I have a matrix of 5 columns and 64 rows, let's call it mat1. All values are
1 or 0. I need to take the values of the elements by row and create a single
numeric element that can be placed its respective slot in a 64-element list,
named list1. For example, mat1[11,1:5] = 0,1,1,0,1 and I must
This seems to work:
apply(mat1,1,function(x){paste(x,collapse=)})
The collapse command inside of paste is (I think) the easiest way to
combine strings.
Michael Weylandt
On Thu, Aug 4, 2011 at 5:45 PM, Wegan, Michael (DNRE) weg...@michigan.gov
wrote:
I have a matrix of 5 columns and
I realize that matrix indexing has been addressed in various flavors, but I'm
stumped and didn't find anything in the archives. It's not clear if it is an
igraph issue or a more general problem. Thanks in advance for your patience.
I am using igraph to read a gml file
Dear all,
I am trying to use the apply family functions to improve the efficiency of
my code, though it is a bit hard for me to find sometimes the solutions with
these functions.
The problem that I want to solve is:
I have a matrix of replicated random generated data, e.g. if I have two
I have a problem with a 3d plot, suppose we have a matrix like this:
v1v2 v3 v4
jan-2010 0.5 0.250.250.3
feb-2010 0.35 0.12 0.120.4
mar-20100.150.250.25 0.1
and i want to plot this matrix in 3d plot where x-axis is
On 05/07/2011 1:36 PM, петрович wrote:
I have a problem with a 3d plot, suppose we have a matrix like this:
v1v2 v3 v4
jan-2010 0.5 0.250.250.3
feb-2010 0.35 0.12 0.120.4
mar-20100.150.250.25 0.1
and i want to plot
On Jul 5, 2011, at 1:36 PM, петрович wrote:
I have a problem with a 3d plot, suppose we have a matrix like this:
v1v2 v3 v4
jan-2010 0.5 0.250.250.3
feb-2010 0.35 0.12 0.120.4
mar-20100.150.250.25 0.1
and i want to
#Hallo again.. Thank you for your answers. To sum up:
#The problem was that we have the matrix m
m-matrix(numeric(length=5*4),nrow=5,ncol=4)
m
# [,1] [,2] [,3] [,4]
# [1,]0000
# [2,]0000
# [3,]0000
# [4,]0000
# [5,]00
Hallo everyone! I have a problem about creating a matrix...
Suppose we have a vector y-c(1,1,1,3,2)
and a zero matrix, m ,with nrows=length(y) and ncol=4.
The matrix would look like this:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
How about:
y - c(1,1,1,3,2)
m - matrix(0, nrow=length(y), ncol=4)
m[y==1, ] - matrix(1:4, nrow=sum(y == 1), ncol=4, byrow=TRUE)
or, depending on your actual problem
y - c(1,1,1,3,2)
m - matrix(0, nrow=length(y), ncol=4)
m[y == 1,] - col(m[y == 1,])
Sarah
On Mon, Jun 20, 2011 at 3:54 PM,
On Jun 20, 2011, at 3:54 PM, Costis Ghionnis wrote:
Hallo everyone! I have a problem about creating a matrix...
Suppose we have a vector y-c(1,1,1,3,2)
and a zero matrix, m ,with nrows=length(y) and ncol=4.
The matrix would look like this:
0 0 0 0
0 0 0 0
Hi,
First of all, I would like to introduce myself as I will probably have many
questions over the next few weeks and want to thank you guys in advance for
your help. I'm a cancer researcher and I need to learn R to complete a few
projects. I have an introductory background in Python.
My
for the
possibility that your numbers have more than one digit.
