[R-sig-Geo] How to calculate climatology in rasterbricks

[Loïc Dutrieux]

This can also be done with zApply:

library(zoo)

sYM <- zApply(s, by = as.yearmon, sum)
sM <- zApply(sYM, by = months, mean)

Cheers,
Loïc

On 06/03/2016 02:02 AM, Vijay Lulla wrote:
> I think the following StackOverflow question has the answer:
> http://stackoverflow.com/questions/16135877/applying-a-function-to-a-multidimensional-array-with-grouping-variable/16136775#16136775
>
> Following the instructions listed on that page for your case might go
> something like below:
>
>> idxYM <- as.integer(strftime(idx,"%Y%m"))
>> idxM <- unique(idxYM)%%100
>> meanYM <- calc(s,fun=function(x) { by(x, idxYM, mean) })
>> meanYM
> class       : RasterBrick
> dimensions  : 20, 20, 400, 360  (nrow, ncol, ncell, nlayers)
> resolution  : 18, 9  (x, y)
> extent      : -180, 180, -90, 90  (xmin, xmax, ymin, ymax)
> coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0
> data source : in memory
> names       : X196101, X196102, X196103, X196104, X196105, X196106,
> X196107, X196108, X196109, X196110, X196111, X196112, X196201,
> X196202, X196203, ...
> min values  :  0.3728,  0.2725,  0.3421,  0.3652,  0.3342,  0.3185,
> 0.3130,  0.3780,  0.3376,  0.3727,  0.3537,  0.3737,  0.3515,  0.3588,
>   0.3334, ...
> max values  :  0.6399,  0.6652,  0.6583,  0.6640,  0.6359,  0.6761,
> 0.6442,  0.6800,  0.6397,  0.6769,  0.6489,  0.6388,  0.6471,  0.6661,
>   0.6255, ...
>
>> meanM <- calc(meanYM, fun=function(x) { by(x, idxM, mean) })
>> meanM
> class       : RasterBrick
> dimensions  : 20, 20, 400, 12  (nrow, ncol, ncell, nlayers)
> resolution  : 18, 9  (x, y)
> extent      : -180, 180, -90, 90  (xmin, xmax, ymin, ymax)
> coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0
> data source : in memory
> names       :     X1,     X2,     X3,     X4,     X5,     X6,     X7,
>     X8,     X9,    X10,    X11,    X12
> min values  : 0.4645, 0.4715, 0.4768, 0.4717, 0.4749, 0.4705, 0.4697,
> 0.4724, 0.4629, 0.4774, 0.4736, 0.4708
> max values  : 0.5274, 0.5275, 0.5293, 0.5259, 0.5285, 0.5276, 0.5269,
> 0.5260, 0.5256, 0.5281, 0.5279, 0.5286
>
>>
>
> I'm not sure how [in]efficient this is for actual (i.e. not toy
> example) data.  Maybe others more experienced, and knowledgeable,
> members can provide better answers.
>
> HTH,
> Vijay.
>
> On Thu, Jun 2, 2016 at 4:30 PM, Thiago V. dos Santos via R-sig-Geo
> <r-sig-geo at r-project.org> wrote:
>> Dear all,
>>
>> I am working with daily time series of meteorological variables. This is an example of the dataset:
>>
>> library(raster)
>>
>> # Create date sequence
>> idx <- seq(as.Date("1961/1/1"), as.Date("1990/12/31"), by = "day")
>>
>> # Create raster stack and assign dates
>> r <- raster(ncol=20, nrow=20)
>> s <- stack(lapply(1:length(idx), function(x) setValues(r, runif(ncell(r)))))
>> s <- setZ(s, idx)
>>
>>
>> Now, let's assume those values represent daily precipitation. What I need to do is to integrate daily to monthly values,
>> and then take a monthly climatology. Climatology in this case means multi-year average of selected months, e.g., an average of the 30 Octobers from 1961 to 1990, an average of the 30 Novembers from 1961 to 1990 and etc.
>>
>> On the other hand, let's assume the raster values represent daily temperature. Integrating daily to monthly temperature doesn't make sense. Hence, instead of integrating daily values, I need to take monthly means (e.g. mean value of all days in every month), and then calculate the climatology.
>>
>> What would be the best approach to achieve that using the raster package?
>>
>>   Greetings,
>>   -- Thiago V. dos Santos
>>
>> PhD student
>> Land and Atmospheric Science
>> University of Minnesota
>>
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