On Mar 28, 2011, at 05:36 , Frank Harrell wrote:
I have a question about the computation of the degrees of freedom in a linear
model:
x - runif(20); y - runif(20)
f - lm(y ~ x)
logLik(f)
'log Lik.' -1.968056 (df=3)
The 3 is coming from f$rank + 1. Shouldn't it be f$rank? This
Hello
I'm using function gam from package mgcv to fit splines. When I try
to make a prediction slightly beyond the original 'x' range, I get
this error:
A = runif(50,1,149)
B = sqrt(A) + rnorm(50)
range(A)
[1] 3.289136 145.342961
fit1 = gam(B ~ s(A, bs=ps), outer.ok=TRUE)
predict(fit1,
Hi all,
I have site data with plant species cover and am looking for trends. I'm
kind of new to this, but have done lots of reading and can't find an answer.
I tried decorana (I know it's been replaced by ca.) and see a trend, but I'm
not sure what it means. Is there a way to get the
Hello!
I have some problem with package rminer function imputation.
For example, i have data frame
data.frame(X1,X2,X3)
X1 X2 X3
1 2002 82 88.53316
2 2001 39 68.41058
3 NA NA NA
but when i use imputation, R gives an error
print(imputation(hotdeck,d, X3))
Error in
Thanks,
the drop=FALSE is bound to come in handy.
And aggregate was indeed what I was looking.
Thanks again
2011/3/27 David Winsemius dwinsem...@comcast.net:
On Mar 27, 2011, at 3:22 AM, peter dalgaard wrote:
On Mar 27, 2011, at 08:25 , David Winsemius wrote:
On Mar 26, 2011, at 10:26
Dear list,
My question to follow is not a pure R question but contains also a
more general statistical/econometrical part, but I was hoping that
perhaps someone knowledgable on this list could offer some help.
I have estimated a binary logistic regression model and would like to
calculate
Dear all,
I would like to know what family of orthogonal polynomials is used in the
function poly(). I am using poly(X, 2) in a PLS regression, and I need the
formula used to calculate the orthogonal polynomials to get back to the effect
of the X variable.
The reference cited in the help page
Dear Ning,
are you referring to the deprecated function garchOxFit() of the package
fGarch, formerly contained in fSeries? If so:
library(sos)
findFn(garchOxFit)
which yields:
http://finzi.psych.upenn.edu/R/library/fGarch/html/00fGarch-package.html
And there you will find at the
Hi,
I am using the quantile function currently and I have just bumped into a little
problem.
I have a very small data frame something like this:
small_df -
c(7,3,4,7,1,10,12,1,12,4,4,8,6,11,9,10,4,13,3,9,6,5,2,10,7,14,2,7,10,10,7,8,2,11,3,10,11,3,11,14,12,7,6,11)
small_df
Now in the next
dear Philip,
I am not able to solve your problem, however the error message you get
does not depends on mgcv::gam, therefore gam(,..outer.ok=TRUE) or
predict.gam(,outer.ok=TRUE) do not make sense.
The error message comes from the function splines::splineDesign which is
called when the option
HI Laszlo,
q-quantile(small_df,probs=0.95)
q[[1]]
[1] 12.85
Regrads
Le 28/03/11 11:37, Bodnar Laszlo EB_HU a écrit :
Hi,
I am using the quantile function currently and I have just bumped into a
little problem.
I have a very small data frame something like this:
small_df-
On Sun, 2011-03-27 at 21:40 -0700, Akane Nishimura wrote:
Hi all,
Hi Akane,
I answered this on the R-forge forum where you first posted it. I have
posted a follow-up there too. More below...
I have site data with plant species cover and am looking for trends. I'm
kind of new to this, but
Hello I have downloaded the fPortfolioSolver package from R-forge but I have
not been able to install it. I don't know exactly where I should place the
file and which commands to give R. Could somebody please help me with this.
Thank you
Felipe Parra
[[alternative HTML version deleted]]
You can get around this by using the 'knots' argument to 'gam' to
specify p-spline knots which span the range over which you want to
predict. Alternatively use the cr, tp or ds bases (splines with
derivative based penalties), which don't have this problem.
best,
Simon
On 28/03/11 06:10,
On 28/03/2011 7:30 AM, Luis Felipe Parra wrote:
Hello I have downloaded the fPortfolioSolver package from R-forge but I have
not been able to install it. I don't know exactly where I should place the
file and which commands to give R. Could somebody please help me with this.
Thank you
A
Good morning,
I am facing a problem very easy to solve with a program, but not too
easy (at least IMHO) with a declarative approach.
