生成n個樣本，R中的拒絕樣本

Question

拒絕采樣

我正在處理截斷正態分布的拒絕采樣，請參見下面的r代碼。 如何使采樣在特定的n處停止？ 例如1000個觀察值。 即我要在接受的樣本數達到n（1000）時停止采樣。

有什么建議么？ 任何幫助是極大的贊賞 ：）

#Truncated normal curve    
curve(dnorm(x, mean=2, sd=2)/(1-pnorm(1, mean=2, sd=2)),1,9)

#create a data.frame with 100000 random values between 1 and 9

sampled <- data.frame(proposal = runif(100000,1,9))
sampled$targetDensity <- dnorm(sampled$proposal, mean=2, sd=2)/(1-pnorm(1, mean=2, sd=2))

#accept proportional to the targetDensity

maxDens = max(sampled$targetDensity, na.rm = T)
sampled$accepted = ifelse(runif(100000,0,1) < sampled$targetDensity / maxDens, TRUE, FALSE)

hist(sampled$proposal[sampled$accepted], freq = F, col = "grey", breaks = 100, xlim = c(1,9), ylim = c(0,0.35),main="Random draws from skewed normal, truncated at 1")
curve(dnorm(x, mean=2, sd=2)/(1-pnorm(1, mean=2, sd=2)),1,9, add =TRUE, col = "red", xlim = c(1,9),  ylim = c(0,0.35))



X <- sampled$proposal[sampled$accepted]

采樣時如何將X的長度設置為特定數字？

Answer 1

睡一覺之后，如果您確定要使用拒絕采樣並僅在經過1000次之前進行采樣，我認為沒有比使用while循環更好的選擇了。 效率遠不如

sampled$accepted = ifelse(runif(100000,0,1) < sampled$targetDensity / maxDens, TRUE, FALSE)
X <- sampled$proposal[sampled$accepted][1:1000]

上面的代碼0.0624001s的時間為0.0624001s 。 下面的代碼花費的時間為0.780005s 。 我將其包括在內是因為它是您所提出的特定問題的答案，但是這種方法效率低下。 如果還有其他選擇，我會使用它。

#Number of samples
N_Target <- 1000
N_Accepted <- 0

#Loop until condition is met
i = 1
sampled$accepted = FALSE
while( N_Accepted < N_Target ){

    sampled$accepted[i] = ifelse(runif(1,0,1) < sampled$targetDensity[i] / maxDens, TRUE, FALSE)
    N_Accepted = ifelse( sampled$accepted[i], N_Accepted + 1 , N_Accepted )
    i = i + 1
    if( i > nrow( sampled ) ) break

}