我想准备我的时间序列，
这里是初始dput()
df=structure(list(group = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L), year = c(1973L, 1974L, 1975L, 1976L, 1977L, 1978L,
1973L, 1974L, 1975L, 1976L, 1977L, 1978L), Jan = c(9007L, 7750L,
8162L, 7717L, 7792L, 7836L, 9007L, 7750L, 8162L, 7717L, 7792L,
7836L), Feb = c(8106L, 6981L, 7306L, 7461L, 6957L, 6892L, 8106L,
6981L, 7306L, 7461L, 6957L, 6892L), Mar = c(8928L, 8038L, 8124L,
7767L, 7726L, 7791L, 8928L, 8038L, 8124L, 7767L, 7726L, 7791L
), Apr = c(9137L, 8422L, 7870L, 7925L, 8106L, 8192L, 9137L, 8422L,
7870L, 7925L, 8106L, 8192L), May = c(10017L, 8714L, 9387L, 8623L,
8890L, 9115L, 10017L, 8714L, 9387L, 8623L, 8890L, 9115L), Jun = c(10826L,
9512L, 9556L, 8945L, 9299L, 9434L, 10826L, 9512L, 9556L, 8945L,
9299L, 9434L), Jul = c(11317L, 10120L, 10093L, 10078L, 10625L,
10484L, 11317L, 10120L, 10093L, 10078L, 10625L, 10484L), Aug = c(10744L,
9823L, 9620L, 9179L, 9302L, 9827L, 10744L, 9823L, 9620L, 9179L,
9302L, 9827L), Sep = c(9713L, 8743L, 8285L, 8037L, 8314L, 9110L,
9713L, 8743L, 8285L, 8037L, 8314L, 9110L), Oct = c(9938L, 9129L,
8466L, 8488L, 8850L, 9070L, 9938L, 9129L, 8466L, 8488L, 8850L,
9070L), Nov = c(9161L, 8710L, 8160L, 7874L, 8265L, 8633L, 9161L,
8710L, 8160L, 7874L, 8265L, 8633L), Dec = c(8927L, 8680L, 8034L,
8647L, 8796L, 9240L, 8927L, 8680L, 8034L, 8647L, 8796L, 9240L
)), .Names = c("group", "year", "Jan", "Feb", "Mar", "Apr", "May",
"Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"), class = "data.frame", row.names = c(NA,
-12L))
所以我就像过去的主题中那样
load_pkgs <- c("forecast", "zoo", "timetk", "tidyverse")
sapply(load_pkgs, function(x) suppressPackageStartupMessages(library(x, character.only = T)))
ld <- split(df[, -1], df$group)
# Tidy-up the splits
ld <- lapply(ld, function(x) {
x %>%
gather(key, value, -year) %>%
unite(date, year, key, sep = "-") %>%
mutate(date = paste0(date, "-01")) %>%
mutate(date = as.Date(date, format = "%Y-%b-%d"))
})
结果我得到：
$`1`
date value
1 <NA> 9007
2 <NA> 7750
3 <NA> 8162
4 <NA> 7717
但实际上我必须得到如下结果：
date value
1 1973-01-01 9007
2 1974-01-01 7750
3 1975-01-01 8162
4 1976-01-01 7717
5 1977-01-01 7792
6 1978-01-01 7836
为什么我找不到它？
此代码在我的控制台中不起作用。我安装了所有库。
答案 0 :(得分：3)
同时，您可以尝试使用lubridate软件包的解决方法：
library(lubridate)
ld <- lapply(ld, function(x) {
x %>%
gather(key, value, -year) %>%
unite(date, year, key, sep = "-") %>%
mutate(date = paste0(date, "-01")) %>%
mutate(date =ymd(date)) # here you use it
})
结果：
$`1`
date value
1 1973-01-01 9007
2 1974-01-01 7750
3 1975-01-01 8162
4 1976-01-01 7717
5 1977-01-01 7792
...
$`2`
date value
1 1973-01-01 9007
2 1974-01-01 7750
3 1975-01-01 8162
4 1976-01-01 7717
5 1977-01-01 7792
...
答案 1 :(得分：1)
使用tidyverse和lubridate的不同（且可能更快）的解决方案：
lapply(ld, function(x) {
x %>%
gather(var, value, -year) %>%
mutate(date = ymd(paste(year, match(var, month.abb), "01", sep = "-"))) %>%
select(-year, -var)
})
$`1`
value date
1 9007 1973-01-01
2 7750 1974-01-01
3 8162 1975-01-01
4 7717 1976-01-01
5 7792 1977-01-01
6 7836 1978-01-01
7 8106 1973-02-01
8 6981 1974-02-01
9 7306 1975-02-01
10 7461 1976-02-01
$`2`
value date
1 9007 1973-01-01
2 7750 1974-01-01
3 8162 1975-01-01
4 7717 1976-01-01
5 7792 1977-01-01
6 7836 1978-01-01
7 8106 1973-02-01
8 6981 1974-02-01
9 7306 1975-02-01
10 7461 1976-02-01
首先，它正在重塑数据。然后，将年份，缩写的月份转换为数字，并将“ 01”粘贴为1，然后通过ymd（）将其转换为日期。