rloopscumulative-frequencylikert

Algorithm to generate all possible populations for a 5-Likert Scale in R (cumulative frequency per level by 0.1)


I would like to generate all possible populations in a 5-likert scale which values are the cumulative frequency by 0.1 in each level), e.g.:

[1] [2]  [3]  [4]  [5]
1    0    0    0    0
0    1    0    0    0
         ...
0    0    0    0    1
0.9  0.1  0    0    0
0.9  0   0.1   0    0
         ...
0.8 0.2   0    0    0
0.8  0    0.2  0    0
         ...
0.8 0.1  0.1   0    0

and so on...

I have tried with some rudimentary loops like:

fin <- NULL
for (i in 1:10) {
  a <- c(1-(i/10),0,0,0,0)
  fin <- c(fin,a)
  for (j in 1:10) {
    b <- c(a[1],(j/10),0,0,0)
    fin <- c(fin,b)
    for (k in 1:10) {
      c <- c(a[1],b[2],k/10,0,0)
      fin <- c(fin,c)
      for (l in 1:10) {
        d <- c(a[1],b[2],c[3],l/10,0)
        fin <- c(fin,d)
        for (m in 1:10) {
          e <- c(a[1],b[2],c[3],d[4],m/10)
          fin <- c(fin,e)
        }
      }
    }
  }
}
dat <- as.data.frame(matrix(fin, ncol = 5, byrow = T))
head(dat)

a <- NULL
for (i in 1:111110) {
  if(rowSums(dat[i,])==1)
  {b <- dat[i,]
  a <- c(a,b)}
  else{
    next
    }
}
dat <- as.data.frame(matrix(fin, ncol = 5, byrow = T))

I know it is not smart neither efficient but rows where sum = 1 are some of the cases I want to have, but it is insufficient.

I really appreciate any help. Thanks in advance.


Solution

  • Create vector containing all allowed values

    values <- seq(0, 1, by=0.1)
    values
    

    Returns:

    [1] 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
    

    Using base R expand.grid to get all possible combinations of these values for five variables:

    df <- expand.grid(A1 = values, A2 = values, A3 = values, A4 = values, A5 = values)
    

    Calculate row wise sum using rowSums:

    df$TestSum <- rowSums(df)
    

    Keep only rows where the TestSum is 1 (and also only keep the first 5 columns, we don't need the TestSum column anymore):

    result <- df[df$TestSum == 1, 1:5]
    
    head(result)
    

    Returns:

        A1  A2 A3 A4 A5
    11 1.0 0.0  0  0  0
    21 0.9 0.1  0  0  0
    31 0.8 0.2  0  0  0
    41 0.7 0.3  0  0  0
    51 0.6 0.4  0  0  0
    61 0.5 0.5  0  0  0