On Apr 20, 2013, at 2:19 PM, Benjamin Caldwell wrote:
> Dear R helpers
>
> Reproducible example:
>
> #warning - this causes a hard freeze on the machines I've tried it on
> matrix.holder<- matrix(rnorm(150), nrow=30, ncol=5)
>
> Out=
> expand.grid(matrix.holder[,1],matrix.holder[,2],matrix.hol
Dear R helpers
Reproducible example:
#warning - this causes a hard freeze on the machines I've tried it on
matrix.holder<- matrix(rnorm(150), nrow=30, ncol=5)
Out=
expand.grid(matrix.holder[,1],matrix.holder[,2],matrix.holder[,3],matrix.holder[,4],
matrix.holder[,5])
Problem:
I'm running an an
On 06-04-2013, at 22:01, Bert Gunter wrote:
> ?chol
>
> ##also
>
Quite true and therefore:
?backsolve
# forwardsolve
and
?qr
# qr.solve
So there's a lot to choose from.
Berend
> -- Bert
>
> On Sat, Apr 6, 2013 at 11:12 AM, Berend Hasselman wrote:
>>
>> On 06-04-2013, at 19:58, An
?chol
##also
-- Bert
On Sat, Apr 6, 2013 at 11:12 AM, Berend Hasselman wrote:
>
> On 06-04-2013, at 19:58, Angelo Scozzarella Tiscali
> wrote:
>
>> Well, I mean to use the elimination method to transform the matrix, for
>> example, of the coefficients of a linear equations system.
>>
>> AS
On 06-04-2013, at 19:58, Angelo Scozzarella Tiscali
wrote:
> Well, I mean to use the elimination method to transform the matrix, for
> example, of the coefficients of a linear equations system.
>
> AS
>
Well if you only need to solve a linear equation system have a look at solve.
If you h
Well, I mean to use the elimination method to transform the matrix, for
example, of the coefficients of a linear equations system.
AS
Il giorno 06/apr/2013, alle ore 19:43, Angelo Scozzarella Tiscali ha scritto:
> Hi everyone,
>
> I asked myself if exists a way to get a rows (or columns) matr
Incomprehensible. Define:"matrix reduction" .
Perhaps:
?qr
?svd
## and links therein.
-- Bert
On Sat, Apr 6, 2013 at 10:43 AM, Angelo Scozzarella Tiscali
wrote:
> Hi everyone,
>
> I asked myself if exists a way to get a rows (or columns) matrix reduction
> with R.
> My goal is for education
Hi everyone,
I asked myself if exists a way to get a rows (or columns) matrix reduction with
R.
My goal is for education.
Thanks in advance
AS
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PLEASE do read the posting
Hi,
Try this:
#mat1 is the data
res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i)
{new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j)
{x1<-rbind(mat1[i,],mat1[j,]);
x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2
Hi,
Just to add:
res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i)
{new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j)
{x1<-rbind(mat1[i,],mat1[j,]);
x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7
HI Elisa,
You can also use:
mat2<- head(mat1)
resNew<-do.call(cbind,lapply(seq_len(nrow(mat2)),function(i)
do.call(rbind,lapply(split(rbind(mat2[i,],mat2[-i,]),1:nrow(rbind(mat2[i,],mat2[-i,]))),function(x)
{x1<-rbind(mat2[i,],x);
x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[
Wednesday, January 16, 2013 2:59 AM
Subject: [R] matrix manipulation with its rows
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample <- list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr=2,nc=3),
matrix(13:18,nr=2,nc=3))
> sample
[[1]
Not a great solution, I don't think, but:
> kronecker(diag(2), matrix(1:6, 2, byrow=TRUE))[c(1,4),]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123000
[2,]000456
So using a function that does this in 'lapply'
should solve the problem you state. I'm gue
ect.org] On Behalf Of Kathryn Lord
> Sent: Wednesday, January 16, 2013 9:00 AM
> To: r-help@r-project.org
> Subject: [R] matrix manipulation with its rows
>
> Dear R users,
>
> I have a question about matrix manipulation with its rows.
>
> Plz see the simple example bel
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample <- list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr=2,nc=3),
matrix(13:18,nr=2,nc=3))
> sample
[[1]]
[,1] [,2] [,3]
[1,]135
[2,]246
[[2]]
[,1] [,2] [,
One solution that does not require matrix multiplication:
Remember that the steady state vector is in the nullspace of I - T.
