Thanks to everyone for the explanations.
Toby
On 27 Aug 2010 at 12:46, Phil Spector wrote:
> Toby -
> Since dat$doy is just a number, the default S3 method
> for format is used, where the second argument is the trim
> parameter. I suspect you are confusing format (which is for
> output) wi
Toby -
Since dat$doy is just a number, the default S3 method
for format is used, where the second argument is the trim
parameter. I suspect you are confusing format (which is for
output) with strptime (which is for input).
For example,
strptime(dat$doy,'%j')
will assume that the dates
On Aug 27, 2010, at 3:29 PM, Toby Gass wrote:
Hello, helpeRs,
I have a vector of numbers from 1-365 (days of the year) that I would
like to convert to a date. There are no NA's and no missing values.
I did not insert leading zero's for numbers less than 100.
Using the syntax:
dat$doy.1 <- a
Toby,
What is it that you're trying to accomplish?
There seems to be several ideas confused in your post.
It sounds like you have input
x <- 1:365
and you want to call some function to return dates?
Which date should be returned for input 1?
January 1, 2010?
Your error is because if you giv
Hello, helpeRs,
I have a vector of numbers from 1-365 (days of the year) that I would
like to convert to a date. There are no NA's and no missing values.
I did not insert leading zero's for numbers less than 100.
Using the syntax:
dat$doy.1 <- as.numeric(format(dat$doy, "%j" ))
I get the fol
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