[R] Sampling procedure

[Bert Gunter]

If I understand what is wanted correctly, this can be a one-liner! -- think
whole objects:

splitup <- function(x,n.groups)
#split x into n.groups mutually exclusive sets
{
  lx <- length(x)
  if(n.groups >= lx) stop("Number of groups greater than vector length")
  x <- x[sample(lx,lx)]
  split(x,seq_len(n.groups))
}

## testit

> splitup(1:71,9)

$`1`
[1] 22 26 38 50 65 60  9 27

$`2`
[1] 24  2 69 28 71 31 41 13

$`3`
[1] 16 47 63 45 23  1  8 32

$`4`
[1] 34 39 64 35  7 19  4 55

$`5`
[1] 54 10 37 68  6 17 70 18

$`6`
[1] 61 11  5 46 33 43 14 56

$`7`
[1] 42 44 12 62 66 48 57 58

$`8`
[1] 21 40 30 29 20 49 52 67

$`9`
[1] 59 15 25 51  3 36 53


Cheers,

Bert Gunter
Genentech Nonclinical Statistics
 

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of David Winsemius
Sent: Thursday, October 15, 2009 7:55 AM
To: Marcio Resende
Cc: r-help at r-project.org
Subject: Re: [R] Sampling procedure


On Oct 15, 2009, at 10:19 AM, Marcio Resende wrote:

>
> I would like to divide a vector in 9 groups in a way that each  
> number is
> present in only one group.
> In a vector of 783 I would like to divide in 9 different groups of 87
>
> Example <- matrix(c(1:783),ncol = 1)


 > Example <- matrix(c(1:783),ncol = 1)
 > Grp1 <- sample(Example, 87, replace=FALSE)
 > Grp2 <- sample(Example[-Grp1], 87, replace=FALSE)
 > Grp3 <- sample(Example[-c(Grp1, Grp2)], 87, replace=FALSE)
# lather, rinse , repeat


> s1 <- as.matrix(sample(Example,87, re = FALSE))
> Example <- Example[-s1]
> s2 <- as.matrix(sample(Example,87, re = FALSE))
> #however I don´t know how to remove the second group from the  
> "Example" to
> continue sampling.
>

#Don't mess up the original

> There is probably an easy and faster way to do this.
> Could anybody help me?
> Thanks
-- 

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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