Double sampling method in R - r

Question

My initial code for double sampling is the following. I did only one sample.
# Data
samples<-matrix(NA,nrow = 12, ncol = 2000)
for (i in 1:12) {
samples[i,]<- rbinom(2000,1,prob = 0.05)
}
# Double Sampling Plan
accept<-rep(0,12)
for (i in 1:12) {
if (sum(samples[i,1:80])<=5){
accept[i]<-1
} else if (sum(samples[i,1:80]<=8) & sum(samples[i,1:80]>5) ) {
if (sum(samples[i,1:160])<=12) {
accept[i]<-1
}
}
}
sum(accept)
Since I generated randomly from Bernoulli, every time you run the code, the results will not be the same.
I want 100 repetitions of this double sample.
My solution:
nm=double(100)
for (j in 1:100){
# Data
samples<-matrix(NA,nrow = 12, ncol = 2000)
for (i in 1:12) {
samples[i,]<- rbinom(2000,1,prob = 0.05)
}
# Double Sampling Plan
accept<-rep(0,12)
for (i in 1:12) {
if (sum(samples[i,1:80])<=5){
accept[i]<-1
} else if (sum(samples[i,1:80]<=8) & sum(samples[i,1:80]>5) ) {
if (sum(samples[i,1:160])<=12) {
accept[i]<-1
}
}
}
nm[j]=sum(accept)
}
mean(nm)
What do you think?

Answer 1

If we follow the proposition of #Onyambu, we can embeded one simulation inside a function and call it in a loop like this :
one_double_sampling <- function(){
# Data
samples<-matrix(NA,nrow = 12, ncol = 2000)
for (i in 1:12) {
samples[i,]<- rbinom(2000,1,prob = 0.05)
}
# Double Sampling Plan
accept<-rep(0,12)
for (i in 1:12) {
if (sum(samples[i, 1:80])<=5){
accept[i]<-1
} else if (sum(samples[i,1:80]<=8) & sum(samples[i,1:80]>5) ) {
if (sum(samples[i,1:160])<=12) {
accept[i]<-1
}
}
}
return(sum(accept))
}
set.seed(123)
# number of sample
n <- 100
# stock the result
res <- rep(0, n)
for(i in 1:n){
res[i] <- one_double_sampling()
}
# mean
mean(res)
Definitly your code is correct. For people interresting by the double sampling method I advise you to see this.
Edit 1
In one line code based on Onyambu advise :
mean(replicate(n, one_double_sampling()))