Thank you A.K. and Ista. The answer that eliminates duplicated rows is
A.K.'s first solution (brilliant idea using which.max!).
The version of Ista's solution without duplicates is
ddply(data, "state", function(x) x[which.max(x$date), ])$value
Thanks again!
Dan
On Thu, Feb 13, 2014 at 8:36 AM, arun <smartpink...@yahoo.com> wrote:
> Hi,
> Try ?which.max() # unique values for the combination.
>
> ddply(data,.(state),summarize,max_date=value[which.max(date)])[,2]
> #or
> ddply(data,.(state),summarize,max_date=value[date == max(date)])[,2]
>
>
> A.K.
>
>
> On Thursday, February 13, 2014 11:15 AM, Dan Murphy <chiefmur...@gmail.com>
> wrote:
> I can do this in multiple steps with summarise, joins, etc., but can't
> help thinking it can be accomplished in one plyr call.
>
> Here's a small example:
>
>> require(plyr)
>> require(lubridate)
>> data <- data.frame(
> + date = rep(as.Date(ymd(20140101 + (0:3) * 100)), 2),
> + state = rep(c("A", "B"), each = 4),
> + value = rnorm(8))
>
> What I want is clearly
>
>> data[c(4, 8), "value"]
> [1] -1.007111 -1.527541
>
> Here are my multiple steps:
>
>> statemaxval <- ddply(data, "state", summarise, max_date = max(date))
>> rslt <- join(data, statemaxval, by = "state")
>> rslt <- subset(rslt, date == max_date)
>> rslt <- rslt[!duplicated(rslt), ]
>> rslt$value
> [1] -1.007111 -1.527541
>
> Is there a one-step way to accomplish this?
> Something like
> ddply(data, "state", summarise,
> "GiveMeTheValueCorrespondingToMaxDateByState!!")
>
> Or is that only possible if there is only one unique value for a given
> combination of state and max(date)?
>
> Thanks,
> Dan
>
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