如何在一个循环中选择最大元素,为r中的循环写一个二重

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我想对θ[j]进行采样,作为j=1,2,…的第一个采样,。。。,71,然后绘制复制的(像1000次(yrep[k]形式Bin(n[j],theta[j](,n[j]是已知的。

那么对于每个θ[j],我有1000 yrep。如果我想在这1000 yrep中选择最大元素。最后,我将有71个最大元素。

R中的My for循环如下:

theta<-NULL
yrep<-NULL
test<-NULL
k=1
for(i in 1:1000){
for(j in 1:71){
theta[j] <- rbeta(1,samp_A+y[j], samp_B+n[j]-y[j])
yrep[k]<-rbinom(1, n[j], theta[j])
k=k+1
}
t<-c(test, max(yrep))
}

我不确定这个是否正确。

#Data
y <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,
2,1,5,2,5,3,2,7,7,3,3,2,9,10,4,4,4,4,4,4,4,10,4,4,4,5,11,12,
5,5,6,5,6,6,6,6,16,15,15,9,4)
n <- 
c(20,20,20,20,20,20,20,19,19,19,19,18,18,17,20,20,20,20,19,19,18,18,25,24,
23,20,20,20,20,20,20,10,49,19,46,27,17,49,47,20,20,13,48,50,20,20,20,20,
20,20,20,48,19,19,19,22,46,49,20,20,23,19,22,20,20,20,52,46,47,24,14)

#Evaluate densities in grid
x <- seq(0.0001, 0.9999, length.out = 1000)

#Compute the marginal posterior of alpha and beta in hierarchical model Use grid
A <- seq(0.5, 15, length.out = 100)
B <- seq(0.3, 45, length.out = 100)
#Make vectors that contain all pairwise combinations of A and B
cA <- rep(A, each = length(B))
cB <- rep(B, length(A))
#Use logarithms for numerical accuracy!
lpfun <- function(a, b, y, n) log(a+b)*(-5/2) +
sum(lgamma(a+b)-lgamma(a)-lgamma(b)+lgamma(a+y)+lgamma(b+n-y)- 
lgamma(a+b+n))
lp <- mapply(lpfun, cA, cB, MoreArgs = list(y, n))
#Subtract maximum value to avoid over/underflow in exponentiation
df_marg <- data.frame(x = cA, y = cB, p = exp(lp - max(lp)))
#Sample from the grid (with replacement)
nsamp <- 100
samp_indices <- sample(length(df_marg$p), size = nsamp,
replace = T, prob = df_marg$p/sum(df_marg$p))
samp_A <- cA[samp_indices[1:nsamp]]
samp_B <- cB[samp_indices[1:nsamp]]
df_psamp <- mapply(function(a, b, x) dbeta(x, a, b),
samp_A, samp_B, MoreArgs = list(x = x)) %>%
as.data.frame() %>% cbind(x) %>% gather(ind, p, -x)

您应该为我们提供一些数据,我没有测试它,但我认为这很有效:

for(j in 1:71){
theta[j] <- rbeta(1,samp_A+y[j], samp_B+n[j]-y[j])
for(i in 1:1000){
yrep[i]<-rbinom(1, n[j], theta[j])}
test = c(test, max(yrep))}

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