[R] Splitting a data column randomly into 3 groups

Avi Gross @v|gro@@ @end|ng |rom ver|zon@net
Sun Sep 5 01:58:53 CEST 2021


Abou,

I believe I addressed this issue in a private message the other day.

As a general rule, truncating can leave a remainder. If 
	M  = length(whatever)/3 

Then M is no longer an integer. It can be a number ending in .333... or .666... as well as 0.

Now R may silently truncate something like 100/3 which you see to use and make it be as if you typed 33. Same for 2*M. In your code, you used integer division and that is a truncation too!

  m1 <- n1 %/% 3
  s1 <- sample(1:n1, n1)
  group1.IDs <- sample1.IDs[s1[1:m1]]
  group2.IDs <- sample1.IDs[s1[(m1+1):(2*m1)]]
  group3.IDs <- sample1.IDs[s1[(m1*2+1):(3*m1)]]

A proper solution accounts for any leftover items. One method is to leave all extra items till the end and have:

MAX <- length(original or whatever)
group3.IDs <- sample1.IDs[s1[(m1*2+1):MAX]]


The last group then might have one or two extra items. Another is to go for  a second sweep and take any leftover items and move one each into whatever groups you wish for some balance.

Or, as discussed, there are packages available that let you specify percentages you want and handle these edge cases too.

-----Original Message-----
From: R-help <r-help-bounces using r-project.org> On Behalf Of AbouEl-Makarim Aboueissa
Sent: Saturday, September 4, 2021 5:13 PM
To: Thomas Subia <tgs77m using yahoo.com>
Cc: R mailing list <r-help using r-project.org>
Subject: Re: [R] Splitting a data column randomly into 3 groups

Dear Thomas:


Thank you very much for your input in this matter.


The core part of this R code(s) (please see below) was written by *Richard O'Keefe*. I had three examples with different sample sizes.



*First sample of size n1 = 204* divided randomly into three groups of sizes 68. *No problems with this one*.



*The second sample of size n2 = 112* divided randomly into three groups of sizes 37, 37, and 38. BUT this R code generated three groups of equal sizes (37, 37, and 37). *How to fix the code to make sure that the output will be three groups of sizes 37, 37, and 38*.



*The third sample of size n3 = 284* divided randomly into three groups of sizes 94, 95, and 95. BUT this R code generated three groups of equal sizes (94, 94, and 94). *Again*, h*ow to fix the code to make sure that the output will be three groups of sizes 94, 95, and 95*.


With many thanks

abou


###########  ------------------------   #############


N1 <- 485
population1.IDs <- seq(1, N1, by = 1)
#### population1.IDs

n1<-204                                        ##### in this case the size
of each group of the three groups = 68
sample1.IDs <- sample(population1.IDs,n1) #### sample1.IDs

####  n1 <- length(sample1.IDs)

  m1 <- n1 %/% 3
  s1 <- sample(1:n1, n1)
  group1.IDs <- sample1.IDs[s1[1:m1]]
  group2.IDs <- sample1.IDs[s1[(m1+1):(2*m1)]]
  group3.IDs <- sample1.IDs[s1[(m1*2+1):(3*m1)]]

groups.IDs <-cbind(group1.IDs,group2.IDs,group3.IDs)

groups.IDs


####### --------------------------


N2 <- 266
population2.IDs <- seq(1, N2, by = 1)
#### population2.IDs

n2<-112                           ##### in this case the sizes of the three
groups are(37, 37, and 38)
                                          ##### BUT this codes generate three groups of equal sizes (37, 37, and 37) sample2.IDs <- sample(population2.IDs,n2) #### sample2.IDs

####  n2 <- length(sample2.IDs)

  m2 <- n2 %/% 3
  s2 <- sample(1:n2, n2)
  group1.IDs <- sample2.IDs[s2[1:m2]]
  group2.IDs <- sample2.IDs[s2[(m2+1):(2*m2)]]
  group3.IDs <- sample2.IDs[s2[(m2*2+1):(3*m2)]]

groups.IDs <-cbind(group1.IDs,group2.IDs,group3.IDs)

groups.IDs


####### --------------------------



N3 <- 674
population3.IDs <- seq(1, N3, by = 1)
#### population3.IDs

n3<-284                           ##### in this case the sizes of the three
groups are(94, 95, and 95)
                                          ##### BUT this codes generate three groups of equal sizes (94, 94, and 94) sample2.IDs <- sample(population2.IDs,n2) sample3.IDs <- sample(population3.IDs,n3) #### sample3.IDs

####  n3 <- length(sample2.IDs)

  m3 <- n3 %/% 3
  s3 <- sample(1:n3, n3)
  group1.IDs <- sample3.IDs[s3[1:m3]]
  group2.IDs <- sample3.IDs[s3[(m3+1):(2*m3)]]
  group3.IDs <- sample3.IDs[s3[(m3*2+1):(3*m3)]]

groups.IDs <-cbind(group1.IDs,group2.IDs,group3.IDs)

groups.IDs

______________________


*AbouEl-Makarim Aboueissa, PhD*

*Professor, Statistics and Data Science* *Graduate Coordinator*

*Department of Mathematics and Statistics* *University of Southern Maine*



On Sat, Sep 4, 2021 at 11:54 AM Thomas Subia <tgs77m using yahoo.com> wrote:

> Abou,
>
>
>
> I’ve been following your question on how to split a data column 
> randomly into 3 groups using R.
>
>
>
> My method may not be amenable for a large set of data but it surely 
> worth considering since it makes sense intuitively.
>
>
>
> mydata <- LETTERS[1:11]
>
> > mydata
>
> [1] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K"
>
>
>
> # Let’s choose a random sample of size 4 from mydata
>
> > random_grp1
>
> [1] "J" "H" "D" "A"
>
>
>
> Now my next random selection of data is defined by
>
> data_wo_random <- setdiff(mydata,random_grp1)
>
> # this makes sense because I need to choose random data from a set 
> which is defined by the difference of the sets mydata and random_grp1
>
>
>
> > data_wo_random
>
> [1] "B" "C" "E" "F" "G" "I" "K"
>
>
>
> This is great! So now I can randomly select data of any size from this set.
>
> Repeating this process can easily generate subgroups of your original 
> dataset of any size you want.
>
>
>
> Surely this method could be improved so that this could be done 
> automatically.
>
> Nevertheless, this is an intuitive method which I believe is easier to 
> understand than some of the other methods posted.
>
>
>
> Hope this helps!
>
>
>
> Thomas Subia
>
> Statistician
>
>
>
>
>
>
>
>
>
>

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