I have a dataframe df with some information about bank branches with a
validity time associated (start date/end date, format -MM-DD) to
some attributes (for
Thanks Duncan, I already installed Rtools but I don't know well how to sort
it out. I tried the command you gave me and got the following error:
install.packages(fPortfolioSolver.tar.gz, type=source, repos=NULL)
Installing package(s) into C:\Users\Hp\Documents/R/win-library/2.12
(as lib is
On 2011-03-28 02:51, Mohamed Lajnef wrote:
HI Laszlo,
q-quantile(small_df,probs=0.95)
q[[1]]
[1] 12.85
Regrads
Or, perhaps more succinctly:
unname(q)
since '95%' is just the name of the vector.
Peter Ehlers
Le 28/03/11 11:37, Bodnar Laszlo EB_HU a écrit :
Hi,
I am using the
Hello
I am trying to make a graph of 10 different lines built each from 4 different
segments and to add a darker line that will represent the average of all graphs
- all in the same plot.Actually each line is a ROC plot
The code I'm using for plotting one line is as follows:
logit.roc.plot -
Dr Hesterberg:
Independent and dependent were used for convenience. A person selling
hot dogs would render HD the Y variable. The real cause of both WL and HD
is likely hormonal derangement associated with diabetes.
One other question. A log negative binomial regression
m.ln-glm.nb( Count ~
On 03/28/2011 04:18 AM, jouba wrote:
Jeremy thanks a lot for your response I have read sem package help
and I currently reading the help of lavaan I see that there is also
an other function called lavaan can do the SEM analysis So I wonder
what is the difference between this function and the
Try this:
dif - na.omit(c(DF$STARTDATE[-1], NA) - DF$ENDDATE)
list(DF$ENDDATE[dif != 1] + 1, DF$ENDDATE[dif != 1] + (dif[dif != 1] - 1))
On Mon, Mar 28, 2011 at 8:56 AM, angelo.lina...@bancaditalia.it wrote:
Good morning,
I am facing a problem very easy to solve with a program, but not
On 28/03/2011 8:04 AM, Luis Felipe Parra wrote:
Thanks Duncan, I already installed Rtools but I don't know well how to sort
it out. I tried the command you gave me and got the following error:
install.packages(fPortfolioSolver.tar.gz, type=source, repos=NULL)
Installing package(s) into
Hello,
I am analyzing a dataset where the response is count data. I have one
two-level factor that is repeated within-subjects and additional
between-subject variables that are either categorical or continuous. I have
previously modeled a comparable dataset (without the within-subjects factor)
(11/03/27 22:49), KH wrote:
(11/03/25 22:40), Nick Sabbe wrote:
2. Which model, I mean lasso or elastic net, should be selected? and
why? Both models chose the same variables but different coefficient values.
You may want to read 'the elements of statistical learning' to find some
info on the
I am doing a density plot in R. Spcifically, I want to do a Poisson
Regression, and plot the fitted values. Anyone knows how this can be sone in
R?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Thank you Peter. I didn't realize that was the convention used.
Frank
Peter Dalgaard-2 wrote:
On Mar 28, 2011, at 05:36 , Frank Harrell wrote:
gt; I have a question about the computation of the degrees of freedom in
a linear
gt; model:
gt;
gt; x lt;- runif(20); y lt;- runif(20)
gt;
Hello, I am trying to install fPortfolioSolver using the following commands
and I am getting the following error:
filename
[1]
C:\\Users\\Hp\\Documents\\R\\win-library\\2.12\\fPortfolioSolver_271.75.tar.gz
install.packages(filename, type=source, repos=NULL)
ERROR: dependencies 'fEcofin',
However, shouldn't _free parameters_ only be counted for degrees of freedom and
for calculation of AIC?
The sigma parameter is profiled out in a least-squares linear regression, so
it's not free, it's not a dimension of the likelihood.
Just wondering ...
Hello,
I'd like to ask you something again.
I have a database and it has a column which looks like this one here:
small_factor - factor(c(d_variable1,d_variable2,d_variable3))
small_factor
Now the thing is that I would like to convert each element of this factorized
column. Basically I want to
Dear all,
I would like to ask your help of how I can convert the loop below to a mclapply
(this is the apply to many cores)
for (i in seq(from=-1,to=1-2/ncol(sr),length=ncol(sr))){
for (j in seq(from=-1,to=1-2/nrow(sr),length=nrow(sr))){
Bodnar Laszlo EB_HU Laszlo.Bodnar at erstebank.hu writes:
I have a database and it has a column which looks like this one here:
small_factor - factor(c(d_variable1,d_variable2,d_variable3))
small_factor
small_factor - factor(gsub(^d_,,as.character(small_factor)))
Hi,
I'm not sure it's the best solution, but I think this should do:
levels(small_factor) - gsub(d_, , levels(small_factor))
small_factor
[1] variable1 variable2 variable3
Levels: variable1 variable2 variable3
HTH,
Ivan
Le 3/28/2011 16:40, Bodnar Laszlo EB_HU a écrit :
small_factor-
On 2011-03-27 21:37, Alex Olssen wrote:
Hi everyone,
I am looking to do some manual maximum likelihood estimation in R. I
have done a lot of work in Stata and so I have been using output
comparisons to get a handle on what is happening.
