Large 3d array manipulation

Sat, Feb 28, 2009 6:20 AM

On Feb 27, 2009, at 10:37 PM, David Winsemius wrote:

On Feb 27, 2009, at 7:49 PM, Duncan Murdoch wrote:

On 27/02/2009 6:15 PM, Vemuri, Aparna wrote:

I have a large 3 dimensional array of size (243,246,768)
The first dimension is Rows, second is columns and the third is  
Time. So for each row and column, I want to calculate the mean of  
time steps
1:8, 2:9, 3:10 and so on and assign the values to a new array. For  
this
I am using the following script.
for(i in 1:243)
{
for(j in 1:246)
{
for(k in 1:768)
{
newVar[i,j,k] <- mean( myVar[i,j,k:k+7])
}
}
}
This works, but needless to mention it take a very long time to loop
over all the rows, columns and time periods. I was wondering if  
there is
a simpler way to do this.

Yes, vectorize it.  I think this would work:

newVar <- array(NA, c(243, 246, 768))

for (k in 1:768)
newVar[,,k] <- apply(myVar, 1:2, function(x) mean(x[k:(k+7)]))

That's rather interesting. I had not realized that one could use  
that construction with apply. I had earlier tried to substitute  
rollmean for the inner loop. For one thing I thought that trying to  
index 768+7 was going to create some problems with out-of-range  
indexing. I got the same error with my effort to insert rollmean in  
the OP's construction as I do in this construction:

myVar <- array(1:(243*246*768), dim=c(243,246,768))

myVar[1,1,1:8]

[1]      1  59779 119557 179335 239113 298891 358669 418447

newVar <-array(,dim=c(243,246,768))
library(zoo)
for (k in 1:768)  newVar[,,k] <- apply(myVar, 1:2, function(x)  
mean(x[k:(k+7)]))
Error in newVar[, , k] <- apply(myVar, 1:2, function(x) rollmean(x,  
8)) :
 number of items to replace is not a multiple of replacement length

I am guessing that at some point the assignment function cannot  
resolve which index in newVar to use. So I tried redimensioning  
newVar to only have 761 as it third dimension and using:

myVar[1,1,1:8]

for (k in 1:761)  {newVar[ , , ] <- apply(myVar, 1:2, function(x)  
mean(x[k:(k+7)])) ; print(k)}

I forgot to mention that at this point I decided that the for loop  
might be entirely superfluous if one were returning a 761 length  
vector from rollmean. So I tested my theory:

Large 3d array manipulation

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