我已经编写了一个函数来迭代预测使用包dyn构建的模型，我想要一些反馈 . 有一个更好的方法吗？是否有人为dyn类（或dynlm类）编写了规范的“预测”方法，或者我是否冒险进入未知领域？
ipredict <-function(model, newdata, interval = "none",
level = 0.95, na.action = na.pass, weights = 1) {
P<-predict(model,newdata=newdata,interval=interval,
level=level,na.action=na.action,weights=weights)
for (i in seq(1,dim(newdata)[1])) {
if (is.na(newdata[i])) {
if (interval=="none") {
P[i]<-predict(model,newdata=newdata,interval=interval,
level=level,na.action=na.action,weights=weights)[i]
newdata[i]<-P[i]
}
else{
P[i,]<-predict(model,newdata=newdata,interval=interval,
level=level,na.action=na.action,weights=weights)[i,]
newdata[i]<-P[i,1]
}
}
}
P_end<-end(P)[1]*frequency(P)+(end(P)[2]-1) #Convert (time,period) to decimal time
P<-window(P,end=P_end-1*frequency(P)) #Drop last observation, which is NA
return(P)
}
用法示例：
library(dyn)
y<-arima.sim(model=list(ar=c(.9)),n=10) #Create AR(1) dependant variable
A<-rnorm(10) #Create independant variables
B<-rnorm(10)
C<-rnorm(10)
Error<-rnorm(10)
y<-y+.5*A+.2*B-.3*C+.1*Error #Add relationship to independant variables
data=cbind(y,A,B,C)
#Fit linear model
model.dyn<-dyn$lm(y~A+B+C+lag(y,-1),data=data)
summary(model.dyn)
#Forecast linear model
A<-c(A,rnorm(5))
B<-c(B,rnorm(5))
C<-c(C,rnorm(5))
y=window(y,end=end(y)+c(5,0),extend=TRUE)
newdata<-cbind(y,A,B,C)
P1<-ipredict(model.dyn,newdata)
P2<-ipredict(model.dyn,newdata,interval="prediction")
#Plot
plot(y)
lines(P1,col=2)
1 回答
R的核心
predict.Arima有n.ahead参数来预测n前面的步骤，看起来这是你正在寻找的与dyn一起但predict.dyn目前不支持该功能 . 为了达到这个效果，必须按照你的方式迭代地调用dyn$whatever.