Achim,
Thanks for pointing me in the right direction! I broke your user-defined
function down into two separate aggregate calls on which.min and length.
library(zoo)
n <- 1000
dly <- zoo( x = rnorm(n), order.by = seq.Date(as.Date('1970-01-01'),
by = 'day', length.out = n) )
year <- as.numeric( format(index(dly), '%Y') )
annual.min <- aggregate( dly, by = year, FUN = min )
ix <- aggregate(dly, year, which.min )
l <- aggregate(dly, year, length)
a <- c(0, as.vector(cumsum(l)[-length(l)]))
b <- as.vector(ix)
annual.min <- dly[a+b]
plot(dly)
points(annual.min, pch = 19, col = 'red')
-----Original Message-----
From: Achim Zeileis [mailto:achim.zeil...@uibk.ac.at]
Sent: Thursday, May 13, 2010 12:29 PM
To: Hutchinson,David [PYR]
Cc: r-help@r-project.org
Subject: Re: [R] aggregate.zoo
On Thu, 13 May 2010, Hutchinson,David [PYR] wrote:
> I have been using aggregate function to determine the annual minimum
> in a daily time series. However, I was also hoping to determine the
> date of occurrence without having to loop through each year to
> identify the date of each minimum returned from aggregate(). Is there
> a more convenient way of doing so?
You could do something like this:
## determine position of minimum and number of observations ix <-
aggregate(dly, year, function(x) c(which.min(x), length(x)))
## expand to overall index
ix <- as.vector(cumsum(c(0, head(ix[,2], -1))) + ix[,1])
## subset original series
dly[ix]
hth,
Z
> Thanks in advance,
>
> Dave
>
> library(zoo)
> n <- 1000
> dly <- zoo( x = rnorm(n), order.by =
seq.Date(as.Date('1970-01-01'),
> by = 'day', length.out = n) )
> year <- as.numeric( format(index(dly), '%Y') ) annual.min <-
> aggregate( dly, by = year, FUN = min )
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
>
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