Question

我想模拟一些数据并返回Beta 1和Beta 2的不同估计值。我想使用复制或类似功能作为lappy，sapply。

beta = matrix(data = rep(NA, 200), ncol = 2)

colnames(beta) = c("Beta 1", "Beta 2")

beta = replicate (100,{

  x = matrix(data = rep(NA, 300), ncol = 3)

  colnames(x) = c("y", "x1", "x2")

  x$y = rnorm(100,2,4)

  x$x1 = rnorm(100, 4,6)

  x$x2 = rnorm(100,  5,7)

  lm = lm(y ~ x1 + x2, data = x)

  coef(lm)[2]  ; coef(lm)[3]  
})

以上仅返回100个重复的一个beta

Answer 1

    beta = replicate (100,{
                      x = matrix(data = rep(NA, 300), ncol = 3)
                      colnames(x) = c("y", "x1", "x2")
                      x = data.frame(y=rnorm(100,2,4),
                                     x1=rnorm(100, 4,6),
                                     x2=rnorm(100, 5,7))
                      lm = lm(y ~ x1 + x2, data = x)

                     beta=c(Beta1=coef(lm)[2],Beta2=coef(lm)[3])

                      })
                    beta=t(beta)

Answer 2

使用sapply：

res <- sapply(1:100, function(i){
  x = matrix(data = rep(NA, 300), ncol = 3)
  colnames(x) = c("y", "x1", "x2")
  x$y = rnorm(100,2,4)
  x$x1 = rnorm(100, 4,6)
  x$x2 = rnorm(100,  5,7)
  lm = lm(y ~ x1 + x2, data = x)

  return(c(coef(lm)[2], coef(lm)[3]))
})

head(t(res))
#               x1           x2
# [1,]  0.01293551 -0.002520876
# [2,]  0.00294546 -0.034689928
# [3,]  0.05614309  0.049577728
# [4,] -0.02323345 -0.025082705
# [5,] -0.01016728  0.023558755
# [6,] -0.05317357  0.024496896

在R中有复制函数返回R中的几个向量或矩阵

2 个答案: