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PLEASE do read the posting guide http://www.R
Regards
karthick
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But the meaning of a 3x4 table is rather different than the meaning of
a 1x12 table.
Regardless, you probably want to start with integer factorization, and
can read more
about implementations in R here (and elsewhere):
http://tolstoy.newcastle.edu.au/R/help/05/01/10007.html
Sarah
On Thu, Jun
-help@r-project.org
Subject: [R] R matrix help
Dear R experts,
I am interested in getting the dimensions for the matrix dynamically,
based
on the the number of elements in a matrix for example. if the number is
12,
I should get dim= 3X4, if it is 20, dim=5X4.
please help me do
. if the number is 12,
I should get dim= 3X4, if it is 20, dim=5X4.
please help me do this.
Thank you
Regards
karthick
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-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of karthicklakshman
Sent: Thursday, June 14, 2012 7:51 AM
To: r-help@r-project.org
Subject: [R] R matrix help
Dear R experts,
I am interested in getting the dimensions for the matrix dynamically
Can you explain why n=12 should result in 3x4 instead of 2x6 or 6x2 or
4x3 or 1x12 ?
On Thu, Jun 14, 2012 at 8:51 AM, karthicklakshman
karthick.laksh...@gmail.com wrote:
Dear R experts,
I am interested in getting the dimensions for the matrix dynamically, based
on the the number of elements
,
I should get dim= 3X4, if it is 20, dim=5X4.
please help me do this.
Thank you
Regards
karthick
--
View this message in context:
http://r.789695.n4.nabble.com/R-matrix-help-tp4633372.html
Sent from the R help mailing list archive at Nabble.com
On Wed, May 18, 2011 at 9:49 PM, jim holtman jholt...@gmail.com wrote:
Is this what you were after:
mdat - matrix(c(1,0,1,1,1,0), nrow = 2, ncol=3, byrow=TRUE,
+ dimnames = list(c(T1, T2),
+ c(sp.1, sp.2, sp.3)))
mdat
sp.1 sp.2 sp.3
T1 1
Is this what you are looking for:
mdat3
sp.1 sp.2 sp.3 sp.4 sp.5
T110010
T210010
T311100
T410111
# create a matrix of when species first appeared
first - apply(mdat3, 2, function(x) (cumsum(x == 1) 0) + 0L)
#
Dear R help,
Apologies for the less than informative subject line. I will do my
best to describe my problem.
Consider the following matrix:
mdat - matrix(c(1,0,1,1,1,0), nrow = 2, ncol=3, byrow=TRUE,
dimnames = list(c(T1, T2),
c(sp.1, sp.2, sp.3)))
Is this what you were after:
mdat - matrix(c(1,0,1,1,1,0), nrow = 2, ncol=3, byrow=TRUE,
+ dimnames = list(c(T1, T2),
+ c(sp.1, sp.2, sp.3)))
mdat
sp.1 sp.2 sp.3
T1101
T2110
# do 'rle' on each column and see if it is
Listers,
I have a simple matrix:
--
m -c(1:7)
m - cbind(m)
m
[1,] 1
[2,] 2
[3,] 3
[4,] 4
[5,] 5
[6,] 6
[7,] 7
---
I want to add a second column using:
[[alternative HTML version deleted]]
On Sun, Feb 20, 2011 at 5:55 PM, Dmitry Berman ravenb...@gmail.com wrote:
Listers,
I have a simple matrix:
--
m -c(1:7)
m - cbind(m)
m
[1,] 1
[2,] 2
[3,] 3
[4,] 4
[5,] 5
[6,] 6
[7,] 7
---
I want to add a second
On Feb 20, 2011, at 5:56 PM, Dmitry Berman wrote:
On Sun, Feb 20, 2011 at 5:55 PM, Dmitry Berman ravenb...@gmail.com
wrote:
Listers,
I have a simple matrix:
--
m -c(1:7)
m - cbind(m)
m
[1,] 1
[2,] 2
[3,] 3
[4,] 4
[5,] 5
[6,] 6
[7,] 7
On Sun, Feb 20, 2011 at 2:56 PM, Dmitry Berman ravenb...@gmail.com wrote:
On Sun, Feb 20, 2011 at 5:55 PM, Dmitry Berman ravenb...@gmail.com wrote:
Listers,
I have a simple matrix:
--
m -c(1:7)
m - cbind(m)
m
[1,] 1
[2,] 2
[3,] 3
[4,] 4
[5,] 5
[6,]
Anyone know how write a function that solves:
(1 + c)x1 +x2 +x3 = 5
x1+(1 + c)x2+x3 = 5 + 2c
x1+x2 +(1 + c)x3= 5 + 3c,
where c is a small constant, for 1000 equidistant values c = (10^-14,
2*10^-14,
I think you want ?solve ...
Remko
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