How to transform a dataframe of characters to the respective dates?

Question

I noticed already a couple of times that working with dates doesn\'t allow for using the usual tricks in R. Say I have a dataframe Data with Dates (see below), and I want to

Answer 1

How about

str(as.data.frame(lapply(Data,as.Date,format="%d %B %Y")))
# 'data.frame':   6 obs. of  4 variables:
#  $ Rep1:Class 'Date'  num [1:6] 12898 12898 13907 13907 13907 ...
#  $ Rep2:Class 'Date'  num [1:6] 13278 13278 14217 14217 14217 ...
#  $ Rep3:Class 'Date'  num [1:6] 13600 13600 14340 14340 14340 ...
#  $ Rep4:Class 'Date'  num [1:6] 13831 13831 14669 14669 14669 ...

Answer 2

I think the most succinct way to do this is:

Data[] <- lapply(Data, as.Date,format="%d %B %Y")

This also nicely generalises to the case where not all columns are dates:

Data[date_col] <- lapply(Data[date_col], as.Date,format="%d %B %Y")

You can also simplify the date parsing with a couple of other packages

library(stringr)
library(lubridate)
Data[] <- lapply(Data, function(x) dmy(str_trim(x)))

which is a little more verbose, but has the advantage that you don't need to figure out the data format yourself.