Thanks!
Antonio
>
>
>> -Christos
>>
>> -Original Message-
>> From: [EMAIL PROTECTED]
>> [mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez
>> Sent: Tuesday, November 07, 2006 2:01 PM
>> To: R-Help
>> Subject: [R] subsetting a matrix an
prompt again. I think I have enough memory (512) and a
pretty good processor (P4, 2.8)
Antonio
> -Christos
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez
> Sent: Tuesday, November 07, 2006 2:01 PM
> To: R-H
: R-Help
Subject: [R] subsetting a matrix and filling other
Hi,
Having a matrix F(189,6575) I want to do this:
z1<-subset(F[,1], F[,1] >= 5 & F[,1] <= 10) .
.
.
z189<-subset(F[,189], F[,189] >= 5 & F[,189] <= 10)
I would prefer to have an empty matrix, say 'z&#
Hi,
Having a matrix F(189,6575) I want to do this:
z1<-subset(F[,1], F[,1] >= 5 & F[,1] <= 10)
.
.
.
z189<-subset(F[,189], F[,189] >= 5 & F[,189] <= 10)
I would prefer to have an empty matrix, say 'z' in order to fill its
columns with the output of subsetting F. But each of the subsets can
dif
006 at 16:29, Camarda, Carlo Giovanni wrote:
Date sent: Mon, 30 Jan 2006 16:29:38 +0100
From: "Camarda, Carlo Giovanni" <[EMAIL PROTECTED]>
To:
Subject: [R] Subsetting a matrix without for-loop
> Dear R-users,
&g
use 'filter':
> x <- matrix(1:100,10)
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]1 11 21 31 41 51 61 71 8191
[2,]2 12 22 32 42 52 62 72 8292
[3,]3 13 23 33 43 53 63 73 8393
[4,]4 14 24 34 4
The result is linear in A so its a matter of finding the matrix to multiply it
by:
matrix(c(rep(1,3), rep(0,7)), 3, 9, byrow = TRUE) %*% A
On 1/30/06, Camarda, Carlo Giovanni <[EMAIL PROTECTED]> wrote:
> Dear R-users,
> I'm struggling in R in order to "squeeze" a matrix without using a
> for-loo
Dear R-users,
I'm struggling in R in order to "squeeze" a matrix without using a
for-loop.
Although my case is a bit more complex, the following example should
help you to understand what I would like to do, but without the slow
for-loop.
Thanks in advance,
Carlo Giovanni Camarda
A <- matrix(1:5
On Tue, 2003-09-30 at 15:51, Jason Turner wrote:
> (Ted Harding) wrote:
>
> > On 30-Sep-03 Rajarshi Guha wrote:
> >
> >>Hi,
> >> I'm trying to take a set of rows and columns out of a matrix. I hve
> >>been using the index aray approach. My overll matrix is X and is 179 x
> >>65. I want to take o
(Ted Harding) wrote:
On 30-Sep-03 Rajarshi Guha wrote:
Hi,
I'm trying to take a set of rows and columns out of a matrix. I hve
been using the index aray approach. My overll matrix is X and is 179 x
65. I want to take out 4 columns and 161 rows.
...
This is documented in "An Introduction to R", u
On 30-Sep-03 Rajarshi Guha wrote:
> Hi,
> I'm trying to take a set of rows and columns out of a matrix. I hve
> been using the index aray approach. My overll matrix is X and is 179 x
> 65. I want to take out 4 columns and 161 rows.
>
> Thus I made a 161 x 2 array I and filled it up with the row,
Rajarshi -
Why not simply subscript your matrix X to return the rows and
columns you want to keep ? For example,
new <- X[16:176, c(3,5,7,9)]
assuming those are the rows and columns you want.
See help("Extract").
- tom blackwell - u michigan medical school - ann arbor -
On Tue, 30 Sep
Hi,
I'm trying to take a set of rows and columns out of a matrix. I hve
been using the index aray approach. My overll matrix is X and is 179 x
65. I want to take out 4 columns and 161 rows.
Thus I made a 161 x 2 array I and filled it up with the row,col indices.
However doing,
X[ I ] gives me a
On 15-Jul-03 Achim Zeileis wrote:
>> >[...]
>> > mu1<-mu[iX1,drop=FALSE]; mu2<-mu[iX2,drop=FALSE];
>> > mu1
>>
>> [1] 1
>>
>> > mu2
>>
>> [1] 2 3
>>
>> So now I still don't get my 1xk matrices, even though mu is a
>> matrix and I've used "drop=FALSE". Why?
>
> Because you are subsetting mu like a
On Tuesday 15 July 2003 12:14, Ted Harding wrote:
> Hi Folks,
>
> People's suggestion of "drop=FALSE" seemed to do the trick
> (preserving "matrix" character when subestting to a row,
> i.e. creating 1xk matrix).
>
> However, I seem to have encountered a case where even this does
>
> not work:
> >
Hi Folks,
People's suggestion of "drop=FALSE" seemed to do the trick
(preserving "matrix" character when subestting to a row,
i.e. creating 1xk matrix).
However, I seem to have encountered a case where even this does
not work:
> mu<-c(1,2,3)
> mu<-matrix(mu,nrow=1)
> mu
[,1] [,2] [,3]
On 14-Jul-03 Adelchi Azzalini wrote:
>> Maybe it is reasonable to propose incorporating "drop" as one of the
>> things you can set with "options"? :--
>> [...]
>
> Perhaps this is a way-out. As usual, everything is feasible when we
> resort entirely on our own code. The danger is that we might ma
> Maybe it is reasonable to propose incorporating "drop" as one of the
> things you can set with "options"? :--
>
> `options' allows the user to set and examine a variety of global
> ``options'' which affect the way in which R computes and displays
> its results.
>
Perhaps this is
On 14-Jul-03 Adelchi Azzalini wrote:
> Personally, I find this automatic conversion to "vector" a somewhat
> confusing feature (although I can see its reasons), and I know of
> many people that would have preferred that drop=FALSE was the
> default behaviour, but surely now is difficult to cha
On Monday 14 July 2003 11:59, you wrote:
> I'd welcome some comments or advice regarding the situation described
> below.
>
> The following illustrates what seems to me to be an inconsistency
> in the behaviour of matrix subsetting:
>
> > Z<-matrix(c(1.1,2.1,3.1,1.2,2.2,3.2,1.3,2.3,3.3),nrow=3)
>
I'd welcome some comments or advice regarding the situation described
below.
The following illustrates what seems to me to be an inconsistency
in the behaviour of matrix subsetting:
> Z<-matrix(c(1.1,2.1,3.1,1.2,2.2,3.2,1.3,2.3,3.3),nrow=3)
> Z
[,1] [,2] [,3]
[1,] 1.1 1.2 1.3
[2
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