Therefore:
require(MASS)
n1 <- Null(t(diag(nrow(T)) - T))
n1 / sum(n1)
On Wed, Dec 12, 2012 at 2:19 AM, annek wrote:
> Hi,
> I have a transition matrix T for which I
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of annek
> Sent: Tuesday, December 11, 2012 11:19 PM
> To: r-help@r-project.org
> Subject: [R] Matrix multiplication
>
> Hi,
> I have a transition matrix
On Dec 12, 2012, at 08:19 , annek wrote:
> Hi,
> I have a transition matrix T for which I want to find the steady state matrix
> for. This could be approximated by taking T^n , for large n.
>
> T= [ 0.8797 0.0382 0.0527 0.0008
> 0.02120.8002 0.0041 0.0143
> 0.09810
, ncol = 4,byrow = T)
T %^% 200
- Original Message -
From: annek
To: r-help@r-project.org
Cc:
Sent: Wednesday, December 12, 2012 12:49 PM
Subject: [R] Matrix multiplication
Hi,
I have a transition matrix T for which I want to find the steady state matrix
for. This could be approximated
Hi,
I have a transition matrix T for which I want to find the steady state matrix
for. This could be approximated by taking T^n , for large n.
T= [ 0.8797 0.0382 0.0527 0.0008
0.02120.8002 0.0041 0.0143
0.09810.0273 0.8802 0.0527
0.00100.1343 0.0630
Many thanks !
Tolga
-Original Message-
From: Martin Maechler [mailto:maech...@stat.math.ethz.ch]
Sent: 06 December 2012 17:19
To: Uzuner, Tolga I
Cc: r-help@r-project.org
Subject: Re: [R] Matrix package will not loead
>>>>> Uzuner, Tolga
>>>>> on
; -Original Message-
> From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
> Sent: 13 November 2012 14:08
> To: Uzuner, Tolga I
> Cc: Prof Brian Ripley; r-help@r-project.org
> Subject: Re: [R] Matrix package will not loead
> On 13/11/2012 8:28 AM,
environment), but
could it be that I am missing something ?
Thanks in advance,
Tolga
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 13 November 2012 14:08
To: Uzuner, Tolga I
Cc: Prof Brian Ripley; r-help@r-project.org
Subject: Re: [R] Matrix package
On 12-12-03 5:00 PM, Christian Hoffmann wrote:
Hi,
I find it cumbersomesome the I have to use \%*\% in .Rd files vs. %*% in
.R files. R CMD check will refuse %*% in .Rd files. I would like to have
%*% in .Rd files to be able to execute expressions with matrix
multiplication from .Rd files direct
Hi,
I find it cumbersomesome the I have to use \%*\% in .Rd files vs. %*% in
.R files. R CMD check will refuse %*% in .Rd files. I would like to have
%*% in .Rd files to be able to execute expressions with matrix
multiplication from .Rd files directly, but ESS (version 5.13) would
refuse to e
So, if you have duplicate row.names, get rid of them. Try
rownames(comb_model0) <- NULL
as.data.frame(comb_model0)
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 11/12/12 7:45 AM, "PavloEs" wrote:
>I have a matri
It's a bit complicated. Is there any shorter way?
Is there possibility to read datas from .csv as matrix, like this which i
want to have?
--
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Sent from the R help mailing list archive at Nabble.com.
_
Is there posiibility to read.table change in matrix?
When i used read.table it gave me:
V1 V2 V3 V4
[1,] "OsobaA" "10,00" "9,00" "8,00"
[2,] "OsobaB" "2,00" "3,00" "1,00"
[3,] "OsobaC" "5,00" "6,00" "4,00"
I want to change it in:
[1,] [2,] [3,] [4,
Not Found'
Warning in download.packages(pkgs, destdir = tmpd, available = available, :
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: 13 November 2012 13:03
To: Duncan Murdoch
Cc: Uzuner, Tolga I; r-help@r-project.org
Subject: Re: [R] Matrix packag
ile(url, destfile, method, mode = "wb", ...) :
cannot open: HTTP status was '404 Not Found'
Warning in download.packages(pkgs, destdir = tmpd, available = available, :
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: 13 November 2012 1
rial.ac.uk/src/contrib/Matrix_1.0-9.tar.gz'
In addition: Warning message:
In download.file(url, destfile, method, mode = "wb", ...) :
cannot open: HTTP status was '404 Not Found'
Warning in download.packages(pkgs, destdir = tmpd, available = available, :
-Origin
You will find the problem and solution in the list archives.
