This gets you close to what you want. The 31/33 are row names that
you can extract and store as actual columns,
> x <- list('1995'=list('31'=1, '33'=2), '2006'=list('31'=3, '33'=4))
> y <- lapply(names(x), function(.L1){
+ cbind(as.numeric(.L1),do.call(rbind, x[[.L1]]))
+ })
> (z <- do.call(r
Hello, I have a list in which each element is a list. I want to
create a matrix indexed by the two indices of the list. I have been
using do.call, but I am not getting what I want. Let me show you:
> l.intercepts #the list that nests another list
$`1995`
$`1995`$`31`
(Intercept)
25.37164
$`199
e it is
not obvious what to return without giving extra information, it is better to
require the extra information through other functions.
From: [EMAIL PROTECTED] on behalf of Olivier Lefevre
Sent: Sat 4/26/2008 9:43 AM
To: [EMAIL PROTECTED]
Subject: Re: [R] m
__
>
> From: [EMAIL PROTECTED] on behalf of Olivier Lefevre
> Sent: Sat 4/26/2008 9:43 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [R] matrix from list
>
>
>
> Olivier Lefevre wrote:
> > Anyway you are right that it would still return the kind of object, only
> >
It's true one may have to set some rules but I think you are blowing it up.
First, it is true one would have to agree for list[[vec]] to always return
a matrix but it is the useful behaviour since you can already get a vector
with unlist(list[vec]).
Second, as to the raggedness, matrix(), array
her functions.
From: [EMAIL PROTECTED] on behalf of Olivier Lefevre
Sent: Sat 4/26/2008 9:43 AM
To: [EMAIL PROTECTED]
Subject: Re: [R] matrix from list
Olivier Lefevre wrote:
> Anyway you are right that it would still return the kind of object, only
> subs
Martin Maechler wrote:
> The difference to as.matrix() is that data.matrix() also
> produces a numeric matrix in the case the data frame contains
> factors.
Thanks, that is useful but it is becoming a little rococo: so may ways to
do this. Also, what if I have a list, not a data frame? read
Olivier Lefevre wrote:
> Anyway you are right that it would still return the kind of object, only
> subsetted, which is not I want.
I mean [] would do that; I know [[]] doesn't. Yet I still don't see why one
accepts vector arguments but not the other: they are both indexing
operators. It is suc
> "OL" == Olivier Lefevre <[EMAIL PROTECTED]>
> on Sat, 26 Apr 2008 01:31:16 +0200 writes:
OL> Greg Snow wrote:
>> The '[[' only returns a single element from a data structure
OL> I know but that is precisely what I find arbitrary.
OL> Anyway you are right that it wou
Greg Snow wrote:
> The '[[' only returns a single element from a data structure
I know but that is precisely what I find arbitrary.
Anyway you are right that it would still return the kind of object, only
subsetted, which is not I want. As someone kindly pointed out to me
offline, besides the u
lthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Olivier Lefevre
> Sent: Thursday, April 24, 2008 4:16 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [R] matrix from list
>
> Yes, unlist is
Yes, unlist is the magic wand I was looking for. Thanks a million!
Having said that, I find it rather arbitrary that you can write mat[1:4]
but not list[[1:2]]; IMO there should be no need for a "magic" operator
like unlist: list[[1:length(list)]] could do the job.
-- O.L.
Hi Olivier,
is this what you want?
x="col1 col2
1 0.1 1.1
2 0.2 1.2"
m=read.table(textConnection(x),header=TRUE)
m1=matrix(unlist(m),ncol=2)
m1
[,1] [,2]
[1,] 0.1 1.1
[2,] 0.2 1.2
HTH,
Jorge
On Thu, Apr 24, 2008 at 6:02 PM, Olivier Lefevre <[EMAIL PROTECTED]> wrote:
> Another possibly
Another possibly simple thing that I cannot get right is how to extract the
data part of a list as a matrix. The data were read from xls, with labels,
and thus are of list mode, e.g.,
col1 col2
1 0.1 1.1
2 0.2 1.2
I want to extract from that just the numeric data part, i.e., (in this
case
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