Thanks All. Your idea is useful!!!
On Tue, Oct 7, 2014 at 1:01 PM, Jim Lemon <jim at bitwrit.com.au> wrote:
On Tue, 7 Oct 2014 11:51:34 AM G?ran Brostr?m wrote:
On 2014-10-07 11:27, Jim Lemon wrote:
On Tue, 7 Oct 2014 10:32:42 AM Frederic Ntirenganya wrote:
Dear All,
How can I change the format of date of day of the year ? for
(i.e. "17 Apr" rather than "108").
The following is the type of the dataset I have
head(Samaru)
Year Start End Length
1 1930 108 288 180
2 1931 118 288 170
3 1932 115 295 180
4 1933 156 294 138
5 1934 116 291 175
6 1935 134 288 154
Hi Frederic,
The easiest method I can think of is this:
Samaru$Start<-format(as.Date(
paste(Samaru$Year,"01-01",sep="-"))+Samaru$Start,
"%b %d")
Samaru$End<-format(as.Date(
paste(Samaru$Year,"01-01",sep="-"))+Samaru$End,
"%b %d")
In the package 'eha' I have a function 'toDate':
> require(eha)
> toDate(1930 + 108/365)
[1] "1930-04-19"
(Interestingly, we are all wrong; the correct answer seems to be
"1930-04-18")
function (times)
{
if (!is.numeric(times))
stop("Argument must be numeric")
times * 365.2425 + as.Date("0000-01-01")
}
The 'inverse' function is 'toTime'.
Sometimes it misses by one day; not very important in my
Hi Goran,
You're correct.
as.Date("Apr 19 1930","%b %d %Y") -
+ as.Date("Jan 1 1930","%b %d %Y")
Time difference of 108 days
The t
Samaru$Start<-format(as.Date(
paste(Samaru$Year,"01-01",sep="-"))+Samaru$Start-1,"%b %d")
Samaru$End<-format(as.Date(
paste(Samaru$Year,"01-01",sep="-"))+Samaru$End-1, "%b %d")
Jim