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" "0030/0开发者_开发问答6/20"
[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!
There are two steps here:
- Parse the data. Your example is not fully reproducible, is the data in a file, or the variable in a text or factor variable? Let us assume the latter, then if you data.frame is called X, you can do
X$newdate <- strptime(as.character(X$date), "%d/%m/%Y")
Now the newdate
column should be of type Date
.
- Format the data. That is a matter of calling
format()
orstrftime()
:
format(X$newdate, "%Y-%m-%d")
A more complete example:
R> nzd <- data.frame(date=c("31/08/2011", "31/07/2011", "30/06/2011"),
+ mid=c(0.8378,0.8457,0.8147))
R> nzd
date mid
1 31/08/2011 0.8378
2 31/07/2011 0.8457
3 30/06/2011 0.8147
R> nzd$newdate <- strptime(as.character(nzd$date), "%d/%m/%Y")
R> nzd$txtdate <- format(nzd$newdate, "%Y-%m-%d")
R> nzd
date mid newdate txtdate
1 31/08/2011 0.8378 2011-08-31 2011-08-31
2 31/07/2011 0.8457 2011-07-31 2011-07-31
3 30/06/2011 0.8147 2011-06-30 2011-06-30
R>
The difference between columns three and four is the type: newdate
is of class Date
whereas txtdate
is character.
nzd$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
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.
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 ...
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.
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.
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