Thanks a lot Jim.
I just switched back to R from SAS, so I might carry over the old style of
creating tons of SAS macro variables...
Another related question is :
Suppose each element in my lst now is a numeric character:
lst=list(a1=rnorm(4), a2=rnorm(4),a3=rnorm(4))
Still the number of elements is not fixed.
I'm trying to run a system of equations (actually just separate regressions)
like the following:
fit1= lm(lst[[1]]~lst[[2]]+lst[[3]])
fit2= lm(lst[[2]]~lst[[1]]+lst[[3]])
fit3= lm(lst[[3]]~lst[[1]]+lst[[2]])
Is there any easy way to do the above for a list of non-fixed length?
Thanks a lot.
Wayne (Yanwei) Zhang
Ttatistical Áesearch
CNA
________________________________
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Monday, June 15, 2009 11:14 AM
To: Zhang,Yanwei
Cc: r-help@r-project.org
Subject: Re: [R] Create R object
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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