R help - Feb 2007 - Working with temporal data [Solved]

Just for the record, here are my steps for producing a date based histogram.
Data is stored in a file where each line only has a date - 2007/02/16
>d<-readLines("filename.dat")
>d<-as.Date(d, format="%Y/%m/%d")
>pdf(yearly.pdf)
>hist(d, "years")
>dev.off()
Instead of "years" you can also use "days",
"weeks", "months", "secs",
"mins", "hours".

One final question, how could I easily filter my dataset if, for
instance, I only wanted to see results from 2006 ?

Thanks to all who helped,
S?rgio Nunes

On 2/15/07, S?rgio Nunes <snunes at gmail.com>
wrote:
> Hi,
>
> I have several files with data in this format:
>
> 20070102
> 20070102
> 20070106
> 20070201
> ...
>
> The data is sorted and each line represents a date (YYYYMMDD). I would
> like to analyze this data using R. For instance, I would like to have
> a histogram by year, month or day.
>
> I've already made a simple Perl script that aggregates this data but I
> believe that R can be much more powerful and easy on this kind of
> work.
>
> Any suggestions on where to start?
>
> Thanks in advance,
> S?rgio Nunes
>

On Fri, 2007-02-16 at 15:39 +0000, S?rgio Nunes wrote:
> Just for the record, here are my steps for producing a date based
histogram.
> Data is stored in a file where each line only has a date - 2007/02/16
> 
> >d<-readLines("filename.dat")
> >d<-as.Date(d, format="%Y/%m/%d")
> >pdf(yearly.pdf)
> >hist(d, "years")
> >dev.off()
> 
> Instead of "years" you can also use "days",
"weeks", "months", "secs",
> "mins", "hours".
> 
> One final question, how could I easily filter my dataset if, for
> instance, I only wanted to see results from 2006 ?
> 
> Thanks to all who helped,
> S?rgio Nunes
The easiest way may be to create your own "Year" extractor function,
since there does not appear to be one by default, unless I missed it
someplace (it is not listed in ?weekdays).

For example, using your data from my prior reply:
> d2
[1] "2006-08-09" "2004-02-11" "2004-06-09"


Years <- function(x) format(x, "%Y")

> Years(d2)
[1] "2006" "2004" "2004"

> Years(d2) == 2006
[1]  TRUE FALSE FALSE

> d2[Years(d2) == 2006]
[1] "2006-08-09"


So, just use something like:

 hist(d2[Years(d2) == 2006], "years")

HTH,

Marc Schwartz