从多变量 tsibble 中删除季节和趋势

Question

我正在使用 stl() 函数从多变量时间序列中删除季节和趋势。然后我想保留 stl 分解的余数项，以便对残差执行 ARIMA() 分析。

我正在尝试以一种整洁的方式对此进行编码，因为我将扩展到大量时间序列。这里我举一个3系列的例子。我的问题是我无法将 stl 应用于 df 除非我 filter 应用于特定的单变量时间序列。

library(fable)
#> Loading required package: fabletools
library(tidyverse)
library(feasts)

df <- tibble(
  X = rnorm(71), Y = rnorm(71), Z = rnorm(71),
  quarter = seq(as.Date("2001/4/1"),
    as.Date("2018/10/1"),
    by = "quarter"
  )
) %>%
  mutate(quarter = tsibble::yearquarter(quarter)) %>%
  gather(-quarter, key = "Series", value = "value") %>%
  as_tsibble(index = quarter, key = Series)

# Detrend into seasonal, trend and remainder components

# Applying stl to a univariate series using filter works fine

df_stl <- df %>%
  filter(Series == "Z") %>%
  stl(s.window = "per", t.window = 1000)

df_stl <- df_stl[["time.series"]][,'remainder'] %>%
  as_tsibble()

# Applying stl to a multivariate series fails
df %>%
  stl(s.window = "per", t.window = 1000)
#> Error in stl(., s.window = "per", t.window = 1000): only univariate series are allowed

^{由 reprex package (v0.3.0)}

于 2019-11-14 创建

Answer 1

使用feasts包中的STL()函数：

df_stl <- df %>%
  STL(value ~ season("periodic") + trend(window=1000))

从多变量 tsibble 中删除季节和趋势

Remove season and trend from multivariate tsibble

r

time-series

forecasting