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我正在尝试为 ARIMA(p,0,q) 的所有组合生成拟合模型，例如，对于 p<=5 和 q<=5。

我的最终目标是根据 BIC、AIC 和 RMSE 等（使用精度计算）选择最佳模型。

使用循环方法，我能够获得所需的输出，

pdq_index_mat=expand.grid(0L:5L,0L,0L:5L)
# access the following link for the csv and run the following code to access the data for replication purposes for the following code/error
# https://drive.google.com/open?id=0ByLULem4QYiEcDB2OTJaejdVUkk
tsdf<-as.ts(read.csv(file.choose()))

#Loop method
ic_result_mat<-c()
reg<-c()
y<-tsdf[,"psoda"] #dependent variable of interest
for(i in 1:nrow(pdq_index_mat)){
  # print(as.numeric(pdq_index_mat[i,])) serve as the c(p,0,q) to indicate which ARIMA of interest
  reg<-arima(tsdf[,"psoda"],order=as.numeric(pdq_index_mat[i,]))
  ic_result_mat<-rbind(ic_result_mat,c(BIC=BIC(reg),AIC=AIC(reg),accuracy(reg)[,c("RMSE","MAE","MPE","MAPE")]))
  # uncomment the following line for the first error
  # colnames(tsdf)[ncol(tsdf)+i]<-paste0("ARIMA(",paste0(pdq_index_mat[i,],collapse=","),")")
}
colnames(ic_result_mat)<-c("BIC","AIC","RMSE","MAE","MPE","MAPE")
rownames(ic_result_mat)<-paste0("ARIMA(",do.call(paste,c(pdq_index_mat[1:nrow(ic_result_mat),],sep=",")),")")

除了 1 个缺陷，我无法在循环内生成行名（例如，第一行的 ARIMA(0,0,0)），并提示以下错误：

Error in `rownames<-`(`*tmp*`, value = c(NA, NA, NA, NA, "ARIMA(1,0,0)" : 
  length of 'dimnames' [1] not equal to array extent

或者

 Error in `rownames<-`(`*tmp*`, value = c("ARIMA(0)", "ARIMA(0)", "ARIMA(0)" : 
    attempt to set 'rownames' on an object with no dimensions

至于使用该apply()方法，我什至无法获得回归列表

#List method
ic_result_mat<-c()
y<-tsdf[,"psoda"] #dependent variable of interest
reg_lst<-apply(pdq_index_mat,1,function(i)arima(y,order=as.integer(pdq_index_mat[i,])))

并提示以下错误：

Error in if (!is.numeric(order) || length(order) != 3L || any(order <  : 
  missing value where TRUE/FALSE needed

对不起，不是专业的编码员。非常感谢专业人士的任何帮助。

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