[R] Sampling procedure
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) 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. There is probably an easy and faster way to do this. Could anybody help me? Thanks -- View this message in context: http://www.nabble.com/Sampling-procedure-tp25909497p25909497.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sampling procedure
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
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sampling procedure
If parsimony is needed, then define a 9-row matrix and send a
randomized indexed version of Example to it:
s-matrix(NA, nrow=9, ncol=length(Example)/9)
s[,] - Example[sample(Example, length(Example) )]
str(s)
int [1:9, 1:87] 503 731 708 23 255 675 163 381 361 412 ...
Or even:
s-matrix(Example[ sample(Example, length(Example) )], nrow=9,
ncol=length(Example)/9)
--
David
On Oct 15, 2009, at 11:22 AM, Bert Gunter wrote:
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
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sampling procedure
... except the matrix approach doesn't work if the length of the vector is
not exactly divisible by the number of groups. That's why I used split.
Cheers,
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, October 15, 2009 8:48 AM
To: Bert Gunter
Cc: 'Marcio Resende'; r-help@r-project.org
Subject: Re: [R] Sampling procedure
If parsimony is needed, then define a 9-row matrix and send a
randomized indexed version of Example to it:
s-matrix(NA, nrow=9, ncol=length(Example)/9)
s[,] - Example[sample(Example, length(Example) )]
str(s)
int [1:9, 1:87] 503 731 708 23 255 675 163 381 361 412 ...
Or even:
s-matrix(Example[ sample(Example, length(Example) )], nrow=9,
ncol=length(Example)/9)
--
David
On Oct 15, 2009, at 11:22 AM, Bert Gunter wrote:
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
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sampling procedure
OK, you're right. I thought it might be simple fix to increase the
number of columns to accommodate, but the recycling conventions trips
up that strategy.
Thanks;
David.
On Oct 15, 2009, at 11:55 AM, Bert Gunter wrote:
... except the matrix approach doesn't work if the length of the
vector is
not exactly divisible by the number of groups. That's why I used
split.
Cheers,
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, October 15, 2009 8:48 AM
To: Bert Gunter
Cc: 'Marcio Resende'; r-help@r-project.org
Subject: Re: [R] Sampling procedure
If parsimony is needed, then define a 9-row matrix and send a
randomized indexed version of Example to it:
s-matrix(NA, nrow=9, ncol=length(Example)/9)
s[,] - Example[sample(Example, length(Example) )]
str(s)
int [1:9, 1:87] 503 731 708 23 255 675 163 381 361 412 ...
Or even:
s-matrix(Example[ sample(Example, length(Example) )], nrow=9,
ncol=length(Example)/9)
--
David
On Oct 15, 2009, at 11:22 AM, Bert Gunter wrote:
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
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
