Question

DT是data.table，带有POSIXct列＆＃39; date_and_time＆＃39;。如何使用data.table语法添加3级因子列，指示每个时间段的日期＆＃39; date_and_time＆＃39;实例属于？

DT <- data.table(date_and_time = as.POSIXct(
  c("1997-07-01 23:00",
    "1999-01-01 02:05",
    "2006-01-01 12:00",
    "2009-01-01 22:30",
    "2012-07-01 03:59"), required.components = 5L)
  )
shifts_delimiter = c("00:00", "08:00", "16:00")

要求的结果：

> DT
         date_and_time shift
1: 1997-07-01 23:00:00     3
2: 1999-01-01 02:05:00     1
3: 2006-01-01 12:00:00     2
4: 2009-01-01 22:30:00     3
5: 2012-07-01 03:59:00     1

Answer 1

您可以使用ITime包提供的data.table类与findInterval结合使用

shifts_delimiter = c("00:00", "08:00", "16:00")
DT[, Shift := findInterval(as.ITime(date_and_time), as.ITime(shifts_delimiter))]
DT
#          date_and_time Shift
# 1: 1997-07-01 23:00:00     3
# 2: 1999-01-01 02:05:00     1
# 3: 2006-01-01 12:00:00     2
# 4: 2009-01-01 22:30:00     3
# 5: 2012-07-01 03:59:00     1

另一种选择是将整数用作有序间隔，并使用内置hour函数和findInterval

shifts_delimiter = c(8L, 16L, 24L)
DT[, Shift2 := findInterval(hour(date_and_time), shifts_delimiter) + 1L]
DT
#          date_and_time Shift Shift2
# 1: 1997-07-01 23:00:00     3      3
# 2: 1999-01-01 02:05:00     1      1
# 3: 2006-01-01 12:00:00     2      2
# 4: 2009-01-01 22:30:00     3      3
# 5: 2012-07-01 03:59:00     1      1

Answer 2

如果您的数据集很小，那么这可能与任何事情一样好。

DT[, shift := 1] # Default assignment
DT[hour(date_and_time) >= 8, shift := 2] # adjust for shift 2 and above
DT[hour(date_and_time) >= 16, shift := 3] # adjust for shift 3

或者，您可以通过创建轮班表和小时列来合并表格来进行滚动连接。

DT[, hour := hour(date_and_time)]
setkey(DT, hour)
shift_table <- data.table(hour = hour(as.ITime(shifts_delimiter)), shift = 1:3, key = "hour")
shift_table[DT, roll = TRUE]

R data.table interval

2 个答案: