Using zoo's yearmon class:
library(zoo)
my.dates[!duplicated(as.yearmon(my.dates))]
or, although you seem to disallow this in your question,
this would be an option:
my.dates[!duplicated(format(my.dates, "%Y-%m"))]
On Feb 4, 2008 10:39 AM, Gavin Simpson <[EMAIL PROTECTED]> wrote:
> hits=-2.6 tests=BAYES_00
> X-USF-Spam-Flag: NO
>
> Dear List,
>
> Say I have the following sequence of dates [*]:
>
> start <- as.Date("2005-01-05", format = "%Y-%d-%m")
> end <- as.Date("2007-10-07", format = "%Y-%d-%m")
> my.dates <- seq(start, end, by = "days")
>
> What I would like to generate is a sequence of dates, by month that goes
> from the first day of 2005 (the year of the start date) to the last day
> of 2007 (the year of the end date), so that the output is a vector of 36
> dates containing all months of the three calendar years that the
> sampling spanned.
>
> I could do it via manipulation of dates as so:
>
> new.dates <- seq(as.Date(paste(format(min(my.dates), format = "%Y"),
> "1", "1", sep = "/")),
> as.Date(paste(format(max(my.dates), format = "%Y"),
> "12", "31", sep = "/")),
> by = "months")
>
> And then manipulate that to get only the month and year parts of the 36
> generated dates.
>
> This doesn't seem very elegant to me. Is there a better/easier way than
> converting back and forth between characters and objects of class
> "Date"?
>
> Many thanks,
>
> G
>
> [*] FWIW, my actual application is similar to my.dates, but not sampled
> every day - indeed there are months where there are no samples - and I
> am trying to do a levelplot of the numbers of observations per month,
> per year. Given the following data
>
> dat <- data.frame(temp = rnorm(length(my.dates)),
> my.dates = my.dates)
> dat$year <- as.numeric(format(dat$my.dates, format = "%Y"))
> dat$month <- format(dat$my.dates, format = "%b")
> dat$month <- factor(dat$month, levels = c("Jan","Feb","Mar","Apr",
> "May","Jun","Jul","Aug",
> "Sep","Oct","Nov","Dec"))
>
> I can get a table of the number of observations per month per year via
>
> (obs.yearmon <- with(dat, table(year, month)))
>
> Which when converted to a vector provides what I need for levelplot()'s
> formula method. Now I just need to generate the sequence of 36 months
> ("Jan", "Feb" etc) and years to go with it. The matrix method of
> levelplot works, but I am having to hard code a lot of details that I
> believe the formula method will handle automagically for me.
> --
> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
> Dr. Gavin Simpson [t] +44 (0)20 7679 0522
> ECRC, UCL Geography, [f] +44 (0)20 7679 0565
> Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
> Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
> UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
>
> ______________________________________________
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>
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