Question

如何解释R的面板数据模型的结果？对于我的数据，我估计了Koenker（2004）关于面板数据的分位数回归方法的改编形式：

   rq.fit.panel <- function(X,Y,s,w,taus,lambda)

   {
require(SparseM)
    require(quantreg)

 K <- length(w)
if(K != length(taus))
stop("length of w and taus must match")
 X <- as.matrix(X)
     p <- ncol(X)
     n <- length(levels(as.factor(s)))
     N <- length(y)
if(N != length(s) || N != nrow(X))
stop("dimensions of y,X,s must match")
     Z <- as.matrix.csr(model.matrix(~as.factor(s)-1))
     Fidelity <- cbind(as(w,"matrix.diag.csr") %x% X,w %x% Z)
     Penalty <- cbind(as.matrix.csr(0,n,K*p),lambda*as(n,"matrix.diag.csr"))
     D <- rbind(Fidelity,Penalty)
     y <- c(w %x% y,rep(0,n))
 a <- c((w*(1-taus)) %x% (t(X)%*%rep(1,N)),
 sum(w*(1-taus)) * (t(Z) %*% rep(1,N)) + lambda * rep(1,n))
 rq.fit.sfn(D,y,rhs=a)

} enter code here

 bdeduc2<-read.table("dados_rq.txt", header=T) 
 z<-c("inter","ne","no","su","co")
 X<-bdeduc2[,z]
 y<-bdeduc2$scoreedu
 s<-bdeduc2$uf
 w<-c(0.1,0.25,0.5,0.25,0.1)
taus<-c(0.1,0.25,0.5,0.75,0.9)
lambda<-1

但我不知道下面的结果：

$coef
 [1]  1.02281339 -0.18750668 -0.13688807 -0.04180458 -0.01367417  1.02872440 -0.18055062 -0.13003224 -0.03829135 -0.01409369  1.03377335 -0.16649845 -0.11669812
[14] -0.03854060 -0.01438620  1.03851101 -0.15328087 -0.10440359 -0.03871744 -0.01465492  1.04330584 -0.14660960 -0.09670756 -0.03465501 -0.01430647 -0.29187982
[27] -0.21831160 -0.11295134 -0.21530494 -0.15664777 -0.13840296 -0.03224749 -0.11692122 -0.11237144 -0.15112171 -0.10385352 -0.08385934 -0.16090525 -0.30349309
[40] -0.16121494 -0.03106264 -0.16299994 -0.03182579 -0.22271685 -0.08251486 -0.29031224 -0.19680023 -0.20004209 -0.05601186 -0.21140762 -0.04254752 -0.01864703

$ierr
[1] 0

$it
[1] 16

$time
[1] 0

##summary rq

 summary(rq)

     Length Class  Mode   
coef 52     -none- numeric
ierr  1     -none- numeric
it    1     -none- numeric
time  1     -none- numeric

Answer 1

看起来你适合回归并保存它，然后试图在没有加载分位数回归包的新会话中查看它（它给你列表摘要，而不是包中的对象摘要）

确保已加载用于创建对象的包，然后再次执行摘要以查看是否为您提供了有意义的输出。

如何解释R的分位数回归面板数据模型的结果

1 个答案: