See the aggregate= argument to read.zoo which will aggregate them as
it reads them in like this:
> library(zoo)
> Lines <- "02/10/1995 4925.5
+ 30/10/1995 4915.9
+ 23/01/1996 4963.5
+ 23/01/1996 5009.2
+ 04/03/1996 5031.9 # here
+ 04/03/1996 5006.5 # here
+ 03/04/1996 5069.2
+ 03/05/1996 5103.7
+ 31/05/1996 5107.1
+ 01/07/1996 5153.1
+ 02/08/1996 5151.7"
>
> tail1 <- function(x) tail(x, 1)
> z <- read.zoo(textConnection(Lines), format = "%d/%m/%Y", aggregate = tail1)
> z
1995-10-02 1995-10-30 1996-01-23 1996-03-04 1996-04-03 1996-05-03 1996-05-31
4925.5 4915.9 5009.2 5006.5 5069.2 5103.7 5107.1
1996-07-01 1996-08-02
5153.1 5151.7
On Tue, Jun 22, 2010 at 10:35 AM, Research <risk2...@ath.forthnet.gr> wrote:
> Hello,
>
> I have a zoo time series read from an excel file which has some dates the
> same, such as the following example:
>
> 02/10/1995 4925.5
> 30/10/1995 4915.9
> 23/01/1996 4963.5
> 23/01/1996 5009.2
> 04/03/1996 5031.9 # here
> 04/03/1996 5006.5 # here
> 03/04/1996 5069.2
> 03/05/1996 5103.7
> 31/05/1996 5107.1
> 01/07/1996 5153.1
> 02/08/1996 5151.7
>
> Is there a simple way to keep the last price of the ones that have the same
> dates?
>
> 04/03/1996 5031.9
> 04/03/1996 5006.5
>
> i.e., keep only the "04/03/1996 5006.5" price and discard the previous
> one... Is there an implicit function that does that or do I need some sort
> of recursive algorithm?
>
> You can try a solution on this example (for convenience):
>
> x.Date <- as.Date("2003-02-01") + c(1, 3, 7, 7, 14) - 1
> x <- zoo(rnorm(5), x.Date)
>
> Zoo object has 2 prices with same dates.
>
> Many thanks in advance,
> Costas
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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