Or this:
brmax <- max(br)
On Fri, Nov 26, 2010 at 7:01 AM, Agustin Lobo <alobolis...@gmail.com> wrote:
> ...well, it's Friday...
>
> it's actually just
>> brmax <- calc(br,max,na.rm=T)
>> show(brmax)
> class : RasterBrick
> filename :
> nlayers : 1
> nrow : 5
> ncol : 5
> ncell : 25
> projection : NA
> min value : 248
> max value : 248
> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax)
> resolution : 0.2, 0.2 (x, y)
>
> that simple.
>
> Agus
>
> 2010/11/26 Agustin Lobo <alobolis...@gmail.com>:
>> Hi!
>>
>> Given a brick object (longitude, latitude, time), I need a (lon,lat)
>> layer with the maximum value of each cell across time.
>> That's the equivalent to
>>> x = array(round(runif(n=3*4*5,min=0,max=10)),dim=c(3,4,5))
>>> apply(x,c(1,2),max,na.rm=T)
>>
>> but for x being a brick object. How can I do it with raster?
>>
>> I've tried
>>
>>> funmax <- function(x) {apply(x,c(1,2),max,na.rm=T)}
>>> show(br)
>> class : RasterBrick
>> filename :
>> nlayers : 3
>> nrow : 5
>> ncol : 5
>> ncell : 25
>> projection : NA
>> min value : ? ? ?
>> max value : ? ? ?
>> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax)
>> resolution : 0.2, 0.2 (x, y)
>>> brmax <- calc(br,funmax)
>>
>> But the result is wrong, as it should be 1 layer and not 3:
>>> show(brmax)
>> class : RasterBrick
>> filename :
>> nlayers : 3
>> nrow : 5
>> ncol : 5
>> ncell : 25
>> projection : NA
>> min value : 232 232 248
>> max value : 248 248 248
>> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax)
>> resolution : 0.2, 0.2 (x, y)
>>
>> I've also tried with stackApply(), but do not think that's the way to go.
>> Also, I'm concerned that apply() is typically very slow in R, and I'm
>> dealing with a large brick object.
>>
>> Thanks
>>
>> Agus
>>
>
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