How to convert Excel date format to proper date in R

Question

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?

Solution 1

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

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

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

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.

Solution 5

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")

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Comments

[count]

This might set you up.

[elksie5000]

Thank you for such a fast reply.

[elksie5000]

Thank you for that. I'm still very new to R. It worked a treat.

[Andrew Brēza]

Quite welcome. I'm editing my response now since I realized the origin is slightly off. Give me a minute before you use this code in your project :-)

[Andrew Brēza]

Actually I just typed 01/12/2016 into Excel and converted it into a number. The value I got was 42381, not 42705. Using 42381 in the as.Date function gets you back to 1/12/2016. Are you sure that you copied the number correctly from Excel?

[elksie5000]

Oh, I did exactly the same thing in my copy of Excel, but had a different number. Is because it's European day-month-year format?

[Andrew Brēza]

That's exactly what happened, I should have asked what date format you were using. It's still early here in the states and I'm obviously not thinking clearly yet.

[elksie5000]

Hey, I just appreciated your swift response. Thank you.

[Robert Alan Greevy Jr PhD]

This is my favorite solution, but one caution mentioned in the documentation for the "as.Date" function is the handling of leap years can vary between programs, e.g. Excel intentionally misclassifies 1900 as a leap year. If your dates cover broad range, double check the conversion for the min and max dates in your range.

[Clément LVD]

In the French version of excel, the best solution is quite fascinating, for a date with hours: as_datetime(Date,origin = "1969-12-31 24:00:00")

[Suren]

convertToDateTime if it is datetime.

[userLL]

convert_to_date() - also from the janitor package - can be used when there is a mix of Excel numeric dates and actual dates

[Mehrad Mahmoudian]

For those who wonder where the 1899-12-30 came from and if it is correct, it seems in some Excel versions Microsoft has used 1900 as the reference date and in some versions 1904! for this reason I suggest you to confirm the date from some other source as well. Source: support.microsoft.com/en-us/office/…