Convert Excel numeric to date

最后都变了- 提交于 2019-12-05 20:24:17

You can simply use as.Date and specify the origin, i.e.

as.Date(date, origin="1899-12-30") 
#[1] "2017-08-16" "2017-09-16" "2017-06-17" "2017-07-17" "2017-08-17"

#or format it to your liking,

format(as.Date(date, origin="1899-12-30"), '%b %Y') 
#[1] "Aug 2017" "Sep 2017" "Jun 2017" "Jul 2017" "Aug 2017"

This link gives quite a bit of information on this matter.

If you want to convert dates from Excel, you can use as.Date() with a specific origin. According to the documentation, '1900-01-01' is used as day ` in Excel.

date <- c(42963,42994,42903,42933,42964)

This is the result of as.Date():

as.Date(date, origin = "1900-01-01")
[1] "2017-08-18" "2017-09-18" "2017-06-19" "2017-07-19" "2017-08-19"

You can then use zoo::as.yearmon()` to get the expected outcome:

zoo::as.yearmon(as.Date(date, origin = "1900-01-01"))
[1] "Aug 2017" "Sep 2017" "Jun 2017" "Jul 2017" "Aug 2017"

Type excel_numeric_to_date to look at the function's code and you'll see it's a wrapper for the line of code used by the other answers to this question: as.Date(date_num, origin = "1899-12-30".

So that's not the issue.

The underlying matter here is confusion about date formatting. You say you expect your first number 42963 to become "Aug 2016", and your last number 42964 to become "Aug 2017". The latter is just one more than the former, which shows up in the conversion - they should be a day apart, not a year apart as you are expecting:

> excel_numeric_to_date(c(42963, 42964))
[1] "2017-08-16" "2017-08-17" # as expected, they are one day apart

Perhaps the day and year fields are switched upstream in your data at the point where these get mapped to integer dates, and it was hard to tell here because of the values chosen.

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