Convert date column with two different structures

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".

	[[alternative HTML version deleted]]

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HI,
You could try:
library(lubridate)

Date1 <- mdy(as.character(df[,1]))
?Date1[is.na(Date1)] <- parse_date_time(paste(1,as.character(df[,1][is.na(Date1)]),sep="-"),"%d-%b-%y")

A.K.

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".

??? [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.