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ben Ganzfried
Sent: Friday, 3 June 2011 4:54 AM
To: r-help@r-project.org
Subject: [R] Matrix Question
Hi
by columns c and d, but the only outputs are
those from column d
- some values from column d are NA
- column d was created with the code:
df$d=rank(df$c, na.last=keep)
#--R Code-#
item=unique(df$a)
n=length(list)
r=matrix(data=NA,nrow=n, ncol=n, dimnames=list(PRR1=item, PRR2=item
: [R] matrix not working
To: r-help@r-project.org
Date: Thursday, May 26, 2011, 12:24 PM
Hello All,
I'm trying to create a matrix from a dataframe (let's call it df):
..a..b.c.d
a inputs output
b inputs output
c inputs output
d inputs output
e inputs
character, list, and
logical.
--
david.
--- On Thu, 5/26/11, Dat Mai dat.d@gmail.com wrote:
From: Dat Mai dat.d@gmail.com
Subject: [R] matrix not working
To: r-help@r-project.org
Date: Thursday, May 26, 2011, 12:24 PM
Hello All,
I'm trying to create a matrix from a dataframe (let's
.
--
david.
--- On Thu, 5/26/11, Dat Mai dat.d@gmail.com wrote:
From: Dat Mai dat.d@gmail.com
Subject: [R] matrix not working
To: r-help@r-project.org
Date: Thursday, May 26, 2011, 12:24 PM
Hello All,
I'm trying to create a matrix from a dataframe (let's call it df
Subject: [R] matrix not working
To: r-help@r-project.org
Date: Thursday, May 26, 2011, 12:24 PM
Hello All,
I'm trying to create a matrix from a dataframe (let's call it df):
..a..b.c.d
a inputs output
b inputs output
c inputs output
d inputs
Hello everyone,
I have a 2 x 5 matrix: say
0.2 0.3 1 -1 3
0.2. 0.4 5 0.5 -1
I want to replace all the values greater than or equal to 1 with 1 and those
less than or equal to 0 with 0. So I should end up with a mtrix looking
like:
0.2 0.3 1 0 1
0.2. 0.4 1 0.5 0
It's very easy to do in two steps:
testmat - matrix(c(.2, .3, 1, -1, 3, .2, .4, 5, .5, -1), byrow=TRUE, nrow=2)
testmat
[,1] [,2] [,3] [,4] [,5]
[1,] 0.2 0.31 -1.03
[2,] 0.2 0.45 0.5 -1
testmat[testmat = 1] - 1
testmat[testmat 0] - 0
testmat
[,1] [,2] [,3] [,4]
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
Hi all,
I have a problem with getting my code to do what I want!
This is the code I have:
create.means.one.size-function(nsample,var,nboot){
mat.x-matrix(0,nrow=nboot,ncol=nsample)
for(i in 1:nboot){
mat.x[i,]-sample(var,nsample,replace=T)
}
mean.mat-rep(0,nboot)
for(i in
hello
I think if you try this:
for(j in 1: length(nsample)){
MEANS[,]-create.means.one.size(j,var,nboot)
}
it will work
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-manipulation-in-for-loop-tp3525849p3525888.html
Sent from the R help mailing list
Hi,
thank you very much, both methods worked perfectly.
Regards
On Fri, Apr 29, 2011 at 4:17 PM, Berend Hasselman b...@xs4all.nl wrote:
David Winsemius wrote:
On Apr 29, 2011, at 4:27 AM, ivan wrote:
Hi All,
I am trying to create a function which evaluates whether the values
Hi All,
I am trying to create a function which evaluates whether the values (which
are equal to one) of a matrix are the same as their mirror values. Consider
the following matrix:
n-matrix(cbind(c(0,1,1),c(1,0,0),c(0,1,0)),3,3)
colnames(n)-cbind(A,B,C);rownames(n)-cbind(A,B,C)
n
A B C
A 0
On Apr 29, 2011, at 4:27 AM, ivan wrote:
Hi All,
I am trying to create a function which evaluates whether the values
(which
are equal to one) of a matrix are the same as their mirror values.
Consider
the following matrix:
n-matrix(cbind(c(0,1,1),c(1,0,0),c(0,1,0)),3,3)
David Winsemius wrote:
On Apr 29, 2011, at 4:27 AM, ivan wrote:
Hi All,
I am trying to create a function which evaluates whether the values
(which
are equal to one) of a matrix are the same as their mirror values.