The current version of Matrix can be installed *from source* on R (>=
2.15.0), as it claims. But if you install it on R >= 2.15.2 then it
uses features of 2.15.2 and hence can only be run on R >= 2.15.2. And
there was a warning abou
On 12-11-13 7:06 AM, Uzuner, Tolga I wrote:
Dear Fellow R Users,
I am having a problem with the Matrix package, in Windows XP on R 2.15.1 .
This is the only package where I experience this. I remove the package first,
re install, and then when trying to load, get a LoadLibrary failure as belo
Dear Fellow R Users,
I am having a problem with the Matrix package, in Windows XP on R 2.15.1 .
This is the only package where I experience this. I remove the package first,
re install, and then when trying to load, get a LoadLibrary failure as below.
Thanks in advance for any assistance.
Reg
avloEs
> Sent: Monday, November 12, 2012 1:09 PM
> To: r-help@r-project.org
> Subject: Re: [R] Matrix to data frame conversion
>
> May be I have not clearly explained my problem. Le me try it again. My
> problem was with the matrix "comb_model0" . I have tried to conve
May be I have not clearly explained my problem. Le me try it again. My
problem was with the matrix "comb_model0" . I have tried to convert it to a
data frame (xx), but could not succeed . As a result I have exported it to a
test.csv file and re-imported it. Data frame test is the product of th
way will be to use
?rbind() comb_model4<-rbind(as.data.frame(mat4),as.data.frame(mat3)) # no need
colnames(comb_model4)[4]<-"Pr(>|t|)"
A.K.
- Original Message -
From: PavloEs
To: r-help@r-project.org
Cc:
Sent: Monday, November 12, 2012 10:45 AM
Subject: [R] M
Hello,
I'm unable to reproduce your error, removing the space in "Std. Error"
and changing "t value" to "t.value" the following read in as data.frame
with no problems.
x <- read.table(text="
Estimate Std.Error t.value Pr(>|t|)
(Intercept)
I have a matrix which I wanted to convert to a data frame. As I could not
succeed and resorted to export to csv and reimport it again. Why did I fail
in the attempt and how can I achieve what I wanted without this
roundabouts?
The original matrix:
> str(comb_model0)
num [1:90, 1:4] 3.5938 0.0
ec1,"-")),matrix(apply(expand.grid(vec1,vec1),1,diff),ncol=7))
#[1] TRUE
A.K.
- Original Message -
From: cleberchaves
To: r-help@r-project.org
Cc:
Sent: Sunday, November 11, 2012 7:07 AM
Subject: Re: [R] matrix of all difference between rows values
Hi Arun,
i don't
Hi Arun,
i don't know exactly the error of yours script.
Maybe when i changed from "10-x" to "dat1[1,2]-x" (because my real matrix
does not start with 10) the error has appeared, the same numbers repeat in
all columns.
Maybe when i change for your second script that error does not appear again.
Ne
t1)<-dimnames(res1)
mat1
# a b c d e
#a 0 -1 -2 -3 -4
#b 1 0 -1 -2 -3
#c 2 1 0 -1 -2
#d 3 2 1 0 -1
#e 4 3 2 1 0
A.K.
- Original Message -
From: cleberchaves
To: r-help@r-project.org
Cc:
Sent: Saturday, November 10, 2012 3:55 PM
Subject: Re: [R] matrix of all differe
r-help@r-project.org
Cc:
Sent: Saturday, November 10, 2012 2:25 PM
Subject: Re: [R] matrix of all difference between rows values
Thank you very much, arun kirshna!
That's it! I only modified the "res1<-apply(toeplitz(dat1[,2]),1,function(x)
10-x)" for "res1<-apply(toepli
Mmmm...
Actually, Rui Barradas is the right!
Arun kirshna, yours script has an error. That repeat the same set of numbers
in all columns...
Anyway, thanks for both!
--
View this message in context:
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Hello,
Try the following.
# Create the dataset
Table1 <- matrix(10:6, ncol = 1)
rownames(Table1) <- letters[1:5]
Table1
t(outer(Table1[,1], Table1[,1], `-`))
Hope this helps,
Rui Barradas
Em 10-11-2012 18:32, cleberchaves escreveu:
Hello all,
i would like to calculate the difference of all r
Thank you very much, arun kirshna!
