具有分组数据 R 的 ARIMAX

Question

我是 R 的新手，我需要比较 ARIMAX 和 ARIMA 的准确性。这是我的数据示例以及我为 ARIMA 模型所做的工作：

library(dplyr)
library(forecast)
library(lubridate)
data<-tibble::tribble(
        ~id, ~day, ~month, ~year, ~value, ~reg1, ~reg2,
         1L,   1L,     1L, 2019L,  4.634, 0.626, 0.684,
         1L,   1L,     2L, 2019L,  2.969, 0.698, 0.049,
         1L,   1L,     3L, 2019L,  1.885,  0.62, 0.155,
         1L,   1L,     4L, 2019L,  2.415, 0.553, 0.959,
         1L,   1L,     5L, 2019L,  2.215, 0.598, 0.065,
         1L,   1L,     6L, 2019L,  1.805, 0.454,  0.07,
         1L,   1L,     7L, 2019L,  4.682, 0.045, 0.376,
         1L,   1L,     8L, 2019L,  4.248, 0.087, 0.094,
         1L,   1L,     9L, 2019L,   0.55, 0.523,  0.86,
         1L,   1L,    10L, 2019L,  0.109, 0.176, 0.591,
         2L,   1L,     1L, 2019L,  2.918, 0.442, 0.956,
         2L,   1L,     2L, 2019L,  3.083, 0.233, 0.388,
         2L,   1L,     3L, 2019L,  3.271, 0.652, 0.946,
         2L,   1L,     4L, 2019L,  2.175, 0.704, 0.902,
         2L,   1L,     5L, 2019L,   4.51, 0.851, 0.533,
         2L,   1L,     6L, 2019L,  4.178, 0.655, 0.614,
         2L,   1L,     7L, 2019L,  1.956, 0.434, 0.977,
         2L,   1L,     8L, 2019L,  3.219, 0.418,   0.4,
         2L,   1L,     9L, 2019L,   2.72, 0.335, 0.096,
         2L,   1L,    10L, 2019L,  4.519, 0.534, 0.388,
         3L,   1L,     1L, 2019L,  2.969, 0.707, 0.752,
         3L,   1L,     2L, 2019L,  2.456, 0.085, 0.651,
         3L,   1L,     3L, 2019L,  0.418, 0.851, 0.399,
         3L,   1L,     4L, 2019L,  2.324, 0.626, 0.317,
         3L,   1L,     5L, 2019L,  3.548, 0.175, 0.081,
         3L,   1L,     6L, 2019L,   3.74, 0.667, 0.691,
         3L,   1L,     7L, 2019L,   4.48, 0.853, 0.259,
         3L,   1L,     8L, 2019L,   0.18, 0.016, 0.489,
         3L,   1L,     9L, 2019L,  3.028,  0.51, 0.741,
         3L,   1L,    10L, 2019L,  4.652, 0.916, 0.953
        )
data<-data %>% 
  mutate(date=as.character(make_date(year,month,day)),YearMonth = tsibble::yearmonth((ymd(date)))) %>%
  as_tsibble(key=id,index = YearMonth)


fit <- data %>% 
  filter(YearMonth <= yearmonth("2019 Aug")) %>%
  model(ARIMA(value ~ PDQ(0,0,0), stepwise=FALSE, approximation=FALSE))

# Now forecast the test set and compute RMSE and MSE
fit %>%
  forecast(h = 2) %>%
  accuracy(data)

现在我需要使用 ARIMAX 执行此操作：

covariates <- c("reg1","reg2")
fit_arimax <- data %>% 
  filter(YearMonth <= yearmonth("2019 Aug")) %>%
  group_by(id) %>% 
  do(autoarima=auto.arima(.$value,xreg = as.matrix(data[,covariates])))

我收到以下错误：

Error in model.frame.default(formula = x ~ xregg, drop.unused.levels = TRUE) : 
  variable lengths differ (found for 'xregg')

In addition: Warning message: In !is.na(x) & !is.na(rowSums(xreg)) : longer object length is not a multiple of shorter object length

我看到了但我做不到，因为我是 R 的初学者。所以我想知道 ARIMA 是否有一些与回归器一起工作的东西或者如何用 auto.arima 解决它，然后比较 ARIMA 和 ARIMAX 中的 ID 的准确性。有谁知道怎么做？谢谢！

Answer 1

您已从使用 tsibble 和 fable 包切换到使用 forecast 包。它们使用不同的数据结构，一般不应混用。

您可以使用寓言轻松地拟合具有 ARIMA 误差的回归模型，如下所示。

fit_arimax <- data %>% 
  filter(YearMonth <= yearmonth("2019 Aug")) %>%
  model(
    ARIMA(value ~ reg1 + reg2 + PDQ(0,0,0))
  ) 

fc <- fit_arimax %>%
  forecast(new_data = filter(data, YearMonth > yearmonth("2019 Aug"))) 
fc %>% accuracy(data)

请注意，这实际上不是 ARIMAX 模型——参见 https://robjhyndman.com/hyndsight/arimax/

具有分组数据 R 的 ARIMAX

ARIMAX with grouped data R

r

time-series

arima