Thank you Bert.
Yes, your suggestion is correct and there is no need to pre-define the matrix
and the sapply function works quite fast. This resolves my issue.
Thank you both againHC
On Sunday, 3 July 2016 11:38 AM, Bert Gunter <bgunter.4...@gmail.com> wrote:
Well, yes, ... but no: there is no need to pre-define the matrix.
The following is still a (interpreted) loop, but it is fast and short.
## ex is the downloaded array, here filled with random numbers
reqX = c(35,35,40,65,95)
reqY = c(2,5,10,112,120)
out <-sapply(seq_along(reqX), function(i)ex[reqX[i],reqY[i],] )
> dim(out)
[1] 365 5
You might find it useful to go through a (web) tutorial or two to
learn more about such R functionality.
Useful suggestions can be found here: https://www.rstudio.com/online-learning/#R
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Jul 2, 2016 at 3:43 PM, Hemant Chowdhary via R-help
<r-help@r-project.org> wrote:
> I am working with a 3-dimensional netCDF file having dimensions of X=100,
>Y=200, T=365.
> My objective is to extract time vectors of a few specific grids that may not
> be contiguous on X and/or Y.
>
> For example, I want to extract a 5x365 matrix where 5 rows are each vectors
> of length 365 of 5 specific X,Y combinations.
>
> For this, I am currently using the following
>
> reqX = c(35,35,40,65,95);
> reqY = c(2,5,10,112,120,120);
> nD = length(reqX)
> for(i in 1:nD){
> idX = ncvar_get(nc=myNC, varid="myVar", start=c(reqX[i],reqY[i]),
> count=c(1,1))
> if(i==1){dX = idX} else {dX = rbind(dX,idX)}
> }
>
> Is there more elegant/faster way other than to using a For Loop like this? It
> seems very slow when I may have to get much larger matrix where nD can be
> more than 1000.
>
> Thank you HC
>
> [[alternative HTML version deleted]]
>
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