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Splitting a data column randomly into 3 groups

5 messages · AbouEl-Makarim Aboueissa, Avi Gross, Bert Gunter

AbouEl-Makarim Aboueissa
# 
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 at yahoo.com> wrote:

            

      
      
      
    

        

  
  


  


    

    

      
      

        


  

        

      Bert Gunter
    

    

      
      #
    

  


  

      
I have a more general problem for you.

Given n items and 2 <=g <<n , how do you divide the n items into g
groups that are as "equal as possible."

First, operationally define "as equal as possible."
Second, define the algorithm to carry out the definition. Hint: Note
that sum{m[i]} for i <=g must sum to n, where m[i] is the number of
items in the ith group.
Third, write R code for the algorithm. Exercise for the reader.

I may be wrong, but I think numerical analysts might also have a
little fun here.

Randomization, of course, is trivial.

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Sat, Sep 4, 2021 at 2:13 PM AbouEl-Makarim Aboueissa

        
<abouelmakarim1962 at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Avi Gross
    

    

      
      #
    

  


  

      
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 at r-project.org> On Behalf Of AbouEl-Makarim Aboueissa
Sent: Saturday, September 4, 2021 5:13 PM
To: Thomas Subia <tgs77m at yahoo.com>
Cc: R mailing list <r-help at 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 at yahoo.com> wrote:

        

            

      
      
      
    

        
______________________________________________
R-help at 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.

  
  


  


      

    

    

      
      

        


  

        

      Bert Gunter
    

    

      
      #
    

  


  

      
In case anyone is still interested in my query, note that if there are
n total items to be split into g groups as evenly as possible, if we
define this as at most two different size groups whose size differs by
1, then:

if n = k*g + r, where 0 <= r < g,
then n = k*(g - r) + (k + 1)*r  .
i.e. g-r groups of size k and r groups of size k+1

So using R's modular arithmetic operators, which are handy to know
about, we have:

r = n %% g and k = n %/% g .

(and note that you should disregard my previous stupid remark about
numerical analysis).

Cheers,
Bert

        
On Sat, Sep 4, 2021 at 3:34 PM Bert Gunter <bgunter.4567 at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      AbouEl-Makarim Aboueissa
    

    

      
      #
    

  


  

      
Hi Bert and All: good morning

I promise this would be the last time to write about this topic.

I come up with this R function (please see below), for sure with your help.
It works for all sample sizes. I also provided three different simple
examples.

with many thanks
abou

##################    Here it is    ###############

Random.Sample.IDs <- function (N,n, ngroups){    #### N = population size,
and n = sample size, ngroups = number of groups

population.IDs <- seq(1, N, by = 1)
sample.IDs <- sample(population.IDs,n)

##### to print sample.IDs in a column format
##### --------------------------------------------------
sample.IDs.in.column<-data.frame(sample.IDs)
print(sample.IDs.in.column)

reminder.n<-n%%ngroups
reminder.n

n.final<-n-reminder.n
n.final

  m <- n %/% 3
  m
  s <- sample(1:n, n)

if (reminder.n == 0) {

  group1.IDs <- sample.IDs[s[1:m]]
  group2.IDs <- sample.IDs[s[(m+1):(2*m)]]
  group3.IDs <- sample.IDs[s[(m*2+1):(3*m)]]

} else if(reminder.n == 1){

  group1.IDs <- sample.IDs[s[1:(m+1)]]
  group2.IDs <- sample.IDs[s[(m+2):(2*m+1)]]
  group3.IDs <- sample.IDs[s[(m*2+2):(3*m+1)]]

} else if(reminder.n == 2){

  group1.IDs <- sample.IDs[s[1:(m+1)]]
  group2.IDs <- sample.IDs[s[(m+2):(2*m+2)]]
  group3.IDs <- sample.IDs[s[(m*2+3):(3*m+2)]]
}
nn<-max(length(group1.IDs),length(group2.IDs),length(group3.IDs))
nn
length(group1.IDs) <- nn
length(group2.IDs) <- nn
length(group3.IDs) <- nn

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

groups.IDs

}


#####  Examples
#####  --------

Random.Sample.IDs (100,12,3)    #### group sizes are equal (n1=n2=n3=4)

Random.Sample.IDs (100,13,3)    #### group sizes are NOT equal (n1=5, n2=4,
n3=4)

Random.Sample.IDs (100,17,3)    #### group sizes are NOT equal (n1=6, n2=6,
n3=5)


______________________


*AbouEl-Makarim Aboueissa, PhD*

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

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

        
On Sun, Sep 5, 2021 at 6:50 PM Bert Gunter <bgunter.4567 at gmail.com> wrote: