It seems that 'calc' does not like this (any more; another thing to
fix) . If your rasters are not very large you can use apply, which
makes it much faster:
library(raster)
#creating some data
r <- raster(nrow=10, ncol=10)
s <- list()
for (i in 1:25) { s[i] <- setValues(r, rnorm(ncell(r), i, 3) ) }
s <- stack(s)
# regression
time <- 1:nlayers(s)
fun <- function(x) { lm(x ~ time)$coefficients[2] }
r[] <- apply(getValues(s), 1, fun)
Robert
On Fri, Nov 26, 2010 at 2:51 PM, Jacob van Etten
<jacobvanet...@yahoo.com> wrote:
> you could try this approach (use calc whenever you can):
>
> (supposing your bricks have 12 layers)
>
> br3 <- stack(brick1, brick2)
> lmS <- function(x) lm(x[1:12] ~ x[13:24)$coefficients[2]
> r <- calc(br3, lmS)
>
> Jacob.
>
> --- On Fri, 26/11/10, steven mosher <mosherste...@gmail.com> wrote:
>
> From: steven mosher <mosherste...@gmail.com>
> Subject: Re: [R-sig-Geo] gridded time series analysis
> To: "Martin" <martin_bra...@gmx.net>
> Cc: r-sig-geo@stat.math.ethz.ch
> Date: Friday, 26 November, 2010, 23:33
>
> that's cool, I'm also interested in a similar problem but just with one
> brick ending up with a slope raster as the output. It may be possible with
> stackApply(). have a look. or maybe robert will chime in
>
>
>
> On Fri, Nov 26, 2010 at 1:35 PM, Martin <martin_bra...@gmx.net> wrote:
>
>>
>> this is what I did to perform a regression between two bricks (each brick
>> represent a time series):
>>
>> r <- raster(brick1)
>> for (i in 1:ncell(r)) {
>> r[i] = lm(as.ts(cellValues(brick1, i)) ~ as.ts(cellValues(brick2,
>> i)))$coefficients[2]
>> }
>>
>> The result will be a slope raster, but it really takes a lot of time, so
>> maybe there is a better solution..
>>
>>
>> --
>> View this message in context:
>> http://r-sig-geo.2731867.n2.nabble.com/gridded-time-series-analysis-tp5775651p5778472.html
>> Sent from the R-sig-geo mailing list archive at Nabble.com.
>>
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