Consider
the following matrix:
Is there an easy way to turn a vector of length n into an n by n matrix, in
which the diagonal equals the vector, the first off diagonal equals the
first order differences, the second... etc. I.e. to do this more
efficiently:
diffmatrix - function(x){
n - length(x);
M - diag(x);
Jeroen Ooms jeroenooms at gmail.com writes:
Is there an easy way to turn a vector of length n into an n by n matrix, in
which the diagonal equals the vector, the first off diagonal equals the
first order differences, the second... etc. I.e. to do this more
efficiently:
diffmatrix -
On Apr 27, 2011, at 7:25 AM, Hans W Borchers wrote:
Jeroen Ooms jeroenooms at gmail.com writes:
Is there an easy way to turn a vector of length n into an n by n
matrix, in
which the diagonal equals the vector, the first off diagonal equals
the
first order differences, the second...
On Wed, Apr 27, 2011 at 11:25:42AM +, Hans W Borchers wrote:
Jeroen Ooms jeroenooms at gmail.com writes:
Is there an easy way to turn a vector of length n into an n by n matrix, in
which the diagonal equals the vector, the first off diagonal equals the
first order differences, the
-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Petr Savicky
Sent: Wednesday, April 27, 2011 11:01 AM
To: r-help@r-project.org
Subject: Re: [R] matrix of higher order differences
On Wed, Apr 27, 2011 at 11:25:42AM +, Hans W Borchers wrote:
Jeroen Ooms jeroenooms at gmail.com writes
@r-project.org
Subject: Re: [R] matrix of higher order differences
On Wed, Apr 27, 2011 at 11:25:42AM +, Hans W Borchers wrote:
Jeroen Ooms jeroenooms at gmail.com writes:
Is there an easy way to turn a vector of length n into an n by n matrix, in
which the diagonal equals the vector
of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins
University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: Wednesday, April 27, 2011 4:59 PM
To: Ravi Varadhan
Cc: R Help
Subject: Re: [R] matrix of higher
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
*0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want to set zero the entire
row.
The desired output looks like the following. In this case row 3 and row 5
are set
On Tue, Apr 26, 2011 at 2:28 PM, Val valkr...@gmail.com wrote:
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
* 0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want to set zero the entire
row.
The desired
Try this:
replace(m, m[,ncol(m)] 0, 0)
On Tue, Apr 26, 2011 at 6:28 PM, Val valkr...@gmail.com wrote:
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
* 0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want
Hi,
Since I installed R 2.13 I cannot use the transpose method t on sparse
matrices inside my package. Outside the package works. Is there
something new that I have to import methods? Can I then import
everything from the Matrix package? The problem is that R tries to use
t.default which of
On Wed, Apr 20, 2011 at 8:37 AM, Tobias Abenius
tobias.aben...@chalmers.se wrote:
Since I installed R 2.13 I cannot use the transpose method t on sparse
matrices inside my package. Outside the package works. Is there something
new that I have to import methods? Can I then import everything
HI,
here is another solution:
int - sample(1:20,10)
int
[1] 10 4 5 2 14 17 9 11 16 13
mat-matrix(11:30,ncol=4)
mat
[,1] [,2] [,3] [,4]
[1,] 11 16 21 26
[2,] 12 17 22 27
[3,] 13 18 23 28
[4,] 14 19 24 29
[5,] 15 20 25 30
mat[apply(mat,1,
Hi all!
I have a vector, let's say for example int - sample(1:20,10);
for now:
now I have a matrix...
M = m x n
where the first column is a feature column and most likely shares at least
one of the int (interesting) numbers.
