Generating "Non-Random" Numbers in R?-CodePudding

I know how to generate 100 random numbers in R (without replacement):

random_numbers = sample.int(100, 100, replace = FALSE)

I was now curious about learning how to generate 100 "non random" numbers (without replacement). The first comes to mind is to generate a random number, and the next number will be the old number 1 with a probability of 0.5 or an actual random number with probability 0.5. Thus, these numbers are not "fully random".

This was my attempt to write this code for numbers in a range of 0 to 100 (suppose I want to repeat this procedure 100 times):

library(dplyr)

all_games <- vector("list", 100) 

for (i in 1:100){

index_i = i
guess_sets <- 1:100 
prob_i = runif(n=1, min=1e-12, max=.9999999999)
    guess_i  = ifelse(prob_i> 0.5, sample.int(1, 100, replace = FALSE), guess_i   1)
    guess_sets_i <- setdiff(guess_sets_i, guess_i)
all_games_i = as.list(index_i, guess_i, all_games_i)
all_games[[i]] <- all_games_i
}

all_games <- do.call("rbind", all_games)

I tried to make a list that stores all guesses such that the range for the next guess automatically excludes numbers that have already been guessed, but I get this error:

Error in sample.int(1, 100, replace = FALSE) : 
  cannot take a sample larger than the population when 'replace = FALSE'

Ideally, I am trying to get the following results (format doesn't matter):

index_1 : 5,6,51,4,3,88,87,9 ...
index_2 77,78,79,2,65,3,1,99,100,4...
etc.

Can someone please show me how to do this? Are there easier ways in R to generate "non-random numbers"?

Thank you!

Note: I think an extra line of logic needs to be added - Suppose I guess the number 100, after guessing the number 100 I must guess a new random number since 100 1 is not included in the original range. Also, if I guess the number 5, 17 then 4 - and after guessing 4, the loop tells me to guess 4 1, this is impossible because 5 has already been guessed. In such a case, I would also have to guess a new random number?

CodePudding user response：

It would be tricky to make your algorithm very efficient in R... it doesn't lend itself nicely to vectorization. Here's how I'd write it directly as a for loop:

semirandom = function(n) {
  safe_sample = function(x, ...) {
    if(length(x) == 1) return(x)
    sample(x, ...)
  }
  
  result = numeric(n)
  result[1] = sample.int(n, size = 1)
  for(i in 2:length(result)) {
    if(runif(1) < .5 && 
       result[i - 1] < n &&
       !((result[i - 1]   1) %in% result)) {
      result[i] = result[i - 1]   1
    } else {
      result[i] = safe_sample(x = setdiff(1:n, result), size = 1)
    }
  }
  result
}

# generate 10 semirandom numbers 5 times
replicate(semirandom(10), n = 5)
#       [,1] [,2] [,3] [,4] [,5]
#  [1,]    6    4    4    2    6
#  [2,]    3    5    5    3    7
#  [3,]    4    3    6    4    5
#  [4,]    5    1    2    5    2
#  [5,]    7    9    3    6    3
#  [6,]    9   10   10    1    1
#  [7,]   10    2    8    9    4
#  [8,]    2    8    1    8   10
#  [9,]    1    7    9   10    9
# [10,]    8    6    7    7    8

CodePudding user response：

You get the error cannot take a sample larger than the population when 'replace = FALSE' because you attempt to extract 100 values from a vector of length one without replacement.

The following draws numbers between 1 and 100, draws each number not more than once, has a 50 percent chance of drawing the previous number 1 and a 50 percent chance of drawing another random number, if the previous number 1 has not been drawn yet, and a 100 percent chance to draw another random number, if the previous number 1 has been drawn.

i <- sample.int(100, 1)
j <- i
for(x in 1:99) {
  if((i   1L) %in% j) {
    i <- sample((1:100)[-j], 1L)
  } else {
    if(runif(1L) > 0.5 || i == 100L) {
      i <- sample((1:100)[-j], 1L)
    } else {
      i <- i   1L
    }
  }
  j <- c(j, i)
}