I have some very simple data in R that needs to have its date format changed:
date midpoint
1 31/08/2011 0.8378
2 31/07/2011 0.8457
3 30/06/2011 0.8147
4 31/05/2011 0.7970
5 30/04/2011 0.7877
6 31/03/2011 0.7411
7 28/02/2011 0.7624
8 31/01/2011 0.7665
9 31/12/2010 0.7500
10 30/11/2010 0.7734
11 31/10/2010 0.7511
12 30/09/2010 0.7263
13 31/08/2010 0.7158
14 31/07/2010 0.7110
15 30/06/2010 0.6921
16 31/05/2010 0.7005
17 30/04/2010 0.7113
18 31/03/2010 0.7027
19 28/02/2010 0.6973
20 31/01/2010 0.7260
21 31/12/2009 0.7154
22 30/11/2009 0.7287
23 31/10/2009 0.7375
Rather than %d/%m/%Y, I would like it in the standard R format of %Y-%m-%d
How can I make this change? I have tried:
nzd$date <- format(as.Date(nzd$date), "%Y/%m/%d")
But that just cut off the year and added zeros to the day:
[1] "0031/08/20" "0031/07/20" "0030/06/20" "0031/05/20" "0030/04/20"
[6] "0031/03/20" "0028/02/20" "0031/01/20" "0031/12/20" "0030/11/20"
[11] "0031/10/20" "0030/09/20" "0031/08/20" "0031/07/20" "0030/06/20"
[16] "0031/05/20" "0030/04/20" "0031/03/20" "0028/02/20" "0031/01/20"
[21] "0031/12/20" "0030/11/20" "0031/10/20" "0030/09/20" "0031/08/20"
[26] "0031/07/20" "0030/06/20" "0031/05/20" "0030/04/20" "0031/03/20"
[31] "0028/02/20" "0031/01/20" "0031/12/20" "0030/11/20" "0031/10/20"
[36] "0030/09/20" "0031/08/20" "0031/07/20" "0030/06/20" "0031/05/20"
Thanks!
You could also use the parse_date_time function from the lubridate package:
library(lubridate)
day<-"31/08/2011"
as.Date(parse_date_time(day,"dmy"))
[1] "2011-08-31"
parse_date_time returns a POSIXct object, so we use as.Date to get a date object. The first argument of parse_date_time specifies a date vector, the second argument specifies the order in which your format occurs. The orders argument makes parse_date_time very flexible.
nzd$date <- format(as.Date(nzd$date), "%Y/%m/%d")
In the above piece of code, there are two mistakes. First of all, when you are reading nzd$date inside as.Date you are not mentioning in what format you are feeding it the date. So, it tries it's default set format to read it. If you see the help doc, ?as.Date you will see
format
A character string. If not specified, it will try "%Y-%m-%d" then "%Y/%m/%d" on the first non-NA element, and give an error if neither works. Otherwise, the processing is via strptime
The second mistake is: even though you would like to read it in %Y-%m-%d format, inside format you wrote "%Y/%m/%d".
Now, the correct way of doing it is:
> nzd <- data.frame(date=c("31/08/2011", "31/07/2011", "30/06/2011"),
+ mid=c(0.8378,0.8457,0.8147))
> nzd
date mid
1 31/08/2011 0.8378
2 31/07/2011 0.8457
3 30/06/2011 0.8147
> nzd$date <- format(as.Date(nzd$date, format = "%d/%m/%Y"), "%Y-%m-%d")
> head(nzd)
date mid
1 2011-08-31 0.8378
2 2011-07-31 0.8457
3 2011-06-30 0.8147
This is really easy using package lubridate. All you have to do is tell R what format your date is already in. It then converts it into the standard format
nzd$date <- dmy(nzd$date)
that's it.
After reading your data in via a textConnection, the following seems to work:
dat <- read.table(textConnection(txt), header = TRUE)
dat$date <- strptime(dat$date, format= "%d/%m/%Y")
format(dat$date, format="%Y-%m-%d")
> format(dat$date, format="%Y-%m-%d")
[1] "2011-08-31" "2011-07-31" "2011-06-30" "2011-05-31" "2011-04-30" "2011-03-31"
[7] "2011-02-28" "2011-01-31" "2010-12-31" "2010-11-30" "2010-10-31" "2010-09-30"
[13] "2010-08-31" "2010-07-31" "2010-06-30" "2010-05-31" "2010-04-30" "2010-03-31"
[19] "2010-02-28" "2010-01-31" "2009-12-31" "2009-11-30" "2009-10-31"
> str(dat)
'data.frame': 23 obs. of 2 variables:
$ date : POSIXlt, format: "2011-08-31" "2011-07-31" "2011-06-30" ...
$ midpoint: num 0.838 0.846 0.815 0.797 0.788 ...
Using one line to convert the dates to preferred format:
nzd$date <- format(as.Date(nzd$date, format="%d/%m/%Y"),"%Y/%m/%d")
I believe that
nzd$date <- as.Date(nzd$date, format = "%d/%m/%Y")
is sufficient.
Source: Stackoverflow.com