I want to extract the rows where int[] = M[,1]
I thought:
Hi:
Here' s one approach:
int - sample(1:20,10)
m - matrix(sample(1:40, 20), nrow = 10)
int
[1] 7 12 4 6 1 19 17 20 15 5
m
[,1] [,2]
[1,]9 15
[2,] 23 32
[3,] 40 14
[4,] 19 38
[5,] 286
[6,] 26 18
[7,] 34 22
[8,]7 35
[9,] 213
On Mon, Mar 28, 2011 at 04:51:00PM +0200, Rosario Garcia Gil wrote:
Hello
I have this matrix which I am trying to invert. I get a message about
reciprocal condition number, what that does mean?
XT_X
[,1] [,2] [,3] [,4] [,5]
[1,]30021
[2,]0201
Hello
I have this matrix which I am trying to invert. I get a message about
reciprocal condition number, what that does mean?
XT_X
[,1] [,2] [,3] [,4] [,5]
[1,]30021
[2,]02011
[3,]00211
[4,]21140
[5,]1
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Rosario Garcia Gil
Sent: Monday, March 28, 2011 7:51 AM
To: r-help@r-project.org
Subject: [R] matrix inverstion
Hello
I have this matrix which I am trying to invert. I get
Daniel Nordlund djnordlund at frontier.com writes:
On Behalf Of Rosario Garcia Gil
I have this matrix which I am trying to invert. I get a message about
reciprocal condition number, what that does mean?
[snip]
Well, it means exactly what the message says. Within the precision of
Thanks, we're almost there. The 3rd statement needs to
satisfy fi_2[r,c]-fi_2[c,r] where rc.
On Tue, Mar 15, 2011 at 10:06 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
fi_2 - diag(1, i)
fi_2[lower.tri(fi_2)] - 1 - runif(sum(lower.tri(fi_2))) ^ .5
fi_2[upper.tri(fi_2)] -
Hello R users,
I would like to reduce the number of for loops in my code. I build these
matrices (5 times). The main diagonal are 1s and the two sides along the
main diagonal mirror each other as follows:
i-5
fi-matrix(0,nrow=i,ncol=i)#floral inhibition matrix for(r in 1:i){ for(c in
1:i){
Try this:
fi_2 - diag(1, i)
fi_2[lower.tri(fi_2)] - 1 - runif(sum(lower.tri(fi_2))) ^ .5
fi_2[upper.tri(fi_2)] - fi_2[lower.tri(fi_2)]
On Tue, Mar 15, 2011 at 7:51 PM, Brian Pellerin
brianpatrickpelle...@gmail.com wrote:
Hello R users,
I would like to reduce the number of for loops in my
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,]
danielepippo wrote:
I'm building a matrix in R with a cycle for like this:
pp_ris2=matrix(NA,6,6)
for(i in 0:6){
.
R is not like c, indexing starts with 1.
Dieter
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-in-R-tp3312748p3312764.html
Sent from the
but if in my function
pp_ris2[i,j]=myfunction}
must be the indexes 0-0,0-1,0-2,0-3, ?
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-in-R-tp3312748p3312780.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi everyone,
I'm building a matrix in R with a cycle for like this:
pp_ris2=matrix(NA,6,6)
for(i in 0:6){
for(j in 0:6){
if(ij){
pp_ris2[i,j]=myfunction}
else if(i==j){
print(c(i,j))
pp_ris2[i,j]=myfunction}
On Fri, Feb 18, 2011 at 06:32:01AM -0800, danielepippo wrote:
but if in my function
pp_ris2[i,j]=myfunction}
must be the indexes 0-0,0-1,0-2,0-3, ?
You'll have to take care of that yourself with a bit of index
arithmetics. It's the same you encounter in C, if you are
modelling
On Feb 18, 2011, at 9:32 AM, danielepippo wrote:
but if in my function
pp_ris2[i,j]=myfunction}
must be the indexes 0-0,0-1,0-2,0-3, ?
From a search of RSiteSearch() that started with with terms: zero
matrix indexing
http://finzi.psych.upenn.edu/R/Rhelp02/archive/39031.html
On Feb 18, 2011, at 9:32 AM, danielepippo wrote:
but if in my function
pp_ris2[i,j]=myfunction}
must be the indexes 0-0,0-1,0-2,0-3, ?
