如何在R中创建月度期间

Question

我有每月的数据，我想为该期间添加另一列。专栏会说M01代表一月，M02代表二月，M03代表三月，依此类推。有办法做到这一点吗？

这就是我所拥有的：

unemployment = data.frame(Month = c("Sept 2002", "Oct 2002", "Nov 2002", "Dec 2002", "Jan 2003", "Feb 2003"), 
Total = c(5.7, 5.7, 5.9, 
6, 5.8, 5.9))
> unemployment
Month Total
1 Sept 2002   5.7
2  Oct 2002   5.7
3  Nov 2002   5.9
4  Dec 2002   6.0
5  Jan 2003   5.8
6  Feb 2003   5.9

这就是我想要的：

Month Period Total
1 Sept 2002 M09      5.7
2  Oct 2002 M10      5.7
3  Nov 2002 M11      5.9
4  Dec 2002 M12      6.0
5  Jan 2003 M01      5.8
6  Feb 2003 M02      5.9

编辑更新代码以显示所有12个月的

structure(list(Month = c("Jan", "Feb", "Mar", "Apr", "May", "June"
), Year = c("2003", "2003", "2003", "2003", "2003", "2003"), 
Unemp_percent = c(5.8, 5.9, 5.9, 6, 6.1, 6.3)), row.names = 5:10, class = "data.frame")

Answer 1

使用dplyr:

unemployment %>% 
mutate(Period = case_when(grepl("Jan",Month) ~ "M01",
grepl("Feb",Month) ~ "M02",
grepl("Mar",Month) ~ "M03",
grepl("Apr",Month) ~ "M04",
grepl("May",Month) ~ "M05",
grepl("June",Month) ~ "M06",
grepl("July",Month) ~ "M07",
grepl("Aug",Month) ~ "M08",
grepl("Sept",Month) ~ "M09",
grepl("Oct",Month) ~ "M10",
grepl("Nov",Month) ~ "M11",
grepl("Dec",Month) ~ "M12"))

Answer 2

您可以使用内置month.abb数据的前3个字母mutate、gsub和match

library(dplyr)
unemployment |> 
mutate(.after = Month,  
Period = paste0("M", match(gsub("(.{3})(.*)", "\1", Month ), month.abb)))

Month Period Total
1 Sept 2002     M9   5.7
2  Oct 2002    M10   5.7
3  Nov 2002    M11   5.9
4  Dec 2002    M12   6.0
5  Jan 2003     M1   5.8
6  Feb 2003     M2   5.9

Answer 3

这里有另一种选择：

unemployment %>% 
mutate(month = gsub("(^.{3}).*", "\1", Month),
Period = paste0("M", as.numeric(factor(x$month, month.abb)))) %>% 
select(Month, Period, Total)

输出：

Month Period Total
1 Sept 2002     M9   5.7
2  Oct 2002    M10   5.7
3  Nov 2002    M11   5.9
4  Dec 2002    M12   6.0
5  Jan 2003     M1   5.8
6  Feb 2003     M2   5.9

Answer 4

只需制作一个月和期间的参考表，然后制作left_join。

library(tidyverse)
months <- data.frame(mo = c("Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sept", "Oct", "Nov", "Dec"),
Period = paste0("M", 1:12))
unemployment %>%
mutate(mo = str_extract(Month, "^[:alpha:]+")) %>%
left_join(months, by = "mo")
Month Total   mo Period
1 Sept 2002   5.7 Sept     M9
2  Oct 2002   5.7  Oct    M10
3  Nov 2002   5.9  Nov    M11
4  Dec 2002   6.0  Dec    M12
5  Jan 2003   5.8  Jan     M1
6  Feb 2003   5.9  Feb     M2

Answer 5

如果Month列的第4个字符是一个提供标准3个月缩写的单词字符，则删除该字符。然后将其转换为yearmon对象，并以所需的方式对其进行格式化。最后，我们将Period列重新定位到Month列之后。如果Period列位于Month列之后并不重要，则省略最后一行代码。

library(dplyr, exclude = c("filter", "lag"))
library(zoo)
unemployment %>%
mutate(Period = format(as.yearmon(sub("^(...)\w?", "\1", Month)), "M%m")) %>%
relocate(Period, .after = Month)

给予：

Month Period Total
1 Sept 2002    M09   5.7
2  Oct 2002    M10   5.7
3  Nov 2002    M11   5.9
4  Dec 2002    M12   6.0
5  Jan 2003    M01   5.8
6  Feb 2003    M02   5.9