27, c(3,3,3))
> > apply(a,2, var)
>       [,1] [,2] [,3]
>  [1,]    1    1    1
>  [2,]    1    1    1
>  [3,]    1    1    1
>  [4,]    1    1    1
>  [5,]    1    1    1
>  [6,]    1    1    1
>  [7,]    1    1    1
>  [8,]    1    1    1
>  [9,]    1    1    1
> > apply(a,2, mean)
> [1] 11 14 17
> > apply(a,2, sd)
>      [,1] [,2] [,3]
> [1,]    1    1    1
> [2,]    1    1    1
> [3,]    1    1    1
> 
> I could not figure out from the documentation how MARGIN 
> argument of function
> "apply" works in case of arrays with dimentions larger than 
> 2, so I created the
> above test code. I still do not know how it suppose to work 
> but I should not get
> the results with different dimentions, while calculating var and sd.
> 
> Hope this helps,
> 
> Jarek
> 
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