3

我想将 R 中的动物园数据按两个、四个或六个月的时间段汇总。这种类型的日期处理只有两个可用的选项，使用：

a) as.yearmon=> 处理按月分组的每日数据

b) as.yearqtr=> 处理按 3 个月的固定组（一月至三月、四月至六月、七月至十月和十月至十二月）分组的每日数据。

一个最小的例子

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)

我期待这样的结果：

                V1         V2         V3
2001 S1  3.538950   1.175599  -2.147845
2001 S2 -1.125309  -0.035416  -0.639698
4

2 回答 2

2

先生，我建议您使用lubridate包来处理自定义日期间隔。应用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
于 2017-12-28T08:17:16.140 回答
2

日期2期间

Date2period输入一个"Date"对象并返回一个表示周期（学期等）的字符串，具体取决于参数的值，period该参数应该是一个除数为 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)
}

要测试上述内容：

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

period2yearmon, period2Date

以下是其他转换功能，但用于另一个方向：

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))

以下是对这些功能的一些测试。由于从 Date 转换为 period 并返回 Date 给出了输入日期所在的期间开始的日期，因此我们aggregate在最后显示了效果。

# 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

这可以通过创建 S3 对象来变得更加复杂，yearmon但出于问题中所示的目的，这并不是真正需要的。

于 2017-12-28T08:18:10.730 回答