Here is the answer for the second part:
> lst=list(a1=matrix(rnorm(4),2,2),
a2=matrix(rnorm(4),2,2),a3=matrix(rnorm(4),2,2))
> l.dim <- sapply(lst, dim)
> l.dim
a1 a2 a3
[1,] 2 2 2
[2,] 2 2 2
> all(l.dim[1,1] == l.dim[1,]) && all(l.dim[2,1] == l.dim[2,])
[1] TRUE
>
For the first part, why don't you just leave it a list and access the list
elements by names? If you want to create new objects, the 'for' is fine.
On Mon, Jun 15, 2009 at 11:58 AM, Zhang,Yanwei <yanwei.zh...@cna.com> wrote:
> Dear R users,
>
> I have two simple questions here, and hope someone can help me on this.
> Thanks in advance.
>
> 1.
> I have a list object lst=list(a1=matrix(rnorm(4),2,2),
> a2=matrix(rnorm(4),2,2),a3=matrix(rnorm(4),2,2)). Here I only use three
> elements for illustration, and in fact the length of lst, n, is unknown in
> advance. I want to define an object for each element of this lst, and the
> objects have names like "object1", "object2",...,"objectn". I could use a
> loop like the following:
> for (i in 1:length(lst)){
> assign(paste("object",i,sep=""),lst[[i]])
> }
>
> I wonder if there's a way to avoid this loop?
>
>
> 2.
> I want to validate whether all of the "object"s have the same dimension,
> that is, whether dim(object1)[1]==dim(object2)[1]==dim(object3)[1]&
> dim(object1)[2]==dim(object2)[2]==dim(object3)[2] is true. How can I do
> that easily since the length of lst is not fixed?
>
> Thanks.
>
> Wayne (Yanwei) Zhang
> Ttatistical *esearch
> > CNA
> >
> >
> >
>
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