Splitting a data column randomly into 3 groups

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:

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:

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

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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