I came across a posting in r-help that called this package blasphemy:
http://cran.r-project.org/web/packages/Oarray/index.html
--
David Winsemius,
On Tue, Feb 15, 2011 at 09:40:54AM -0800, Alaios wrote:
Thank you very much for your help again.
One more question is that after I have that list of matrices what is the
easiest way to find the minimun and maximum values of ALL the matrices in
that list?
Try the following.
# get a list
On Thu, Feb 10, 2011 at 11:54:50PM -0800, Alaios wrote:
Dear all I have a few matrices that I would like to store alltogether under a
bigger object.
My matrixes with the same name were calculated inside a loop like
for (few times){
estimatedsr- this is my matrix
/11/11, Petr Savicky savi...@praha1.ff.cuni.cz wrote:
From: Petr Savicky savi...@praha1.ff.cuni.cz
Subject: Re: [R] Matrix of Matrices?
To: r-help@r-project.org
Date: Friday, February 11, 2011, 12:22 PM
On Thu, Feb 10, 2011 at 11:54:50PM
-0800, Alaios wrote:
Dear all I have a few matrices
On Fri, Feb 11, 2011 at 06:17:16AM -0800, Alaios wrote:
Thanks that did the work. Once I have that list what is the easiest way to
export the structure as well as the contents (numbers) into a file.
The purpose is to share that file with a colleague and ask him to load that
variable with
Dear all I have a few matrices that I would like to store alltogether under a
bigger object.
My matrixes with the same name were calculated inside a loop like
for (few times){
estimatedsr- this is my matrix
savematrixasimagefile();
}
which means that I was losing all that instances.
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I want to use the package tgp with its sens() function to cunduct a
sensitivity analysis of an ecological simulation model and six
independent input parameter.
I conducted 10.000 simulations based on a Latin Hypercube design to
sample the whole
R Version 2.12.1 (2010-10-15) vs 2.12.0 has slowed down 8 fold for dual core
and 17 fold for dual-core-dual-processor Macs. I have checked this result on 3
different macs using the following R-script:
Using Version 2.12.0 on a dual core dual processor Mac:
You'll need to ask the person who built R (you haven't told us). If
this was a binary CRAN build, you are asked to discuss that only on
R-sig-mac, and you will find plenty of discussion on that list's
archives. Note that
- this is Mac-specific (not mentioned in your subject line)
- it even
Dear R members,
I have a quite large of function that are named like that
f11,f12,...f15
f21,f22,...f25
..
f51,f52,...f52
These are static (hard-coded) functions that the only common they have is that
they take the same number and type of input fij(a,b,c,d). As you might
understand this is
Seems funny to me:
f - list (mean, sd, median, sum)
dim (f) - c (2, 2)
or in one line:
f - structure (.Data=list (mean, sd, median, sum), dim = c(2,2))
f
[,1] [,2]
[1,] ??
[2,] ??
f [1,1]
[[1]]
function (x, ...)
UseMethod(mean)
environment: namespace:base
f [[1,1]] (1:3)
[1]
2011 16:33
To: R-help@r-project.org
Subject: [R] Matrix' with Functions
Dear R members,
I have a quite large of function that are named like that
f11,f12,...f15
f21,f22,...f25
..
f51,f52,...f52
These are static (hard-coded) functions that the only common they have is that
they take the same number
-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Alaios
Sent: 03 February 2011 16:33
To: R-help@r-project.org
Subject: [R] Matrix' with Functions
Dear R members,
I have a quite large of function that are named like that
f11,f12,...f15
f21,f22,...f25
To: R-help@r-project.org
Subject: [R] Matrix' with Functions
Dear R members,
I have a quite large of function that are named like that
f11,f12,...f15
f21,f22,...f25
..
f51,f52,...f52
These are static (hard-coded) functions that the only common they have is
that they take the same number
Dear all,
I would like to thank you for your contribution.
I think I got more that one good solutions to work on.
