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
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=
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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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Incomprehensible. Define:matrix reduction .
Perhaps:
?qr
?svd
## and links therein.
-- Bert
On Sat, Apr 6, 2013 at 10:43 AM, Angelo Scozzarella Tiscali
angeloscozzare...@tiscali.it wrote:
Hi everyone,
I asked myself if exists a way to get a rows (or columns) matrix reduction
with R.
My
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) matrix
On 06-04-2013, at 19:58, Angelo Scozzarella Tiscali
angeloscozzare...@tiscali.it 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
?chol
##also
-- Bert
On Sat, Apr 6, 2013 at 11:12 AM, Berend Hasselman b...@xs4all.nl wrote:
On 06-04-2013, at 19:58, Angelo Scozzarella Tiscali
angeloscozzare...@tiscali.it wrote:
Well, I mean to use the elimination method to transform the matrix, for
example, of the coefficients of a
On 06-04-2013, at 22:01, Bert Gunter gunter.ber...@gene.com 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 b...@xs4all.nl wrote:
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);
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,]);
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,]);
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]
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 below
sample - list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr
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
:
Sent: 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
,byrow = T)
T %^% 200
- Original Message -
From: annek an...@ifm.liu.se
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
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
0.09810.0273
-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 T for which I want to find
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 an...@ifm.liu.se wrote:
Hi,
I have a transition matrix
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
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
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, Uzuner, Tolga I wrote:
Many thanks for your advice and assistance
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 tolga.i.uzu...@jpmorgan.com
on Thu, 6 Dec 2012 16
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
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
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 nici...@gmail.com 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 below.
Thanks in advance for any assistance.
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
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 about
= 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 package will not loead
You will find the problem and solution in the list archives
...@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 package will not loead
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
-project.org
Subject: Re: [R] Matrix package will not loead
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
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,]
[1,] A 10,00
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?
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-in-R-tp4649426p4649429.html
Sent from the R help mailing list archive at Nabble.com.
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
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) 3.593793
]-Pr(|t|)
A.K.
- Original Message -
From: PavloEs nici...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Monday, November 12, 2012 10:45 AM
Subject: [R] Matrix to data frame conversion
I have a matrix which I wanted to convert to a data frame. As I could not
succeed and resorted to export
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
: 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 convert it
to a
data frame (xx), but could not succeed
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.
,diff),ncol=7))
#[1] TRUE
A.K.
- Original Message -
From: cleberchaves clebercha...@gmail.com
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 know exactly the error of yours script
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
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:
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
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:
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(toeplitz(dat1[,2]),1,function(x) dat1[1,2]-x) and
worked very well!
Thanks again!
--
View
-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 clebercha...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Saturday, November 10, 2012 3:55 PM
Subject: Re: [R] matrix of all difference between rows values
Mmmm...
Actually, Rui
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:n){
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: [R] matrix algebra for constructing a special matrix?
Dear R-users,
would it be a better way
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??
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 convert
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
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 -
Thanks a lot!
--
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Obrigado Rui, é isso mesmo ;)
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__
R-help@r-project.org mailing list
...@gmail.com
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!
--
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Sent from the R help mailing list
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, when I do just
in context:
http://r.789695.n4.nabble.com/R-matrix-help-tp4633372.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R
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
Regards
karthick
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__
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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
--
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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 1
0
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
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 1,
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’
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
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
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
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)
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
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 have the following
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 gunda...@gmail.com wrote:
I have the following matrix:
dat
[,1] [,2]
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[x10]
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]]
__
R-help@r-project.org
a[NROW(a):1, ]
Michael
On May 11, 2012, at 3:00 PM, Trying To learn again
tryingtolearnag...@gmail.com 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
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
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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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
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 way?
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 =
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
I just do not understand what these parameters that must pass the heat map
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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 think it worked here, the data of 25 families are wrong, I'll pack up and
post already! earned
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last doubts, how do I remove these trace that sits on top of colors?
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R-help@r-project.org
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))
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,
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 to
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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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
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
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 lpchaparro...@gmail.com wrote:
Dear R people.
I惴 facing a big problem.
I
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
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)
I would
Thank you sire, this is exactly what I was looking for.
Many thanks,
Phil
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