R：按一天中的时间缩短日期时间

Question

我有一个 data_frame 和 POSIXct 日期时间。我现在想创建一个变量，将这些日期时间切割成时间段：1 -- [00:00:00, 08:00:00), 2 -- [08:00:00, 17:00:00) , 3 -- [17:00:00, 18:30:00), 4 -- [18:30:00, 00:00:00).

这是一些示例数据：

df_times = data_frame(
  datetime = seq.POSIXt(
    from = as.POSIXct(strftime("2016-01-01 00:00:00", format = "%Y-%m-%d :%H:%M:%S")),
    by = "min",
    length.out = 100000
  ),
  value = rnorm(100000)
)

这是预期的输出：

> df_times
# A tibble: 100,000 × 3
              datetime      value  band
                <dttm>      <dbl> <dbl>
1  2016-01-01 00:00:00  0.5855288     1
2  2016-01-01 00:01:00  0.7094660     1
3  2016-01-01 00:02:00 -0.1093033     1
4  2016-01-01 00:03:00 -0.4534972     1
5  2016-01-01 00:04:00  0.6058875     1
6  2016-01-01 00:05:00 -1.8179560     1
7  2016-01-01 00:06:00  0.6300986     1
8  2016-01-01 00:07:00 -0.2761841     1
9  2016-01-01 00:08:00 -0.2841597     1
10 2016-01-01 00:09:00 -0.9193220     1
# ... with 99,990 more rows

我试过 cut.POSIXt 但坚持要跟踪日期。理想的解决方案将使用 dplyr::recode 或 forcats::.

Answer 1

您可以创建一个 hour 列，然后将其剪切：

df_times$hour = as.numeric(df_times$datetime) %% (24*60*60) / 3600
df_times$band = cut(df_times$hour, breaks=c(0,8,17,18.5,24), include.lowest=TRUE, 
                    right=FALSE)

Answer 2

这是我认为直接将问题的意图转化为代码的解决方案：

set.seed(12345)

# create a dataset
df_times = data_frame(
  datetime = seq.POSIXt(
    from = as.POSIXct("2016-01-01 00:00:00", format = "%Y-%m-%d %H:%M:%S"),
    by = "min",
    length.out = 100000
  ),
  value = rnorm(100000)
) %>% 
  mutate(
    time = times(format(datetime, "%H:%M:%S")),
    cut(
      time,
      breaks = times(c(
        "00:00:00", 
        "08:00:00", 
        "17:00:00",
        "18:30:00",
        "23:59:59"
      )),
      labels = c(
        "1",
        "2",
        "3",
        "4"
      ),
      include.lowest = TRUE,
      right = FALSE
    )
  )

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