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
differ in
] 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' in order to fill its columns
with the output of subsetting F
) 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-Help
Subject: [R] subsetting a matrix and filling other
Hi,
Having a matrix F
Of antonio rodriguez
Sent: Tuesday, November 07, 2006 2:01 PM
To: 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
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(1:54,
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-loop.
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 44
at 16:29, Camarda, Carlo Giovanni wrote:
Date sent: Mon, 30 Jan 2006 16:29:38 +0100
From: Camarda, Carlo Giovanni [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Subsetting a matrix without for-loop
Dear R-users,
I'm
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
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
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,col
(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,
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 out 4 columns and
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:
mu-c(1,2,3)
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 vector not like a matrix. The
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,]
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 change
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 make
use of
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