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-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Winsemius
Sent: Thursday, October 15, 2009 7:55 AM
To: Marcio Resende
Cc: r-help@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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