How to convert Excel date format to proper date in R
RExcelDateDatetimeLubridateR Problem Overview
I'm working with a csv which unfortunately has logged datetimes using the number format of 42705 although it should be 01/12/2016.
I'd like to convert it to the right format in R using lubridate or some other package. Is there a function that will handle it?
R Solutions
Solution 1 - R
You don't need to use lubridate
for this, the base function as.Date
handles this type of conversion nicely. The trick is that you have to provide the origin, which in Excel is December 30, 1899.
as.Date(42705, origin = "1899-12-30")
# [1] "2016-12-01"
If you want to preserve your column types, you can try using the read_excel
function from the readxl
package. That lets you load an XLS or XLSX file with the number formatting preserved.
Solution 2 - R
Here is another way to do it using janitor and tibble packages:
install.packages("janitor")
install.packages("tibble")
library(tibble)
library(janitor)
excel_numeric_to_date(as.numeric(as.character(YourDate)), date_system = "modern")
Solution 3 - R
openxlsx
package also allows xls
date
conversion:
openxlsx::convertToDate(42705)
[1] "2016-12-01"
And as suggested by @Suren, convertToDateTime
allows datetime
conversion:
openxlsx::convertToDateTime(42705.5)
[1] "2016-12-01 12:00:00"
Solution 4 - R
As it was said, very good options:
as.Date(42705, origin = "1899-12-30")
openxlsx::convertToDate(42705)
Another way also could be:
format(as.Date(as.Date("1899-12-30") + 42705, "%d-%m-%Y"), "%d-%m-%Y")
Note you can change the output format where it's written %d-%m-%Y
(first of all, convert as.numeric
if it's imported as character!,or converting in the formula:
format(as.Date(as.Date("1899-12-30") + as.numeric( number formatted as character), "%d-%m-%Y"), "%d-%m-%Y")
Solution 5 - R
If you work with the data.table package you could use as.IDate()
for that:
require(data.table)
as.IDate(42705, origin = "1899-12-30")
# [1] "2016-12-01"
Works like base::as.Date()
here.