Lubridate：如何减去一个月的最后一次观察

Question

我有一个时间序列，想获取每个月最后一次观察的信息。这个问题不是关于生成新的时间序列，而是在现有时间序列中找到每个月的最后一次观察。最后一次观察可能不是一个月的最后一天。下面只是一个小例子，

date <- c(ymd(20010129, 20010228, 20010330, 20010429), ymd(20010501) + days(1:90))

# "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29" "2001-05-02" "2001-05-03" "2001-05-04" "2001-05-05"
# "2001-05-06" "2001-05-07" "2001-05-08" "2001-05-09" "2001-05-10" "2001-05-11" "2001-05-12" "2001-05-13"
# "2001-05-14" "2001-05-15" "2001-05-16" "2001-05-17" "2001-05-18" "2001-05-19" "2001-05-20" "2001-05-21"
# "2001-05-22" "2001-05-23" "2001-05-24" "2001-05-25" "2001-05-26" "2001-05-27" "2001-05-28" "2001-05-29"
# "2001-05-30" "2001-05-31" "2001-06-01" "2001-06-02" "2001-06-03" "2001-06-04" "2001-06-05" "2001-06-06"
# "2001-06-07" "2001-06-08" "2001-06-09" "2001-06-10" "2001-06-11" "2001-06-12" "2001-06-13" "2001-06-14"
# "2001-06-15" "2001-06-16" "2001-06-17" "2001-06-18" "2001-06-19" "2001-06-20" "2001-06-21" "2001-06-22"
# "2001-06-23" "2001-06-24" "2001-06-25" "2001-06-26" "2001-06-27" "2001-06-28" "2001-06-29" "2001-06-30"
# "2001-07-01" "2001-07-02" "2001-07-03" "2001-07-04" "2001-07-05" "2001-07-06" "2001-07-07" "2001-07-08"
# "2001-07-09" "2001-07-10" "2001-07-11" "2001-07-12" "2001-07-13" "2001-07-14" "2001-07-15" "2001-07-16"
# "2001-07-17" "2001-07-18" "2001-07-19" "2001-07-20" "2001-07-21" "2001-07-22" "2001-07-23" "2001-07-24"
# "2001-07-25" "2001-07-26" "2001-07-27" "2001-07-28" "2001-07-29" "2001-07-30"

我想继续观察 "2001-01-29"、"2001-02-28"、"2001-03-30"、"2001-04-29"、"2001-05-31"、"2001-06-30" 和 "2001-07-30"。有办法实现吗？

Answer 1

您可以按月对日期进行分组并计算最大值：

library(lubridate)
unique(ave(date, month(date), FUN = max))

# [1] "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29"
# [5] "2001-05-31" "2001-06-30" "2001-07-30"

Answer 2

我们可以使用data.table。将 'date' 向量转换为 data.table，按 'date' 的 year 和 month 分组，我们得到 'date' 的 max。

library(data.table)
as.data.table(date)[, .(Date = max(date)), .(Year = year(date), Month = month(date))]
#   Year Month       Date
#1: 2001     1 2001-01-29
#2: 2001     2 2001-02-28
#3: 2001     3 2001-03-30
#4: 2001     4 2001-04-29
#5: 2001     5 2001-05-31
#6: 2001     6 2001-06-30
#7: 2001     7 2001-07-30

或者使用 base R 和基于 tapply 的直接方法，而不是获取与原始向量长度相同的向量，然后采用 unique.

do.call("c", tapply(date, list(month(date), year(date)), 
                FUN = function(x) list(max(x))))
#[1] "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29" "2001-05-31" 
#[6] "2001-06-30" "2001-07-30"

或者简明扼要

 unname(as.Date(tapply(date, substr(date, 1,7), FUN = max), origin = "1970-01-01"))
 #[1] "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29" "2001-05-31" 
 #[6] "2001-06-30" "2001-07-30"

此外，我们可以通过检查相邻元素（假设它是有序的）来获得没有任何分组的输出，它应该是非常有效的。

v1 <- substr(date, 1, 7)
date[c(v1[-1]!= v1[-length(v1)], TRUE)]
[1] "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29" "2001-05-31" 
[6] "2001-06-30" "2001-07-30"

基准

date1 <- c(ymd(20010129, 20010228, 20010330, 20010429), ymd(20010501) + days(1:1e6))
system.time(as.data.table(date1)[, .(Date = max(date1)), 
      .(Year = year(date1), Month = month(date1))])
#   user  system elapsed 
#   5.53    0.05    5.58  


system.time({
 v1 <- substr(date1, 1, 7)
 date1[c(v1[-1]!= v1[-length(v1)], TRUE)]
})
# user  system elapsed 
#  10.25    0.23   10.49

基于上述性能，data.table方法非常有效，虽然base R相邻元素之间的比较也没有那么落后，但闪光的不是金子。

system.time(unique(ave(date1, year(date1), month(date1), FUN = max)))
#   user  system elapsed 
# 242.35  120.80  364.55

Answer 3

endpoints xts 包中的一个函数完全符合其名称的含义：

> date[endpoints(date,on='months')]
[1] "2001-01-29" "2001-02-28" "2001-03-30" "2001-04-29" "2001-05-31"
[6] "2001-06-30" "2001-07-30”

参数的有效值包括：“us”（微秒）、“微秒”、“ms”（毫秒）、“毫秒”、“secs”（秒）、“秒”、“分钟”（分钟）、“分钟”、“小时”、“天”、“周”、“月”、“季度”和“年”。

Lubridate：如何减去一个月的最后一次观察

Lubridate: How to subtract the last observation of a month

r

time-series

lubridate

基准