How to convert a string such as 20100502 into date format %m/%d/%Y
I tried to convert using
as.Date(20100502, origin="1970-01-01", format = "%Y-%m-%d") But, the result is "57003-04-23" instead of 05/02/2010
2 Answers
You can use as.Date and format
format(as.Date("20100502", format = "%Y%m%d"), "%m/%d/%Y") #[1] "05/02/2010" The format argument in as.Date refers to the format in which you provide your data to the function. Then in the function format you provide the format that you desire.
from ?as.Date
as.Date will accept numeric data (the number of days since an epoch), but only if origin is supplied.
A workaround as suggested by @Wimpel in comments is to wrap x in as.character.
X$new_Date <- format(as.Date(as.character(X$Date), format = "%Y%m%d"), "%m/%d/%Y") 
4 Comments
amu1
Thanks. I tried the code to call X$Date from my .csv file that has multiple dates. But, I am getting an error 'origin' must be supplied. When I supply with origin = lubridate::origin, I am getting the result as "04-24-57003".
markus
@GN Can you add the output of
dput(head(X$Date)) at the end of your question?
Wimpel
@GN
X$Date has to be character-format. If not, make it so using as.character().amu1
Sure, it's c(20100502L, 20100502L, 20100502L, 20100502L, 20100502L, 20100502L
If you already know regular expressions then that provides a very generalisable way of manipulating string variables:
date <- "20100502" gsub("(\\d{4})(\\d{2})(\\d{2})", "\\2/\\3/\\1", date) #[1] "05/02/2010" 
Dateobject and then back to a string with a different format. In the first step you have to specify the format of the input string (not the one you want eventually):x<-as.Date("20100502", format = "%Y%m%d"). Then withformatyou specify the output format:format(x,format="%m/%d/%Y").