如何在for循环中设置条件以中断

Question

我尝试调整AR(1) ols小样本偏差，所以我写了一个代码然后在循环函数中设置break但总是失败，我如何在for循环中设置条件？

n = 31 # simulate regression samples

yt <- c()

ANS <- rep(NA,1000) # create a place for beta loop

for (M in c((1:3)/100)) { # after 1000 beta loop, I want simulate AR(1)+0.01 (EX:0.91、0.92、0.93...) Data for next loop in beta
  
  for (i in 1:1000){ # Monte Carole to estimate beta

yt[1] <- rnorm(1,0,1)

  for (z in 2:n) { # simulate AR(1)=0.9+M loop 
  
  yt[z] <- (0.9 + M) * yt[z-1] + rnorm(1,0,1)
  
}

yt

X <- yt[-n]

beta <- solve(t(X) %*% X) %*% t(X) %*% yt[-1] # Estimate AR(1) coeff(beta)

ANS[i] <- sort(beta) 

cat(M," ","\n")

### if (beta[500] - 0.9 <= 0.01) break ### I want set this condition to stop loop but failed , get error message : Error in if (beta[500] - 0.9 <= 1) break , missing value where TRUE/FALSE needed
}}

ifelse(abs(quantile(ANS , probs = c(0.5)) - 0.9) <= 0.01, TRUE, FALSE ) # Because I didn't know how to set break in it, I had to operate it manually

.

下面我尝试添加“if (beta[500] - 0.9

n = 31 # simulate regression samples

yt <- c()

ANS <- rep(NA,1000) # create a place for beta loop

for (M in c((1:3)/100)) { # after 1000 beta loop, I want simulate AR(1)+0.01 (EX:0.91、0.92、0.93...) Data for next loop in beta
  
  for (i in 1:1000){ # Monte Carole to estimate beta

yt[1] <- rnorm(1,0,1)

  for (z in 2:n) { # simulate AR(1)=0.9+M loop 
  
  yt[z] <- (0.9 + M) * yt[z-1] + rnorm(1,0,1)
  
}

yt

X <- yt[-n]

beta <- solve(t(X) %*% X) %*% t(X) %*% yt[-1] # Estimate AR(1) coeff(beta)

ANS[i] <- sort(beta) 

cat(M," ","\n")

if (beta[500] - 0.9 <= 0.01) break 
}}

有谁知道我设置条件时为什么会出错？ # if (beta[500] - 0.9

Answer 1

break 会跳出最近的for 循环，但不会跳出它之外的任何东西。在这种情况下，它位于内循环 for (1 in 1:1000) 内，因此它会跳出内循环，但外循环 (for (M in ...)) 将继续。

如果您希望内部循环中的条件但希望它从两者中分离出来，请考虑使用某种预定义的标志以不中断，但在需要时设置它。我将在这里使用breakout：

breakout <- FALSE
n = 31 # simulate regression samples
yt <- c()
ANS <- rep(NA,1000) # create a place for beta loop
for (M in c((1:3)/100)) {
  for (i in 1:1000){
    yt[1] <- rnorm(1,0,1)
    for (z in 2:n) {
      yt[z] <- (0.9 + M) * yt[z-1] + rnorm(1,0,1)
    }
    yt
    X <- yt[-n]
    beta <- solve(t(X) %*% X) %*% t(X) %*% yt[-1] # Estimate AR(1) coeff(beta)
    ANS[i] <- sort(beta) 
    cat(M," ","\n")
    if (beta[500] - 0.9 <= 0.01) {
      breakout <- TRUE
      break
    }
  }
  if (breakout) break
}

你还没有分享你的错误，所以我会显示：

Error in if (beta[500] - 0.9 <= 0.01) break (from #12) : 
  missing value where TRUE/FALSE needed

这是为什么？当beta 只是一个数字时，您正在引用beta[500]。您应该手动跟踪代码以查找错误发生的位置，因为这是缩小错误范围的最快方法。

M <- 0.01
i <- 1
    yt[1] <- rnorm(1,0,1)
    for (z in 2:n) { # simulate AR(1)=0.9+M loop 
      yt[z] <- (0.9 + M) * yt[z-1] + rnorm(1,0,1)
    }
    X <- yt[-n]
    beta <- solve(t(X) %*% X) %*% t(X) %*% yt[-1] # Estimate AR(1) coeff(beta)
beta
#          [,1]
# [1,] 0.890663
    ANS[i] <- sort(beta) 
    cat(M," ","\n")
# 0.01   
    if (beta[500] - 0.9 <= 0.01) break 
# Error in if (beta[500] - 0.9 <= 0.01) break : 
#   missing value where TRUE/FALSE needed
#     x

那为什么会这样呢？

beta[500] - 0.9 <= 0.01
# [1] NA
beta[500]
# [1] NA
beta
#          [,1]
# [1,] 0.890663

那为什么会这样呢？好吧，即使i 是 1000，beta 也是矩阵缩减为 1x1 矩阵。你的条件beta[500] 的前提被打破了。我不知道你认为它应该与什么进行比较，但那是行不通的。

修正数学背后的逻辑，然后逐步运行代码以确保尺寸/长度等是您想象的那样。