Rereading your email, still not sure what the question is -- perhaps you
could give a better code example to illustrate the difference between a[[2]]
and mat1 -- but, since you mentioned briefly lists of lists, have you looked
at unlist(, recursive = F)? If applied to a list of lists, it won't unlist
the sub-lists and that gives you a list of matrices, which, as I said
before, are matrices when subsetted with `[[`, but not `[`.
Michael
On Wed, Aug 24, 2011 at 2:37 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> I'm not sure I understand your question: a[[2]] is a matrix.
>
> > a <- list(matrix(1:6,2),matrix(5:10,2))
> > is.matrix(a[[2]])
> TRUE
> x = a[[2]]
> > is.matrix(x)
> TRUE
> > x+2
> [,1] [,2] [,3]
> [1,] 7 9 11
> [2,] 8 10 12
> > a[[2]] + 2
> [,1] [,2] [,3]
> [1,] 7 9 11
> [2,] 8 10 12
>
> What else do you need?
>
> Michael Weylandt
>
>
> On Wed, Aug 24, 2011 at 2:28 PM, Chee Chen <chee.c...@yahoo.com> wrote:
>
>> Dear All,
>> As always, I appreciate all your help.
>> I would like to know the easiest way to convert each of the homogeneous
>> elements of a numeric list into a matrix. Each element of this list is also
>> a list such that when displayed, looks like a 2-by-3 matrix , I would like
>> to convert each of them into a matrix, without changing the double index of
>> each entry.
>>
>> Suppose:
>> a<- vector("list",2)
>> > a[[1]]
>> [,1] [,2] [,3]
>> [1,] 1 2 0
>> [2,] 3 4 5
>>
>> > a[[2]]
>> [,1] [,2] [,3]
>> [1,] 5 6 9
>> [2,] 7 8 10
>>
>> >is.list(a[[1]])
>> True
>>
>> Target: I would like to convert a[[2]] into a matrix, keeping the double
>> index, into
>> > mat1
>> [,1] [,2] [,3]
>> [1,] 5 6 9
>> [2,] 7 8 10
>>
>> The list I have is huge and so is each of its elements, and do not want to
>> use unlist because I do not understand it fully.
>> Thank you,
>> Chee
>>
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
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>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
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