R 根据行数为列分配时间序列间隔

需要根据“cnt_行”列按“名称”分组来指定时间间隔。i、 e如果计数在
左右 96表示时间间隔为15分钟。因此，如果计数为94，则时间间隔应停止在
晚上11:15（基于行数），如果行数正好是96，则应在每天晚上11:45结束。
间隔5分钟也一样。间隔时间不得超过一天

cnt_rows = c("94","94",".",".","94","286","286",".",".",".","286","96","96",".",".","96")
Name = c("Alan","Alan",".",".","Alan","Steve","Steve",".",".",".","Steve","Mike","Mike",".",".","Mike")
Values = c("10","10",".",".","45","91","35",".",".",".","46","34","5",".",".","34")

输入表

  df = data.frame(cnt_rows,Name,Values)

输出表

dt = c("2019-12-01 00:00:00","2019-12-01 00:15:00",".",".","2019-12-01 23:15:00","2019-12-01 00:00:00","2019-12-01 00:05:00",".",".",".","2019-12-01 23:45:00","2019-12-01 00:00:00","2019-12-01 00:15:00",".",".","2019-12-01 23:45:00")

df_out = data.frame(cnt_rows,Name,Values,dt)

提前谢谢。

也许你可以试试：

library(dplyr)
date <- as.POSIXct('2019-12-01')

df %>%
  mutate(breaks = ifelse(cnt_rows %in% c(94, 96), '15 min', '5 min')) %>%
  group_by(Name) %>%
  mutate(dt = seq(date, by = first(breaks), length.out = n()))

库（dplyr）
日期%
突变（中断=ifelse（在%c（94,96），‘15分钟’、‘5分钟’）中的cnt_行百分比）%>%
分组单位（名称）%>%
突变（dt=seq（日期，by=first（breaks），length.out=n（））

shaw，谢谢您的回复。创建偶数秒的间隔需要创建精确的间隔minute@san1对不起，我不明白。你是什么意思？@Shaw，我试过使用94行计数的实际数据，预期的时间序列是15分钟，而我使用上面的代码intervel column（dt）运行数据，生成时间为偶数秒，如果行数为94，则时间序列应在晚上11:15购买。谢谢。我不清楚你的问题。你能解释一下什么时候数据必须以15分钟的间隔分割，什么时候数据必须以5分钟的间隔分割吗？如果cnt_rows列为94或96，那么15分钟，如果cnt_rows列为286，那么5分钟。谢谢@ronak，数据中只有94、96和286个值吗？没有其他内容？是的，数据中只有94、96和286个值