Hello,
Try the following.
idx <- grep("[[:alpha:]]", df$Date)
Date <- as.Date(df$Date, "%m/%d/%y")
Date[idx] <- as.Date(paste("01", df$Date[idx]), "%d %b-%y")
Hope this helps,
Rui Barradas
Em 05-11-2013 16:00, Abraham Mathew escreveu:
Let's say I have the following data frame and the date column has two
different ways in which date is presented. How can I use as.Date or the
lubridate package to have one date structure for the entire colum
df = data.frame(Date=c("5/1/13","8/1/13","9/1/13","Apr-10",
"Apr-11","Apr-12","Apr-13"))
It's "month/date/year" and "month-year".
An alternative approach would be to perform some conditional statements
where if the date is "Apr-11", then populate it with "04/01/2011".
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