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R 中的 rmse function 问题

[英]rmse function issue in R

I have an R code that contains some nested bracket for loop within which I used rmse() function from Metrics package.我有一个R代码,其中包含一些用于循环的嵌套括号,我在其中使用了来自Metrics package 的rmse() function。 I tried it without the function and it worked, but inside my nested R code it does not.我在没有 function 的情况下尝试了它并且它有效,但在我的嵌套R代码中它没有。

Here is what I desire to do with R这是我想用R做的事情

  1. I have generated a 50-time series dataset.我已经生成了 50 个时间序列数据集。
  2. I lice the same time series dataset into chunks of the following sizes: 2,3,...,48,49 making me have 48 different time series formed from step 1 above.我将相同的时间序列数据集分成以下大小的块: 2,3,...,48,49使我从上面的步骤 1 中形成了 48 个不同的时间序列。
  3. I divided each 48-time series dataset into train and test sets so I can use rmse function in Metrics package to get the Root Mean Squared Error (RMSE) for the 48 subseries formed in step 2.我将每个 48 个时间序列数据集划分为train集和test集,因此我可以使用Metrics rmse中的 rmse function 来获得在步骤 2 中形成的 48 个子序列的均方根误差 (RMSE)。
  4. The RMSE for each series is then tabulated according to their chunk sizes然后根据它们的块大小将每个系列的 RMSE 制成表格
  5. I obtained the best ARIMA model for each 48 different time series data set.我为每个 48 个不同的时间序列数据集获得了最好的ARIMA model。

My R code我的 R 代码

 # simulate arima(1,0,0)
 library(forecast)
 library(Metrics)
 n <- 50
 phi <- 0.5
 set.seed(1)
 wn <- rnorm(n, mean=0, sd=1)
    ar1 <- sqrt((wn[1])^2/(1-phi^2))
 for(i in 2:n){
   ar1[i] <- ar1[i - 1] * phi + wn[i]
 }
 ts <- ar1

 t<-length(ts)# the length of the time series
 li <- seq(n-2)+1 # vector of block sizes(i.e to be between 1 and n exclusively)

 RMSEblk<-matrix(nrow = 1, ncol = length(li))#vector to store block means
 colnames(RMSEblk)<-li
 for (b in 1:length(li)){
     l<- li[b]# block size
     m <- ceiling(t / l) # number of blocks
     blk<-split(ts, rep(1:m, each=l, length.out = t)) # divides the series into blocks
     singleblock <- vector() #initialize vector to receive result from for loop
     for(i in 1:10){
         res<-sample(blk, replace=T, 100) # resamples the blocks
         res.unlist<-unlist(res, use.names = F) # unlist the bootstrap series
         # Split the series into train and test set
         train <- head(res.unlist, round(length(res.unlist) * 0.6))
         h <- length(res.unlist) - length(train)
         test <- tail(res.unlist, h)

        # Forecast for train set
        model <- auto.arima(train)
        future <- forecast(test, model=model,h=h)
        nfuture <- as.numeric(out$mean) # makes the `future` object a vector
        # use the `rmse` function from `Metrics` package
        RMSE <- rmse(test, nn)
        singleblock[i] <- RMSE # Assign RMSE value to final result vector element i
    }
    #singleblock
    RMSEblk[b]<-mean(singleblock) #store into matrix
 }
 RMSEblk

The error I got我得到的错误

#Error in rmse(test, nn): unused argument (nn)
#Traceback:

But when I wrote但是当我写

library(forecast)

train <- head(ar1, round(length(ar1) * 0.6))
h <- length(ar1) - length(train)
test <- tail(ar1, h)
model <- auto.arima(train)
#forecast <- predict(model, h)
out <- forecast(test, model=model,h=h)
nn <- as.numeric(out$mean)
rmse(test, nn)

It did work它确实有效

Please point out what I am missing?请指出我错过了什么?

I am able to run your code after making two very small corrections in your for loop.在您的 for 循环中进行两次非常小的更正后,我能够运行您的代码。 See the two commented lines:请参阅两条注释行:

 for (b in 1:length(li)){
     l<- li[b]
     m <- ceiling(t / l)
     blk<-split(ts, rep(1:m, each=l, length.out = t))
     singleblock <- vector()
     for(i in 1:10){
         res<-sample(blk, replace=T, 100)
         res.unlist<-unlist(res, use.names = F)
         train <- head(res.unlist, round(length(res.unlist) * 0.6))
         h <- length(res.unlist) - length(train)
         test <- tail(res.unlist, h)

        model <- auto.arima(train)
        future <- forecast(test, model=model,h=h)
        nfuture <- as.numeric(future$mean) # EDITED: `future` instead of `out`
        RMSE <- rmse(test, nfuture) # EDITED: `nfuture` instead of `nn`
        singleblock[i] <- RMSEi
    }
    RMSEblk[b]<-mean(singleblock)
 }

It is possible that these typos did not result in errors because nn and out were defined in the global environment while you ran the for loop.这些拼写错误可能不会导致错误,因为在您运行 for 循环时, nnout是在全局环境中定义的。 A good debugging tip is to restart R and try to reproduce the problem.一个好的调试技巧是重新启动 R 并尝试重现问题。

Your code does not define nn.您的代码没有定义 nn。 Other code that works has nn.其他有效的代码有 nn。 To start code with clean slate use this line as first executable line:要使用干净的 slate 开始代码,请将此行用作第一个可执行行:

rm(list=ls())

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