That's it! I only modified the "res1<-apply(toeplitz(dat1[,2]),1,function(x)
10-x)" for "res1<-apply(toeplitz(dat1[,2]),1,function(x) dat1[1,2]-x)" and
worked very well!
Thanks again!
--
View this message in context:
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Hello all,
i would like to calculate the difference of all row values and the others
row values from my matrix (table 1). The output (table 2) would be a matrix
with input matrix's row names on row names and colums names, thereby the
difference values among two of the row names could be bether foun
l Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Camarda, Carlo Giovanni
> Sent: Friday, October 26, 2012 8:13 AM
> To: r-h...@stat.math.ethz.ch
> Subject:
Dear R-users,
would it be a better way to construct the matrix below without using any
for-loop or model.matrix? preferably with some matrix algebra?
Thanks in advance,
Carlo Giovanni Camarda
## dimensions
m <- 3
n <- 4
mn <- m*n
k <- m+n-1
## with a for-loop
X <- matrix(0, mn, k)
for(i in 1:
2 1 1 2 1 2 1 1 1
# $ X5: Factor w/ 2 levels "0","1": 1 1 2 2 2 2 1 1 2 2
A.K.
- Original Message -
From: brunosm
To: r-help@r-project.org
Cc:
Sent: Friday, October 19, 2012 10:15 AM
Subject: Re: [R] Matrix to data.frame with factors
Thanks a lot!
--
View th
Obrigado Rui, é isso mesmo ;)
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https:
Thanks a lot!
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On Oct 19, 2012, at 16:07 , Rui Barradas wrote:
> Hello,
>
> Try the following.
>
> x <- matrix(sample(0:1, 12, TRUE), ncol = 4)
> y <- data.frame(apply(x, 2, factor))
> str(y)
>
> Hope this helps,
Another way, possibly more easily generalized:
x <- matrix(sample(0:1, 12, TRUE), ncol = 4)
y
Well, strictly speaking, this is still doing it "one variable at a
time." The interpreted loop is hidden, but it's still happening.
A loop free but clumsier approach is:
y <- data.frame(matrix(as.character(x),nrow = nrow(x)))
## Note also that the original column names will be lost and will have
Hello,
Try the following.
x <- matrix(sample(0:1, 12, TRUE), ncol = 4)
y <- data.frame(apply(x, 2, factor))
str(y)
Hope this helps,
Rui Barradas
Em 19-10-2012 12:04, brunosm escreveu:
Hi all,
I have a matrix with 100 variables: each variable as a value of 0 or 1.
What i want to do is conver
Hi all,
I have a matrix with 100 variables: each variable as a value of 0 or 1.
What i want to do is convert this matrix to a data.frame but convert all the
variables to factors (0 and 1) also.
I know i can do this one variable a time but i have 100 variables...
Any easy way of doing this??
Th
I have the following matrix operation
A %*% B %*% A
Where these matrices have the following dimensions and class attributes.
> dim(A)
[1] 5764 5764
> class(A)
[1] "dgCMatrix"
attr(,"package")
[1] "Matrix"
> dim(B)
[1] 5764 5764
> class(B)
[1] "dgCMatrix"
attr(,"package")
[1] "Matrix"
Now, wh
On Thu, Jun 14, 2012 at 02:24:20PM -0400, cowboy wrote:
> thank you, Petr.
> This is exactly what I'm looking for in my post.
> An related question can be how to get an arbitrary weight, say if row1
> and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
> row 2 have 2 common value
thank you, Petr.
This is exactly what I'm looking for in my post.
An related question can be how to get an arbitrary weight, say if row1
and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
row 2 have 2 common value 1, then assign a weight 12. I'm not so sure
how to expand your me
On Thu, Jun 14, 2012 at 01:11:45PM +, G. Dai wrote:
> Dear Rlisters,
> I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
>
> To elaborate, let's consider an example. Assume we have the following
> 3 by 4 matrix A with elements either 0 or 1,
> 0 1 1 0
> 1 0 1
X4, 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.
>
>___
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
wrote:
> 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.
Message-
> 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 get
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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Sent from the R help mailing list archive at Nabble.com
PDT)
> 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,
> based
> on the the number of elements in a matrix for example. if the number is
> 12,
> I should get dim= 3X4,
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 14,
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.