Best Regards
Alex
--- On Thu, 2/3/11, Bert Gunter gunter.ber...@gene.com wrote:
From: Bert Gunter gunter.ber...@gene.com
Subject: Re: [R] Matrix' with Functions
To: Samuel Le samuel
Hi
I have this function and this matrix:
function(x,y) x+y/x
m-matrix(c(1,2,4,2,10,8),3,2)
m
[,1] [,2]
[1,]12
[2,]2 10
[3,]48
each row represent a point (x,y) in a chart and I want via my fucntion to
calculate the image in order to get this results:
for point
On Feb 2, 2011, at 9:12 AM, ADias wrote:
Hi
I have this function and this matrix:
function(x,y) x+y/x
m-matrix(c(1,2,4,2,10,8),3,2)
m
[,1] [,2]
[1,]12
[2,]2 10
[3,]48
each row represent a point (x,y) in a chart and I want via my
fucntion to
calculate the
Hi
r-help-boun...@r-project.org napsal dne 02.02.2011 16:05:21:
On Feb 2, 2011, at 9:12 AM, ADias wrote:
Hi
I have this function and this matrix:
function(x,y) x+y/x
m-matrix(c(1,2,4,2,10,8),3,2)
m
[,1] [,2]
[1,]12
[2,]2 10
[3,]48
there is no need for 'apply' here, because R can handle vectors.
ord-m[,1]+m[,2]/m[,1]
Am 02.02.2011 15:12, schrieb ADias:
Hi
I have this function and this matrix:
function(x,y) x+y/x
m-matrix(c(1,2,4,2,10,8),3,2)
m
[,1] [,2]
[1,]12
[2,]2 10
[3,]48
If one wants to take an ordinary r matrix and reorder it in the manner
you describe:
mtx2 - mtx[ nrow(mtx):1, ]
Whether that is an efficient way to get at the sokution your you
programming task I cannot say. It sounds as though it has gotten too
convoluted. I was not able to comprehend
Hello
I would like to ask you if it is possible In R Cran to change the default way
of addressing a matrix.
for example
matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row numbering) # not having R at
this pc
will create something like the following
1 2
3 4
the way R address this matrix is from
On Jan 25, 2011, at 4:50 PM, Alaios wrote:
Hello
I would like to ask you if it is possible In R Cran to change the
default way of addressing a matrix.
for example
matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row numbering) # not
having R at this pc
will create something like the
Regards
Alex
--- On Wed, 1/26/11, David Winsemius dwinsem...@comcast.net wrote:
From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] MAtrix addressing
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org
Date: Wednesday, January 26, 2011, 2:54 AM
On Jan 25, 2011, at 4:50 PM
On Mon, Jan 17, 2011 at 10:37:42PM +, Monica Pisica wrote:
Hi,
I've got 2 very good solutions, thank you very much. One, from Henrique
Dallazuanna using the library reshape and one line of code - although it will
take me quite some time to understand it. Here it is what he sent:
Hi,
I am having some difficulties with matrix operations. It is a little hard to
explain it so please bear with me. I have a very large data set, large enough
that it needs to be split in parts in order to deal with. I can work things on
these parts but the problem lies in adding together
Monica -
Perhaps this small example can demonstrate how factors can
solve your problem:
d1 =
data.frame(cat=sample(c('cat2','cat5','cat6'),100,replace=TRUE),group=sample(c('land','water'),100,replace=TRUE))
d2 =
Try this:
library(reshape)
xtabs(rowSums(cbind(value.x, value.y), na.rm = TRUE) ~ X1 + X2,
merge(melt(m1), melt(m2), by = c('X1', 'X2'), all = TRUE), exclude = FALSE)
On Mon, Jan 17, 2011 at 5:59 PM, Monica Pisica pisican...@hotmail.comwrote:
Hi,
I am having some difficulties with matrix
-project.org
