Set standard format for Date

Question

I have table,

>dt
ID     Date
001   13-Jan-2016
002   2016-01-10
003   15-Jan-2016
004   2016-01-09
005   2016-01-11

and, I want to make my table like this

>dt
    ID     Date
    004   09-Jan-2016
    002   10-Jan-2016
    005   11-Jan-2016
    001   13-Jan-2016
    003   15-Jan-2016

I want to set default for the Date variable and make it order. What should I do? Thanks.

checkout the `lubridate` package, it provides various methods for standardizing dates — Imran Ali, Apr 13 '16 at 23:49

alistaire · Accepted Answer · 2016-04-14T00:13:13.070

3

If you have exactly the two formats you show, you can use grepl to look for letters, and use ifelse to pass as.Date a different format based on the result:

as.Date(dt$Date, ifelse(grepl('[a-z]', dt$Date), '%d-%b-%Y', '%Y-%m-%d'))
# [1] "2016-01-13" "2016-01-10" "2016-01-15" "2016-01-09" "2016-01-11"

If you overwrite dt$Date, you can use order to sort:

dt$Date <- as.Date(dt$Date, ifelse(grepl('[a-z]', dt$Date), '%d-%b-%Y', '%Y-%m-%d'))
dt[order(dt$Date),]

#    ID       Date
# 4 004 2016-01-09
# 2 002 2016-01-10
# 5 005 2016-01-11
# 1 001 2016-01-13
# 3 003 2016-01-15

edited Apr 14 '16 at 00:13

answered Apr 14 '16 at 00:07

alistaire

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2

After all this time I didn't realise you could specify a vector of formats to `as.Date()`. You learn something every day I suppose. – thelatemail Apr 14 '16 at 00:21
1

@alistaire - vector recycling - it is trying to format against `c('%d-%b-%Y', '%F')[c(1,2,1,2,1)]` – thelatemail Apr 14 '16 at 00:30
@alistaire thanks, that's so helpful. – Ameerah Apr 14 '16 at 03:35

Set standard format for Date

1 Answers1