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 =＆gt;处理按月分组的每日数据
b）as.yearqtr =＆gt;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
我想定义一个类似的函数：
as.yearperiod = function(x, period = 6) {...} # convert dates in semesters
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
答案 0 :(得分：2)
先生，我建议您使用lubridate包来处理自定义日期间隔。使用floor_date可以轻松完成您的任务，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
答案 1 :(得分：2)
Date2period输入一个"Date"对象并返回一个表示句点（学期等）的字符串，具体取决于参数period的值，该值应该是一个除数的数字12.在内部，它转换为yearmon，然后提取年份和周期，即月份，并从中生成所需的字符串。
Date2period <- function(x, period = 6, sep = " S") {
ym <- as.yearmon(x)
paste(as.integer(ym), (cycle(ym) - 1) %/% period + 1, sep = sep)
}
测试上述内容：
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
以下是其他转换功能，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会在输入日期所在的期间开头给出日期，因此我们会在aggregate结尾处显示效果。
# 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，但是出于问题中显示的目的并不是真正需要的。