data<-c(50,82,115,126,155,161,243,294,340,384,457,680,855,877,974,1193,1340,1884,2558,3476)
A2<-function(y,lambda){
func<-function(y,lambda){
f1<-(1-exp(-(y-50)*lambda))}
F<-func(y,lambda)
F_n<-ecdf(y)
Fn<-c(0,F_n(y))
a<-numeric()
for(i in 1:(length(y)-1)){
a[i]<-((1-Fn[order(y[i])])^2)*(log(1-F[order(y[i])])-
log(1-F[order(y[i+1])]))
}
b<-numeric()
for(j in 2:(length(y))){
b[j]<-((Fn[order(y[j])])^2)*(log(F[order(y[j+1])])-log(F[order(y[j])]))
}
TEST_STATISTICS<-(-length(y)-1)+(length(y)-1)*sum(a)+(length(y)-1)*sum(b)
print(TEST_STATISTICS)
a
}
A2(data,lambda)
Related
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?
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()))
I have functions and would like to save from the IF function the variable "sick [i]" to take advantage of it
for(i in 1:licznik){
print_func <- function(a, b)
{
if(a > b)
{
print('wspolczynnik jest wiekszy' )
print(sick[i])
}
}
print_func(a[i], b[i])
}
How to do it
?
Try this from R for Data Science.
out <- vector("list", length(licznik))
for (i in seq_along(licznik)) {
if(a[[i]] > b[[i]]) {
print('wspolczynnik jest wiekszy')
print(sick[[i]])
out[[i]] <- sick[[i]]
}
}
str(unlist(out))
I am trying to generate a vector of random numbers based on a finite random variable X
With probGen function I generate a variable X, l1 is the first line and l2 is the second one.
And at this point if(sum1 >= U) I recive this error Error in if (sum1 >= U) { : argument is of length zero
This is my code:
probGen=function(n)
{
v=vector()
k=sample(1:n,1)
v=rep(0,k)
for(i in 1:n)
{
aux=sample(1:k,1)
v[aux]=v[aux]+1
}
vfinal=vector()
klen=0
for(i in 1:k)
{
if(v[i]!=0) klen=klen+1
}
for(i in 1:k)
{
if(v[i]!=0)
vfinal=c(vfinal,rep(1/(klen*v[i]),v[i]))
}
vfinal=sample(vfinal)
return (vfinal)
}
n=22
l1=c(1:n)
l2=probGen(n)
l1
l2
simVar=function(l1,l2)
{
variante=vector()
U=runif(1,0,1)
for(i in 1:length(l1))
{
sum1=1-1
for(j in 1:i-1)
{
if(i-1>=1)
{
sum1=sum1+l2[j]
}
}
sum2=0.0
for(j in 1:i)
{
sum2=sum2+l2[j]
}
if(sum1 >= U)
{
if(U<sum2)
{
variante=c(variante,l1[i])
}
}
}
return (variante)
}
varR=simVar(l1,l2)
varR
Any idea?
Thanks!
The for(j in 1:i-1) near the top of the code for simVar is evaluating as (1:i)-1, resulting in a zero j which produces a NA value of sum1. Use for(j in 1:(i-1)) instead.
I am trying to build a matrix model which ends if certain conditions are invoked - however for some reason the break() command isn't working, although stop() does. Unfortunately stop() is not what I need as I need to run the model a number of times.
The first break command in the model works, but I have left it in with dth>100 so that you can see for yourselves
n.steps <- 200
ns <- array(0,c(14,n.steps))
ns[13,1]<-rpois(1,3)
ns[14,1] <- 1
k<-0
for (i in 1:n.steps){
k<-k+1
ns[13,1]<-rpois(1,2)
death<-sample(c(replicate(1000,
sample(c(1,0), prob=c(surv.age.a, 1-surv.age.a), size = 1))),1)
ns[14,k] <- death
if (death == 0) {
dth <- sample(1:100, 1)
if (dth > 100) {
ns[14,k]<-0
print("stop.1")
break()
} else {
while (death == 0) {
if (ns[13, k] > 0) {
rep.vec[i]<-ns[13,k]
ns[13, k] <- ns[13, k] - 1
ns[14,k+1]<-1
print("replace")
} else {
if (ns[13, k] == 0) {
print("stop.2")
ns[14,k+1]<-0
break()
}
}
}
}
}
}
Try this (only showing the relevant portions):
for (i in 1:n.steps){
# ...
break.out.of.for <- FALSE
while (death == 0) {
if (ns[13, k-1] > 0) {
# ...
} else {
if (ns[13, k] == 0) {
# ...
break.out.of.for = TRUE
break
}
}
if (break.out.of.for) {
break
}
}
vector <- numeric()
for (i in 1:dim(myCor)[1]-1){
for (j in i+1:dim(myCor)[2]){
if(abs(myCor[i,j]-1)<0.1){
vector<-append(vector,j)
}
}
}
I don't know why this happens. I am a very new learner in R.