And yes, I ignored Genotype, but for the example data none of the stand_ID
values are present in more than one Genotype, so it doesn't matter. If that's
not true in general, then constructing the grp variable is a little more
complex, but the principle is the same.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
Lab cell 925-724-7509
On 8/27/18, 4:10 PM, "R-help on behalf of MacQueen, Don via R-help"
<r-help-boun...@r-project.org on behalf of r-help@r-project.org> wrote:
You could start with split()
grp <- rep('', nrow(mydata) )
grp[mydata$stand_ID %in% c(7,9,67)] <- 'A-training'
grp[mydata$stand_ID %in% c(3,18,20,21,32)] <- 'B-testing'
split(mydata, grp)
or perhaps
grp <- ifelse( mydata$stand_ID %in% c(7,9,67) , 'A-training', 'B-testing' )
split(mydata, grp)
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
Lab cell 925-724-7509
On 8/27/18, 3:54 PM, "R-help on behalf of Ahmed Attia"
<r-help-boun...@r-project.org on behalf of ahmedati...@gmail.com> wrote:
I would like to partition the following dataset (dataGenotype) based
on two variables; Genotype and stand_ID, for example, for Genotype
H13: stand_ID number 7 may go to training and stand_ID number 18 and
21 may go to testing.
Genotype stand_ID Inventory_date stemC mheight
H13 7 5/18/2006 1940.1075 11.33995
H13 7 11/1/2008 10898.9597 23.20395
H13 7 4/14/2009 12830.1284 23.77395
H13 18 11/3/2005 2726.42 13.4432
H13 18 6/30/2008 12226.1554 24.091967
H13 18 4/14/2009 14141.68 25.0922
H13 21 5/18/2006 4981.7158 15.7173
H13 21 4/14/2009 20327.0667 27.9155
H15 9 3/31/2006 3570.06 14.7898
H15 9 11/1/2008 15138.8383 26.2088
H15 9 4/14/2009 17035.4688 26.8778
H15 20 1/18/2005 3016.881 14.1886
H15 20 10/4/2006 8330.4688 20.19425
H15 20 6/30/2008 13576.5 25.4774
H15 32 2/1/2006 3426.2525 14.31815
U21 3 1/9/2006 3660.416 15.09925
U21 3 6/30/2008 13236.29 24.27634
U21 3 4/14/2009 16124.192 25.79562
U21 67 11/4/2005 2812.8425 13.60485
U21 67 4/14/2009 13468.455 24.6203
And the desired output is the following;
A-training
Genotype stand_ID Inventory_date stemC mheight
H13 7 5/18/2006 1940.1075 11.33995
H13 7 11/1/2008 10898.9597 23.20395
H13 7 4/14/2009 12830.1284 23.77395
H15 9 3/31/2006 3570.06 14.7898
H15 9 11/1/2008 15138.8383 26.2088
H15 9 4/14/2009 17035.4688 26.8778
U21 67 11/4/2005 2812.8425 13.60485
U21 67 4/14/2009 13468.455 24.6203
B-testing
Genotype stand_ID Inventory_date stemC mheight
H13 18 11/3/2005 2726.42 13.4432
H13 18 6/30/2008 12226.1554 24.091967
H13 18 4/14/2009 14141.68 25.0922
H13 21 5/18/2006 4981.7158 15.7173
H13 21 4/14/2009 20327.0667 27.9155
H15 20 1/18/2005 3016.881 14.1886
H15 20 10/4/2006 8330.4688 20.19425
H15 20 6/30/2008 13576.5 25.4774
H15 32 2/1/2006 3426.2525 14.31815
U21 3 1/9/2006 3660.416 15.09925
U21 3 6/30/2008 13236.29 24.27634
U21 3 4/14/2009 16124.192 25.79562
I tried the following code;
library(caret)
dataPartitioning <-
createDataPartition(dataGenotype$stand_ID,1,list=F,p=0.2)
train = dataGenotype[dataPartitioning,]
test = dataGenotype[-dataPartitioning,]
Also tried
createDataPartition(unique(dataGenotype$stand_ID),1,list=F,p=0.2)
It did not produce the desired output, the data are partitioned within
the stand_ID. For example, one row of stand_ID 7 goes to training and
two rows of stand_ID 7 go to testing. How can I partition the data by
Genotype and stand_ID together?.
Ahmed Attia
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______________________________________________
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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 -- To UNSUBSCRIBE and more, see
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.
