[R] multidimensional lists

[Gabor Grothendieck]

Try this:

AA <- matrix(list(A, 10*A, 100*A, 1000*A), 2, byrow = TRUE)
AA[1,2]


On 9/27/06, Evan Cooch <cooch17 at verizon.net> wrote:
> In the process of moving a number of my scripts from MATLAB -> R, I've
> discovered that there is no 'pure' equivalent of MATLAB's cell arrays,
> which I use quite often. Basically, I create matrices (as a cell array)
> where each element of the matrix is itself a matrix (e.g., 2x2 cell
> array where each element of the array is another matrix). I pass these
> cell arrays to various functions which then do clever things with the
> various matrices (of course) - basically, I need to be able to pass
> collections of matrices to functions to do various things, and I need to
> be able to control the dimensionality of the cell array to preserve some
> structural relationships among the matrices in the array. The cell array
> in MATLAB handles this with aplomb.
>
> So far, in R, I've used lists. Given (say) 4 matrices (A,B,C,D), in
> MATLAB I can use
>
> test={A,B,C,D} for a row vector cell array, or
>
> test={A;B;C;D} for a column vector cell array.
>
> In R, I get more or less the same thing using
>
> test=list(A,B,C,D)
>
> but this only gives me a row list. For a bunch of technical reasons, I
> need to be able to control the orientation (as noted)- this is
> especially true for n-dimensional cell arrays. In MATLAB, for example, I
> could generate a (say) 2x2 cell array using
>
> test={A B;C D}
>
> The only way I can figure out how to do this in R is using something like
>
> test=list(A,B,C,D);
> dim(test) < c(2,2);
>
> This seems to work, but defaults to bycolumn (in other words, instead of
>
> A  B
> C  D
>
> I get
>
> A  C
> B  D
>
> )
>
> So, I follow with
>
> test=t(test) as needed to flip the thing around to byrow.
>
> OK, so the question is - is there a better way? This *seems* to work,
> but I'm discovering that R is a lot like working with LaTeX (something I
> know much more about) - you can do most things, but there is often a
> more elegant way if you can figure out how to find out about it.
>
> Thanks in advance...
>
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