Converting excel DateTime serial number to R DateTime
Use the function convertToDateTime. It's straight forward. Here's an example:
library(openxlsx)
convertToDateTime(helpData$ExcelNum, origin = "1900-01-01")
Let me know how it works.
The time data is still there, it's just not displayed - see:
as.numeric(newDateTime)
#[1] 15037.25 15044.33 15046.33 15047.00 etc etc
If you are wishing to work with parts of days, you are probably best using POSIXct representations though. To do so, you can convert to Date, then convert to POSIXct, though this does bring into play timezone issues if you want to do a direct comparison to your DateTime column.
helpData$newDate <- as.POSIXct(as.Date(helpData$ExcelNum,origin="1899-12-30"))
attr(helpData$newDate,"tzone") <- "UTC"
helpData
# ID DateTime ExcelNum newDate
#1 1 3/4/2011 6:00 40606.25 2011-03-04 06:00:00
#2 2 3/11/2011 7:55 40613.33 2011-03-11 07:54:59
#3 3 3/13/2011 7:55 40615.33 2011-03-13 07:54:59
#4 4 3/14/2011 0:00 40616.00 2011-03-14 00:00:00
#5 5 3/14/2011 10:04 40616.42 2011-03-14 10:03:59
#6 6 3/14/2011 7:55 40616.33 2011-03-14 07:54:59
#7 7 3/15/2011 19:55 40617.83 2011-03-15 19:54:59
#8 8 3/17/2011 7:55 40619.33 2011-03-17 07:54:59
#9 9 3/18/2011 4:04 40620.17 2011-03-18 04:03:59
#10 10 3/18/2011 4:04 40620.17 2011-03-18 04:03:59
Your number is counting days. Convert to seconds, and you're all set (less a rounding error)
helpData[["ExcelDate"]] <-
as.POSIXct(helpData[["ExcelNum"]] * (60*60*24)
, origin="1899-12-30"
, tz="GMT")
# ID DateTime ExcelNum ExcelDate
# 1 1 3/4/2011 6:00 40606.25 2011-03-04 06:00:00
# 2 2 3/11/2011 7:55 40613.33 2011-03-11 07:54:59
# 3 3 3/13/2011 7:55 40615.33 2011-03-13 07:54:59
# 4 4 3/14/2011 0:00 40616.00 2011-03-14 00:00:00
# 5 5 3/14/2011 10:04 40616.42 2011-03-14 10:03:59
# 6 6 3/14/2011 7:55 40616.33 2011-03-14 07:54:59
# 7 7 3/15/2011 19:55 40617.83 2011-03-15 19:54:59
# 8 8 3/17/2011 7:55 40619.33 2011-03-17 07:54:59
# 9 9 3/18/2011 4:04 40620.17 2011-03-18 04:03:59
# 10 10 3/18/2011 4:04 40620.17 2011-03-18 04:03:59