Converting Lists into Data Frames-CodePudding

I am trying to make a data frame with 3 columns (and 100 rows) containing random numbers such that the random numbers in each row add to 72.

I am using this code to generate the random numbers:

random_numbers <- diff(c(0, sort(sample(72, 2)), 72))

Although, I can't "fit" these random numbers into a data frame because of the format. For example:

i <- 1:100
d <- data.frame(i)
d$rand <- diff(c(0, sort(sample(72, 2)), 72))

Error in `$<-.data.frame`(`*tmp*`, rand, value = c(37, 21, 14)) : 
  replacement has 3 rows, data has 100

I had another idea in which at least I can create all 100 random numbers:

results <- list()

for (i in 1:100) {
  r_i <- diff(c(0, sort(sample(72, 2)), 72))
  results[[i]] <- r_i
}

results[1]
# [[1]]
# [1]  3 19 50

results[2]
# [[1]]
# [1] 16 11 45

But I am not sure how I can I can create a data frame with 3 columns and 100 rows from this data.

I know how to do this in the "classical" sense:

i <- 1:100
r_1 <- rnorm(5, 5, 100)
r_2 <- rnorm(5, 5, 100)
r_3 <- rnorm(5, 5, 100)

d <- data.frame(i, r_1, r_2, r_3)

d = data.frame(i, r_1, r_2, r_3)

But of course, in the above data frame, these 3 numbers will most certainly not add to 72.

Is it possible to take the 100 random numbers results that I generated above and then place them into a data frame?

CodePudding user response：

We may use replicate with n specified as the number of rows of 'd' and assign new columns from the matrix output

d[paste0("r_", 1:3)] <-  t(replicate(nrow(d), 
     diff(c(0, sort(sample(72, 2)), 72))))

-testing for equality

> all(rowSums(d[-1]) == 72)
[1] TRUE