>
> __
> R-help@r-project.org maili
Dear Rlisters,
I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
To elaborate, let's consider an example. Assume we have the following
3 by 4 matrix A with elements either 0 or 1,
0 1 1 0
1 0 1 1
0 0 0 1
>From the original matrix A, I'd like to generate a new
context:
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PLEASE do read the posting guide http://www.R
Hi R users all ,
I have a clean install of R-2.15.0 in a win32 machine and a problem loading
Matrix 1.0-6 that looks like this:
> library(Matrix)
Loading required package: lattice
Error : object ‘kronecker’ is not exported by 'namespace:methods'
Error: package/namespace load failed for ‘Matrix’
>
ITS OK. And thanks. Its working !
From: Ãzgür Asar [via R] [mailto:ml-node+s789695n4631684...@n4.nabble.com]
Sent: Tuesday, May 29, 2012 6:04 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: Matrix Header Name Select
Sorry, I misunderstood your question.
colnames(mymatrix)
woul
Hi,
i have a matrix with headers. How can i get header name in that marix ?
i tried in R, with
names(MyMatrix[1]) - its getting exactly.
But when i try in eclips, its not receiving
ColHdr <- names(MyMatrix[1])
- it getting NULL value
- could you please help me ?
- Thanks
Antony.
--
View this
Sorry, I misunderstood your question.
colnames(mymatrix)
would select column names of your matrix.
Best
ozgur
-
Ozgur ASAR
Research Assistant
Middle East Technical University
Department of Statistics
06531, Ankara Turkey
Ph: 90-312-2105309
http://www.s
Hi,
You can try:
colnames(mymatrix)<-NULL
Best
Ozgur
-
Ozgur ASAR
Research Assistant
Middle East Technical University
Department of Statistics
06531, Ankara Turkey
Ph: 90-312-2105309
http://www.stat.metu.edu.tr/people/assistants/ozgur/
--
View this mess
Hello,
> A = matrix(0, 3,3)
> rownames(A) = c("A", "B", "C")
> A
[,1] [,2] [,3]
A000
B000
C000
HTH,
Thanks,
Paolo
On 15 May 2012 10:19, Gundala Viswanath wrote:
> I have the following matrix:
> > dat
>
> [,1] [,2] [,3]
example:
> x<-c(1,1,1)
> y<-c(2,2,2)
> m<-rbind(x,y)
> m
[,1] [,2] [,3]
x111
y222
> dimnames(m)
[[1]]
[1] "x" "y"
[[2]]
NULL
> dimnames(m)[[1]]<-c("a","b")
> m
[,1] [,2] [,3]
a111
b222
Am 15.05.2012 um 11:19 schrieb Gundala Viswanath:
> I ha
I have the following matrix:
> dat
[,1] [,2] [,3][,4]
foo 0.7574657 0.2104075 0.02922241 0.002705617
foo 0.000 0.000 0. 0.0
foo 0.000 0.000 0. 0.0
foo 0.000 0.000 0. 0.0
foo 0.000 0
a[NROW(a):1, ]
Michael
On May 11, 2012, at 3:00 PM, Trying To learn again
wrote:
> I all,
>
> I have a matrix like this
>
> a=
>
> 1 4
> 2 7
> 3 6
>
>
> I want to create a new matrix
>
> b=
>
> 3 6
> 2 7
> 1 4
>
> Anyone knows if there is a "reverse" function?
> I can do it with loop
I all,
I have a matrix like this
a=
1 4
2 7
3 6
I want to create a new matrix
b=
3 6
2 7
1 4
Anyone knows if there is a "reverse" function?
I can do it with loops if no exits.
[[alternative HTML version deleted]]
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R-help@r-project.org
Hi
>
> how do I plot only the data below 10? everything is white for the 0-10
and
> 10-90 is black ..
What data below 10? I do not see any. You posted some mails before but I
do not keep all mails from R. Only those which helped me somehow.
Basically
x <- sample(1:100, 100, raplace=TRUE)
x[x
how do I plot only the data below 10? everything is white for the 0-10 and
10-90 is black ..
those functions which do this?
was bad for such basic questions, but I started tinkering with R is 6 days
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Hi
what is wrong with
heatmap(as.matrix(test), col=my.colors(25))
with test from your dput
Regards
Petr
> The heat map generated the correct result:
>
> library(gplots)
> arq <-read.table("l")
> matrix_l <-data.matrix(arq)
> my.colors <-
> colorRampPalette(c
>
("gray0","gray10","gray20","gr
as was follows:
library(gplots)
arq <-read.table("r")
matrix_l <-data.matrix(arq)
pdf("heatmap.pdf", height = 10 , width=10)
#paleta de 10 cores - sentido branco -> preto
my.colors <-
colorRampPalette(c("gray100","gray90","gray80","gray70","gray60","gray50","gray40","gray30","gray20","gray10"))
h
last doubts, how do I remove these trace that sits on top of colors?
