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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