Use a list to store the partial matrices:
z - vector(list, 189)
for(i in 1:189)
z[[i]] - subset(...)
-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]
Christos Hatzis escribió:
Use a list to store the partial matrices:
z - vector(list, 189)
for(i in 1:189)
z[[i]] - subset(...)
Hi Christos,
I've tried it but it died unexpectedly with a (Killed) message and the
linux console prompt again. I think I have enough memory (512) and a
antonio rodriguez escribió:
Christos Hatzis escribió:
Use a list to store the partial matrices:
z - vector(list, 189)
for(i in 1:189)
z[[i]] - subset(...)
Hi Christos,
I've tried it but it died unexpectedly with a (Killed) message and the
linux console prompt again. I think I
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
Hi
matrix(colSums(embed(A,3)[1:3,]),3,6, byrow=T)[3:1,]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]6 33 60 87 114 141
[2,]9 36 63 90 117 144
[3,] 12 39 66 93 120 147
will do it. However I am not sure if it is quicker than your for
loop.
HTH
Petr
On 30 Jan 2006
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
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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