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I think it worked here, the data of 25 families are wrong, I'll pack up and
post already! earned
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The heat map generated the correct result:
library(gplots)
arq <-read.table("l")
matrix_l <-data.matrix(arq)
my.colors <-
colorRampPalette(c("gray0","gray10","gray20","gray30","gray40","gray50","gray60","gray80","gray90","gray100
I just do not understand what these parameters that must pass the heat map
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I would leave my table as a heatmap where darker colors represent higher
similarity, and the lighter colors represent less level of similarity.
I'm using version 2.11 of R.
I once used this code, maybe it will help you:
#dendogram
plot(dendro15, labels = cellType) ### I first made a dendrogra
Hi
>
> >arq <-read.table("file")
> > arq_matrix <-data.matrix(arq)
Are you sure that arg_matrix is numeric? Did you check it somehow?
> >dput(arq)
You forgot to include dput(arg) result. Without that only you know what
arg is.
> > arq_heatmap <- heatmap(arq_matrix, Rowv = NA, Colv = NA,col
>arq <-read.table("file")
> arq_matrix <-data.matrix(arq)
>dput(arq)
> arq_heatmap <- heatmap(arq_matrix, Rowv = NA, Colv = NA,col = heat.colors
> (256), scale = "column", margins =c(5,10))
dput done with this command, but still gave the same ..
I do it before generating the heatmap?
would be
this would be the same, like, I need color higher for larger numbers, type,
variations of 5 om 5, for example, 0 is white, a little darker 1-5, 6-10
darker still, and so on ...
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Sent from the R hel
I used only these three command lines in R
>arq <-read.table("file")
> arq_matrix <-data.matrix(arq)
> arq_heatmap <- heatmap(arq_matrix, Rowv = NA, Colv = NA,col = cm.colors
> (256), scale = "column", margins =c(5,10))
Excuse me if poorly written, because I'm Brazilian
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Hello,
Please read the posting guide, like this it's difficult for us to give a
sensible an answer. In particular,
1. Use R syntax, your "table" seems to be a "matrix" or "data.frame". Which
is it?
2. Post your data using dput(). Just copy it's output and paste it in here,
then,
we'll beble t
I would like to organize my data as follows:
I have a table that contains various data, and the numbers represent a level
of similarity between these data,
eg RF00013 has 100% similarity with the data RF00014.
I would leave my table as a heatmap where darker colors represent higher
similarity, a
If all you want are binary 0/1, then look at the 'bit' package which
will let you create a vector of bits. Even for your matrix, you will
need almost 1GB of memory to store a copy.
On Mon, May 7, 2012 at 5:23 PM, Lucas wrote:
> Dear R people.
> I惴 facing a big problem.
> I need to create a matri
See ?sparseMatrix from package Matrix.
Eloi
On 12-05-07 02:23 PM, Lucas wrote:
> Dear R people.
> I´m facing a big problem.
> I need to create a matrix with 10.000 columns and 750.000 rows.
> matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
> as you can see, the data frame has huge
Dear R people.
I´m facing a big problem.
I need to create a matrix with 10.000 columns and 750.000 rows.
matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
as you can see, the data frame has huge dimesions. I was able to find out
about thr "L" in data, this way I´m telling that my data
Thank you sire, this is exactly what I was looking for.
Many thanks,
Phil
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Hello,
Filoche wrote
>
> Hi everyone.
>
> I want to transpose a data frame. Lets say the following DF:
>
> df = data.frame(matrix(ncol=4, nrow = 10))
>
> df[,1] = c(rep(1,5),rep(2,4), 3)
> df[,2] = c(rep('a',4),rep('b',3),rep('c',3))
> df[,3] = c(letters[c(5:13,13)])
> df[,4] = runif(10)
>
>
Hi everyone.
I want to transpose a data frame. Lets say the following DF:
df = data.frame(matrix(ncol=4, nrow = 10))
df[,1] = c(rep(1,5),rep(2,5))
df[,2] = c(rep('a',4),rep('b',3),rep('c',3))
df[,3] = c(letters[5:14])
df[,4] = runif(10)
I would like to form a data frame with each line correspo
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