[R] Subarray specification problem
Michael Friendly
friendly at yorku.ca
Fri Dec 17 14:58:26 CET 2010
Use aperm() to make time the first dimension
Reshape to a matrix (all other dimensions combined)
Do your selection on X[1,]
aperm() to Permute back
On 12/16/2010 11:00 AM, Roy Shimizu wrote:
> Hi. I'm new to R, and I'm still learning R's system for addressing
> subsets of data structures. I'm particularly interested in the
> problem of selecting subarrays based on complex criteria involving the
> dimnames (as opposed to the values of the cells) of the array. Here's
> an example of such a problem.
>
> Suppose I have an array x of unknown dimensions (it may have been
> passed as the argument to a function I'm coding), but I know that one
> of its dimensions is called "time", and has values that are (or can be
> meaninfully coerced into) integers. To make this specification
> clearer, here's one possible example of such an array x:
>
>> (x<- array(runif(2*5*2), dim=c(2,5,2), dimnames=list(NULL, time=round(100*runif(5)), NULL)))
> , , 1
>
> time
> 84 69 61 16 77
> [1,] 0.4020976 0.8250189 0.3402749 0.09754860 0.2189114
> [2,] 0.5309967 0.5414850 0.9431449 0.08716723 0.5819100
>
> , , 2
>
> time
> 84 69 61 16 77
> [1,] 0.6238213 0.1210083 0.7823269 0.5004058 0.5474356
> [2,] 0.2491087 0.7449411 0.9561074 0.6685954 0.3871533
>
>
> Now, here's the problem: I want to write an R expression that will
> give me the subarray y of x consisting of the cells whose "time"
> dimensions are greater than 20 but less than 80.
>
> For the example x given above, the desired expression would evaluate
> to this array:
>
> , , 1
>
> time
> 69 61 77
> [1,] 0.8250189 0.3402749 0.2189114
> [2,] 0.5414850 0.9431449 0.5819100
>
> , , 2
>
> time
> 69 61 77
> [1,] 0.1210083 0.7823269 0.5474356
> [2,] 0.7449411 0.9561074 0.3871533
>
>
>
> How can I write such an expression in the general array x as described above?
>
> Remember, the x shown above is just an example. In the general case
> all I know is that one of x's dimensions is called "time", and that
> its values are [or can be coerced meaningfully] into integers. I
> *don't* know where among x's dimensions it is. Hence, the following
> is *not* a solution to the problem, even though it produces the right
> answer for the example above:
>
>> t<- as.integer(dimnames(x)$time)
>> y<- x[,which(t> 20& t< 80),]
>
> This solution does not work in general, because the expression
> "x[,which(t> 20& t< 80),]" relies on the prior knowledge that the
> "time" dimension is the second one of three.
>
> Any ideas?
>
> Thanks!
>
> Roy
>
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele Street Web: http://www.datavis.ca
Toronto, ONT M3J 1P3 CANADA
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