6

我正在尝试存储下面代码的结果，但是我只能想出一个解决方案来保存具有最小残差平方和的模型结果。这在结果处于 c 和 gamma 范围内之前很有用，因此我需要评估其他点的特征。为此，我需要存储每次迭代的结果。有谁知道在这种情况下该怎么做？

提前致谢！

dlpib1 <- info$dlpib1
scale <- sqrt(var(dlpib1))
RSS.m <- 10

for (c in seq(-0.03,0.05,0.001)){
  for (gamma in seq(1,100,0.2))
    {
    trans <- (1+exp(-(gamma/scale)*(dlpib1-c)))^-1
    grid.regre <-lm(dlpib ~ dlpib1 + dlpib8 + trans + trans*dlpib1 + 
                  + I(trans*dlpib4) ,data=info) 
coef <- grid.regre$coefficients
RSS <- sum(grid.regre$residuals^2)

if (RSS < RSS.m){
  RSS.m <- RSS
  gamma.m <- gamma
  c.m <- c
  coef.m <- coef
  }
 }
}
grid <- c(RSS=RSS.m,gamma=gamma.m,c=c.m,coef.m)
grid`
4

3 回答 3

10

通过迭代存储模型结果的最简单方法是list：

List = list()
for(i in 1:100)
    {
       LM = lm(rnorm(10)~rnorm(10))
       List[[length(List)+1]] = LM
     }
于 2013-09-27T18:39:32.543 回答
3

您可能可以for完全避免循环。但是，至于如何完成您的任务，您只需要索引存储值的任何对象。例如，

# outside the for loop
trans <- list()

# inside the for loop
trans[[paste(gamma, c, sep="_")]] <- ...
于 2013-09-27T18:40:11.390 回答
1

我很确定会保存 RSS 的所有迭代，您可以执行以下操作：

dlpib1 <- info$dlpib1
    scale <- sqrt(var(dlpib1))
    RSS.m <- rep(0,N)
    coef <- rep(0,N)
    i <- 0

    for (c in seq(-0.03,0.05,0.001)){
      for (gamma in seq(1,100,0.2))
        {
        trans <- (1+exp(-(gamma/scale)*(dlpib1-c)))^-1
        grid.regre <-lm(dlpib ~ dlpib1 + dlpib8 + trans + trans*dlpib1 + 
                      + I(trans*dlpib4) ,data=info) 
    coef <- grid.regre$coefficients
    RSS.m[i] <- sum(grid.regre$residuals^2)
    i=i+1


      }
     }
    }
于 2013-09-27T18:53:44.740 回答