r - R中多个时间序列的霍尔特预测
问题描述
我正在尝试对多个时间序列进行 Holt 的预测,并将它们与我的原始 data.frame 结合起来。考虑以下data.frame,其中我有两个人口组:
library("forecast")
d <- data.frame(SEX = c("MALE","MALE","MALE","FEMALE","FEMALE","FEMALE"),
EDUCATION = c("01","01","01","01","01","01"),
TIME = c("2000","2001","2002","2000","2001","2002"),
VALUE = c(120,150,140,90,75,60))
然后我正在对两个时间序列进行 Holt 预测:
male <- ts(as.numeric(d[1:3,]$VALUE),start=c(2000))
female <- ts(as.numeric(d[4:6,]$VALUE),start=c(2000))
forecastmale <- holt(male,h = 3,damped = FALSE)
forecastfemale <- holt(female,h = 3,damped = FALSE)
然后我保存结果并与我的原始 data.frame 结合:
forecastmale <- data.frame(forecastmale[["mean"]])
forecastfemale <- data.frame(forecastfemale[["mean"]])
forecastmale$SEX <- c("MALE","MALE","MALE")
forecastmale$EDUCATION <- c("01","01","01")
forecastmale$TIME <- c("2003","2004","2005")
colnames(forecastmale)[1] <- "VALUE"
forecastmale <- forecastmale[, c(2,3,4,1)]
forecastfemale$SEX <- c("FEMALE","FEMALE","FEMALE")
forecastfemale$EDUCATION <- c("01","01","01")
forecastfemale$TIME <- c("2003","2004","2005")
colnames(forecastfemale)[1] <- "VALUE"
forecastfemale <- forecastfemale[, c(2,3,4,1)]
d <- rbind(d,forecastmale,forecastfemale)
这在我只有两个时间序列时有效。但是,如果我有 100 个需要预测的时间序列,那么这不是一种非常有效的方法。任何人都可以帮助提高编码器的效率,例如,如果我在我的 data.frame 中包含一个额外的人口组,那么我在代码中没有任何更改?
解决方案
这就是fable
包的设计目的。这是一个使用与您拥有的相同数据结构的示例。
library(dplyr)
library(tsibble)
library(fable)
# Artifical data
df <- expand.grid(
education = 1:3,
sex = c("male","female"),
year = 1990:2002
) %>%
as_tsibble(index=year, key=c(sex,education)) %>%
mutate(value = rnorm(78))
# Fit Holt's method to each series and forecast 3 years ahead
df %>%
model(holt = ETS(value ~ trend("A"))) %>%
forecast(h=3)
#> # A fable: 18 x 6 [1Y]
#> # Key: sex, education, .model [6]
#> sex education .model year value .mean
#> <fct> <int> <chr> <dbl> <dist> <dbl>
#> 1 male 1 holt 2003 N(0.14, 1.7) 0.137
#> 2 male 1 holt 2004 N(0.17, 1.7) 0.171
#> 3 male 1 holt 2005 N(0.21, 1.7) 0.205
#> 4 male 2 holt 2003 N(-0.75, 1.5) -0.749
#> 5 male 2 holt 2004 N(-0.84, 1.8) -0.837
#> 6 male 2 holt 2005 N(-0.93, 2) -0.926
#> 7 male 3 holt 2003 N(0.51, 0.7) 0.514
#> 8 male 3 holt 2004 N(0.53, 0.7) 0.530
#> 9 male 3 holt 2005 N(0.55, 0.7) 0.546
#> 10 female 1 holt 2003 N(0.44, 0.98) 0.445
#> 11 female 1 holt 2004 N(0.47, 0.98) 0.470
#> 12 female 1 holt 2005 N(0.5, 0.98) 0.495
#> 13 female 2 holt 2003 N(0.13, 0.89) 0.127
#> 14 female 2 holt 2004 N(0.15, 0.89) 0.148
#> 15 female 2 holt 2005 N(0.17, 0.89) 0.168
#> 16 female 3 holt 2003 N(0.78, 1.8) 0.781
#> 17 female 3 holt 2004 N(0.88, 1.8) 0.880
#> 18 female 3 holt 2005 N(0.98, 1.8) 0.978
由reprex 包(v0.3.0)于 2020 年 9 月 5 日创建
推荐阅读
- sql - 使用分页显示数据 - 一次获取整个数据或获取每个页面的一部分?
- javascript - 不能在 Selenium (Javascript) 中使用 findElement
- javascript - 具有特定类的元素数组在 setTimeout 中变得未定义
- amazon-web-services - SNS 是否允许基于 String.array 中存在的多个值进行过滤
- ruby-on-rails - 这个验证有什么作用?
- c++ - Haversine 公式 - 数学略有偏差,不确定原因
- file - 如何下载链接中的所有文件
- apache-spark - 如何在 virtualenv 中为 pyspark 运行 spark-submit?
- python - Azure 功能:我可以实施我的体系结构吗?如何最大限度地降低成本?
- xml - XSLT 1.0 (xsltproc) - 无法解析巨大的 XML