Subject: Re: [R] matrix manipulations
Monica -
Perhaps this small example can demonstrate how factors can
solve your problem:
d1 =
data.frame(cat=sample(c('cat2','cat5','cat6'),100,replace=TRUE),group=sample(c('land','water'),100,replace=TRUE))
d2 =
data.frame(cat=sample(c
Hi,
I have a matrix with numbers and character. I want to evaluate each cell and
change the value of the cell before it depending on the evaluation. My
evaluation: if a cell had the word down change the cell preceding it to a
negative number by multiplying that value by a -1. I am have trouble
Try this:
x
V1 V2 V3 V4
1 PROBE 1 2.5 UP
2 PROBE 2 1.0 UP
3 PROBE 3 1.4 DOWN
4 PROBE 4 2.0 UP
5 PROBE 5 1.3 DOWN
x$V3[which(x$V4 == DOWN) - 1] - x$V3[which(x$V4 == DOWN) - 1] * -1
On Mon, Dec 27, 2010 at 4:11 PM, R_novice kaso...@battelle.org wrote:
Hi,
I have a matrix
Thank you for the quick response :-). I've applied your suggestion to my
code, but I still receive an error:
CEM1_PARTIAL$V3[which(CEM1_PARTIAL$V4 == DOWN) - 1] -
CEM1_PARTIAL$V3[which(CEM1_PARTIAL$V4 == DOWN) - 1] * -1
Warning message:
In Ops.factor(CEM1_PARTIAL$V3[which(CEM1_PARTIAL$V4 ==
Convert your column to numeric:
CEM1_PARTIAL$V3 - as.numeric(as.character(CEM1_PARTIAL$V3))
On Mon, Dec 27, 2010 at 4:28 PM, R_novice kaso...@battelle.org wrote:
Thank you for the quick response :-). I've applied your suggestion to my
code, but I still receive an error:
That just converts my values in V3 to NA
--
View this message in context:
http://r.789695.n4.nabble.com/matrix-looping-accessing-previous-column-tp3165308p3165340.html
Sent from the R help mailing list archive at Nabble.com.
__
Using which() improves my code, but now I'm receiving some data.frame error
and still does not convert the values to negative.
#TAKE PART OF MATRIX
CEM1_PARTIAL - CEM1[1:10,1:10]
#CEM1_PARTIAL=apply(CEM1_PARTIAL,2,as.character)
#CEM1_PARTIAL - as.numeric(as.character(CEM1_PARTIAL))
On Wed, Dec 22, 2010 at 2:57 AM, Phil Spector spec...@stat.berkeley.edu wrote:
To make your loop work, you need to learn about the get function.
I'm not going to give you the details because there are better
approaches available.
First, let's make some data that will give values which can be
Thank you both for your suggestions. I have another question - is there a
specific way to access the individual elements of a 'list' variable? i.e.Â
dmi = matrix(rnorm(20),4,5)
soi = matrix(rnorm(20),4,5)
pe = matrix(rnorm(20),4,5)
y - list(dmi, soi, pe)
y[[1]]Â Â gives
[,1]Â Â Â Â Â Â
On Wed, Dec 22, 2010 at 6:39 PM, govin...@msu.edu wrote:
Thank you both for your suggestions. I have another question - is there a
specific way to access the individual elements of a 'list' variable? i.e.
dmi = matrix(rnorm(20),4,5)
soi = matrix(rnorm(20),4,5)
pe = matrix(rnorm(20),4,5)
y
Have you consulted R's extensive documentation? -- in particular, An
Introduction to R, which would seem like an obvious place for R
newbies to start. If you had done so, you would have found your
question answered there in section 6.1 on lists.
-- Bert Gunter
On Wed, Dec 22, 2010 at 9:39 AM,
Thank you all once again .. Yeah, its working now.
--
Regards,
Mahalakshmi
Graduate Student
#20, Department of Geography
Michigan State University
East Lansing, MI 48824 Quoting Liviu Andronic landronim...@gmail.com:
On Wed, Dec 22, 2010 at 6:39 PM, govin...@msu.edu wrote:
Thank you both
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