Converting excel DateTime serial number to R DateTime
When excel tables are imported as xy points in ArcGIS I continue to lose my correct DateTime stamp for each point. Thus, I have formatted the DateTime serial number, created the .shp, and read the .shp into R using readOGR().
Once in R I can convert to the correct date using as.Date()
and the origin = "1899-12-30"
argument, but the time is left out. While I have seen examples with a sole Date, I have not seen worked examples with DateTime. I have been using as.Date()
as well as as.POSIXct()
but this seemingly simple task as been a bit frustrating, thus the post…
I have created a sample data set with 10 rows of the correct DateTime format as well as the excel serial number.
*Thanks Richard and thelatemail for their keen eye on an earlier hindrance. I have corrected the data and re-posted here.
Here is my sample data
helpData <- structure(list(ID = 1:10, DateTime = structure(c(9L, 1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 8L), .Label = c("3/11/2011 7:55", "3/13/2011 7:55",
"3/14/2011 0:00", "3/14/2011 10:04", "3/14/2011 7:55", "3/15/2011 19:55",
"3/17/2011 7:55", "3/18/2011 4:04", "3/4/2011 6:00"), class = "factor"),
ExcelNum = c(40606.25, 40613.32986, 40615.32986, 40616, 40616.41944,
40616.32986, 40617.82986, 40619.32986, 40620.16944, 40620.16944
)), .Names = c("ID", "DateTime", "ExcelNum"), class = "data.frame", row.names = c(NA,
-10L))
head(helpData)
The DateTime is GMT. The time is a 24 hr clock (i.e. not AM/PM). I am working on Windows 7, have the most recent R, and ArcGIS 10.
The code below gets the correct Date, but the time is still missing.
newDateTime <- as.Date(helpData[ , "ExcelNum"], origin = "1899-12-30")
head(newDateTime)
Thanks in advance!
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
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