我想将R中的动物园数据汇总两个月，四个月或六个月。此类日期处理只有两个可用选项，使用：
a）as.yearmon =>处理按月分组的每日数据
b）as.yearqtr =>处理按3个月周期的固定组（jan-mar，apr-jun，jul-set和oct-dec）分组的每日数据。
library(zoo)
# creating a vector of Dates
dt = as.Date(c("2001-01-01","2001-01-02","2001-04-01","2001-05-01","2001-07-01","2001-10-01"),
"%Y-%m-%d")
# the original dates
dt
[1] "2001-01-01" "2001-01-02" "2001-04-01" "2001-05-01" "2001-07-01" "2001-10-01"
# conversion to monthly data
as.yearmon(dt)
[1] "jan 2001" "jan 2001" "abr 2001" "mai 2001" "jul 2001" "out 2001"
# conversion to quarterly data
as.yearqtr(dt)
[1] "2001 Q1" "2001 Q1" "2001 Q2" "2001 Q2" "2001 Q3" "2001 Q4"
set.seed(0)
# irregular time series
daily_db = zoo(matrix(rnorm(3 * length(dt)),
nrow = length(dt),
ncol = 3),
order.by = dt)
daily_db
2001-01-01 1.2629543 -0.928567035 -1.1476570
2001-01-02 -0.3262334 -0.294720447 -0.2894616
2001-04-01 1.3297993 -0.005767173 -0.2992151
2001-05-01 1.2724293 2.404653389 -0.4115108
2001-07-01 0.4146414 0.763593461 0.2522234
2001-10-01 -1.5399500 -0.799009249 -0.8919211
# data aggregated by month
aggregate(daily_db,as.yearmon,sum)
V1 V2 V3
jan 2001 0.9367209 -1.223287482 -1.4371186
abr 2001 1.3297993 -0.005767173 -0.2992151
mai 2001 1.2724293 2.404653389 -0.4115108
jul 2001 0.4146414 0.763593461 0.2522234
out 2001 -1.5399500 -0.799009249 -0.8919211
# data aggregated by quarter
aggregate(daily_db,as.yearqtr,sum)
V1 V2 V3
2001 Q1 0.9367209 -1.2232875 -1.4371186
2001 Q2 2.6022286 2.3988862 -0.7107260
2001 Q3 0.4146414 0.7635935 0.2522234
2001 Q4 -1.5399500 -0.7990092 -0.8919211
我想定义一个函数，如：
as.yearperiod = function(x, period = 6) {...} # convert dates in semesters
要使用这种方式：
# data aggregated by semester
aggregate(base_dados_diaria, as.yearperiod, period = 6, sum)
我期待像这样的结果：
V1 V2 V3
2001 S1 3.538950 1.175599 -2.147845
2001 S2 -1.125309 -0.035416 -0.639698
先生，我建议你使用lubridate包来处理自定义日期间隔。使用floor_date可以轻松完成任务，如下所示：
six_m_interval <- lubridate::floor_date( dt , "6 months" )
# [1] "2001-01-01" "2001-01-01" "2001-01-01" "2001-01-01" "2001-07-01" "2001-07-01"
aggregate( daily_db , six_m_interval , sum )
# V1 V2 V3
# 2001-01-01 3.538950 1.17559873 -2.1478445
# 2001-07-01 -1.125309 -0.03541579 -0.6396977
Date2period输入一个"Date"对象并返回一个表示句点（学期等）的字符串，这取决于参数period的值，该值应该是一个除数为12的数字。在内部它转换为yearmon然后提取年份和周期，即月份，并从那些生成所需的字符串。
Date2period <- function(x, period = 6, sep = " S") {
ym <- as.yearmon(x)
paste(as.integer(ym), (cycle(ym) - 1) %/% period + 1, sep = sep)
}
为了测试上述情况：
library(zoo)
# inputs
period <- 6
dt <- as.Date(c("2001-01-01","2001-04-01","2001-07-01","2001-10-01"))
Date2period(dt)
## [1] "2001 S1" "2001 S1" "2001 S2" "2001 S2"
aggregate(daily_db, Date2period, sum)
## V1 V2 V3
## 2001 S1 0.9367209 -1.125309 2.39888622
## 2001 S2 2.6022286 -1.223287 -0.03541579
以下是其他转换函数，但另一方面：
period2yearmon <- function(x, period = 6) {
year <- as.numeric(sub("\\D.*", "", x))
cyc <- as.numeric(sub(".*\\D", "", x))
as.yearmon(year + period * (cyc - 1) / 12)
}
period2Date <- function(x, period = 6) as.Date(period2yearmon(x, period))
以下是这些功能的一些测试。由于从日期转换为期间并返回日期，因此在输入日期所在的期间的开头给出日期，我们在结尾处显示aggregate中的效果。
# create a period string
d <- Date2period(dt)
## [1] "2001 S1" "2001 S1" "2001 S2" "2001 S2"
period2yearmon(d)
## [1] "Jan 2001" "Jan 2001" "Jul 2001" "Jul 2001"
period2Date(d)
## [1] "2001-01-01" "2001-01-01" "2001-07-01" "2001-07-01"
aggregate(daily_db, function(x) period2Date(Date2period(x)), sum)
## V1 V2 V3
## 2001-01-01 0.9367209 -1.125309 2.39888622
## 2001-07-01 2.6022286 -1.223287 -0.03541579
通过创建诸如yearmon之类的S3对象可以使这更加复杂，但是出于问题所示的目的并不是真正需要的。