Question

给定比利时电力系统 2010-01-04 至 2016-10-30 期间限价的小时时间序列。

我的数据有以下信息：

datetime_utc：记录日期和时间 Generation_BE：比利时计划发电 Generation_FR：法国计划发电 Prices.BE：比利时假期 OTS 电力BE：比利时国定假日（1 如果是，0 如果不是）

>

训练数据：

> dput(head(data1))
structure(list(datetime_utc = c("2010-01-04 00:00:00", "2010-01-04 01:00:00", 
"2010-01-04 02:00:00", "2010-01-04 03:00:00", "2010-01-04 04:00:00", 
"2010-01-04 05:00:00"), Generation_BE = c(13143.7, 13143.7, 13143.7, 
13143.7, 13143.7, 13143.7), Generation_FR = c(63599, 62212, 62918, 
62613, 62432, 63411), Prices.BE = c(37.15, 33.47, 28, 21.29, 
16.92, 28), holidaysBE = c(0L, 0L, 0L, 0L, 0L, 0L)), row.names = c(NA, 
6L), class = "data.frame")

测试数据：

> dput(head(data2))
structure(list(datetime_utc = c("2016-10-24 00:00:00", "2016-10-24 01:00:00", 
"2016-10-24 02:00:00", "2016-10-24 03:00:00", "2016-10-24 04:00:00", 
"2016-10-24 05:00:00"), Generation_BE = c(9615.7075, 9626.865, 
9648.0025, 9668.42, 9681.805, 9688.425), Generation_FR = c(45605L, 
44629L, 44073L, 44359L, 44056L, 44799L), Prices.BE = c(44.6, 
40.92, 37.39, 36.4, 33.01, 37.89), holidaysBE = c(0L, 0L, 0L, 
0L, 0L, 0L)), row.names = c(NA, 6L), class = "data.frame")

我已经为 Prices.BE 时间序列构建了时间序列交叉验证，我想预测 Prices.BE 7 天的时间段，因此对于 168 次观察，因为我们有每小时数据。

代码：

> k <- 168 
> n <- length(data1)
> mae1 <- mae2 <- matrix(NA,n-k,12)
Error in matrix(NA, n - k, 12) : invalid 'nrow' value (< 0)
> st <- tsp(data1)[1]+(k-2)/12
> 
> for(i in 1:(n-k))
+ {
+   xshort <- window(data1, end=st + i/12)
+   xnext <- window(data1, start=st + (i+1)/12, end=st + (i+12)/12)
+   fit1 <- Arima(xshort, order=c(3,0,1), seasonal=list(order=c(0,1,1), period=12),
+                 include.drift=TRUE, lambda=0, method="ML")
+   fcast1 <- forecast(fit1, h=12)
+   fit2 <- ets(xshort,model="MMM",damped=TRUE)
+   fcast2 <- forecast(fit2, h=12)
+   mae1[i,1:length(xnext)] <- abs(fcast1[['mean']]-xnext)
+   mae2[i,1:length(xnext)] <- abs(fcast2[['mean']]-xnext)
+ }
Error in attr(x, "tsp") <- c(1, NROW(x), 1) : 
  invalid time series parameters specified
> 
> plot(1:12, colMeans(mae1,na.rm=TRUE), type="l", col=2, xlab="horizon", ylab="MAE",
+      ylim=c(0.65,1.05))
Error in h(simpleError(msg, call)) : 
  error in evaluating the argument 'y' in selecting a method for function 'plot': object 'mae1' not found
> lines(1:12, colMeans(mae2,na.rm=TRUE), type="l",col=3)
> legend("topleft",legend=c("ARIMA","ETS"),col=2:3,lty=1)

正如您所看到的代码显示了一些错误，我无法找到有关它们的解决方案，欢迎任何帮助！

时间序列预测的交叉验证

0 个答案: