Question

首先，我将所有假期日期分别存储，并希望通过执行以下操作来确定数据集中哪些日期是假期：

publicHolidays <- as.Date(c("2019-01-01", "2019-01-15", "2019-01-26", "2019-03-04", "2019-03-21", "2019-04-06"))

sampledata <- data.frame(
   sid = c (1:5), 
   DOJ = c("21/03/2019", "26/1/2019", "1/03/2019", "12/03/2019", "1/1/2019"),
   stringsAsFactors = FALSE
)

sampledata$isholiday <- as.numeric(as.Date(sampledata$DOJ,'%d/%m/%Y') %in% publicHolidays)
#sampledata$isholiday

str(sampledata)

接下来，我想为每个日期查找到最近的假期有多少天（向前或向后）。我该怎么办？

Answer 1

比排序全天和所有假期更有效的是利用排序；这是通过data.table的滚动联接完成的：

library(data.table)
setDT(sampledata)

sampledata[ , DOJ := as.IDate(DOJ, '%d/%m/%Y')]
setkey(sampledata, DOJ)

holidays = data.table(date = as.IDate(publicHolidays))
holidays[ , I := .I]
setkey(holidays)

sampledata[ , nearest_holiday := {
  idx = holidays[copy(.SD), I, roll = 'nearest']
  holidays$date[idx]
}]
sampledata[]
#    sid        DOJ nearest_holiday
# 1:   5 2019-01-01      2019-01-01
# 2:   2 2019-01-26      2019-01-26
# 3:   3 2019-03-01      2019-03-04
# 4:   4 2019-03-12      2019-03-04
# 5:   1 2019-03-21      2019-03-21

有了这个，很容易计算距离：

sampledata[ , days_to_nearest := nearest_holiday - DOJ][]
#    sid        DOJ nearest_holiday days_to_nearest
# 1:   5 2019-01-01      2019-01-01               0
# 2:   2 2019-01-26      2019-01-26               0
# 3:   3 2019-03-01      2019-03-04               3
# 4:   4 2019-03-12      2019-03-04              -8
# 5:   1 2019-03-21      2019-03-21               0

Answer 2

使用sapply的基本R方法是检查min和DOJ之间的publicHolidays绝对绝对值

sampledata$nearest_holiday <- sapply(as.Date(sampledata$DOJ, "%d/%m/%Y"),
                        function(x) min(abs(x - publicHolidays)))

sampledata
#  sid        DOJ isholiday nearest_holiday
#1   1 21/03/2019         1               0
#2   2  26/1/2019         1               0
#3   3  1/03/2019         0               3
#4   4 12/03/2019         0               8
#5   5   1/1/2019         1               0

如果您想在dplyr链中使用它，我们可以翻译相同的逻辑

library(dplyr)
library(lubridate)
library(purrr)

sampledata %>%
    mutate(nearest_holiday =  map_dbl(dmy(DOJ), ~min(abs(. - publicHolidays))))

假期天数

2 个答案: