在R中生成统一采样的时间序列对象

Question

Hi I am looking to generate a uniformly sampled time series at 30 minute interval from a particular start date to some end date. However the constraint is that on each day the 30 minute interval begins at 7:00 and ends at 18:30 ie I need the time series object to be something like

c('2016-08-19 07:00:00', 
  '2016-08-19 07:30:00',  
   ..., 
  '2016-08-19 18:30:00', 
  '2016-08-20 07:00:00',
   ...,
  '2016-08-20 18:30:00',
   ...
   '2016-08-31 18:30:00')

Without the constraints it can be done with something like

seq(as.POSIXct('2016-08-19 07:00:00'), as.POSIXct('2016-08-21 18:30:00'), by="30 min")

But I dont want the times between '2016-08-20 18:30:00' and '2016-08-21 07:30:00' in this case. Any help will be appreciated. Thanks!

Answer 1

Using the example series you created:

  ts <- seq(as.POSIXct('2016-08-19 07:00:00'), 
              as.POSIXct('2016-08-21 18:30:00'), by="30 min")

Pull out the hours from your series using strftime :

hours <- strftime(ts, format="%H:%M:%S")

> head(hours)
[1] "07:00:00" "07:30:00"
[3] "08:00:00" "08:30:00"
[5] "09:00:00" "09:30:00"

You can then convert it back to POSIXct :

hours <- as.POSIXct(hours, format="%H:%M:%S")

This will retain the times of the day but it will make the date today's date:

> head(hours)
[1] "2016-09-11 07:00:00 EDT"
[2] "2016-09-11 07:30:00 EDT"
[3] "2016-09-11 08:00:00 EDT"
[4] "2016-09-11 08:30:00 EDT"
[5] "2016-09-11 09:00:00 EDT"
[6] "2016-09-11 09:30:00 EDT"
> tail(hours)
[1] "2016-09-11 16:00:00 EDT"
[2] "2016-09-11 16:30:00 EDT"
[3] "2016-09-11 17:00:00 EDT"
[4] "2016-09-11 17:30:00 EDT"
[5] "2016-09-11 18:00:00 EDT"
[6] "2016-09-11 18:30:00 EDT"

You can then create a TRUE/FALSE vector based on the condition you want:

condition <- hours > "2016-09-11 07:30:00 EDT" & 
             hours < "2016-09-11 18:30:00 EDT"

Then filter your original series based on this condition:

ts[condition]

Answer 2

Here is my short and handy solution with package lubridate

library("lubridate")

list <- lapply(0:2, function(x){

  temp <- ymd_hms('2016-08-19 07:00:00') + days(x) 

  result <- temp + minutes(seq(0, 690, 30))

  return(strftime(result))

})

do.call("c", list)

I have to use strftime(result) to remove the timezone and to have the right times.