I estimated a simple linear model in R with lm() and
Dear all ,
I would like to extend a list structure to a structure that can hold much more
date.
Please find a small example of my code and my questions regarding it:
lst - list()
for (k in c(1:mmax)){ # 1 ..max as the f.fxy gets as input order-1.
for (l in c(1:nmax)){
On 28.03.2011 15:58, Luis Felipe Parra wrote:
Hello, I am trying to install fPortfolioSolver using the following commands
and I am getting the following error:
filename
[1]
C:\\Users\\Hp\\Documents\\R\\win-library\\2.12\\fPortfolioSolver_271.75.tar.gz
install.packages(filename,
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jack Tanner
Sent: Sunday, March 27, 2011 9:14 PM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] altering a call variable from quote()
Jack Tanner ihok at hotmail.com writes:
On Mar 28, 2011, at 11:32 AM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jack Tanner
Sent: Sunday, March 27, 2011 9:14 PM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] altering a call variable from
Thank you. The problem is that I tryed to use the repository provided in
R-forge and it didn't work. Do you know by any chance a repository where I
could find this package? thank you
2011/3/28 Uwe Ligges lig...@statistik.tu-dortmund.de
On 28.03.2011 15:58, Luis Felipe Parra wrote:
Hello, I
Hi,
I am new to R and I want to plot points on the Map of New York . I also
want to divide map of New York into small grids(not fixed) .I want that
these point should be plotted from a file.How can I do it?Any help would be
greatly appreciated.
Thanks
Jaimin
[[alternative HTML version
Are you sure that 1.78 is not the estimate of sigma and 3.14 the
estimate of sigma^2.
Best Regards
John
On Monday, 28 March 2011, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-03-27 21:37, Alex Olssen wrote:
Hi everyone,
I am looking to do some manual maximum likelihood estimation in R.
Hi all,
I am trying to write a script that will compute Kendall's tau for a 75 time
series (using the Kendall package) and will then write the tau and p values
from the Kendall test to a text file table that can be read into Excel.
I am having no problem calculating Kendall's tau and the
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
Hi, guys. Thanks for all your help.
I tried Gabors methods and they seem to work fine - robust as well. I wish
I had though of those a few days ago! I'll try and give the other methods a
try later.
In the end though this drove me so nuts that I've managed to query the
database which outputed
Rubén Roa rroa at azti.es writes:
However, shouldn't _free parameters_ only be counted for degrees of
freedom and for calculation of AIC?
The sigma parameter is profiled out in a least-squares
linear regression, so it's not free, it's not a
dimension of the likelihood.
Just wondering
Dear all
My dataframe has 80,000 variables which I can not everytime load into R
using *.txt files (read.table option), cost me time and sometime computer
decomes not responsive. So I need a way to save my dataframe in my
workdirectory as such.
Execuse me if the problem is too simple. I tried
Dear all ,
I am trying to run sem by an example with my data but i have problme with an
exogen variable x1 so my examlpe is below
when i add i the equation we have no pboblem but i donât know why ??
x1 -x1, sigmma7, NA
for me this an exogen variable and i am not obliged to specify this
Cleber N. Borges klebyn at yahoo.com.br writes:
Hello All,
I am trying to learn about the GUI in R (with GTK+Glade+RGtk2) and in my
test I don't get sucess in to make
an callback to destroy the application...
When I try to define an function for delet-event callback, I get the
error
Hello,
I've seen various threads on people reporting:
step factor 0.000488281 reduced below `minFactor' of 0.000976563
While I know how to set the minFactor, what I'd like to have happen is for nls
to return to me, the last or closest fitted parameters before it errors out.
In other words,
Hi all,
I am trying to calculating derivatives for vectors using R. I have a
numerical vector, which is in time series manner and contains NAs. I
want to know the changing rate throughout the whole time range. I am
wondering if there is function in R can do this. In R references I
only found
Thank you Jim and David for your help.
The 'levels' call is not a misdirection, in my actual dataset it is
necessary because the flows aren't symmetrical. So while your solution is
quite elegant David, it doesn't apply to my actual data, just the example.
Too bad, it's quite nice!
I do
hi, i have some data, a subset of which is pasted at the end of this message.
i am trying to understand how to do repeated measures as our study
design consists
of a subject and up to 2 siblings.
thus far, my model looks like this--with family_id indicating a
sibling relationship:
formula = y ~
Hello Everyone,
I am using R for creating heatmap from a square matrix. Below is my script
to do so
my_map - read.csv(Desktop/input.csv, sep=,)
my_matrix - data.matrix(my_map)
my_heatmap - heatmap(my_matrix, Rowv=NA, Colv=NA, col = cm.colors(256),
scale=column, margins=c(5,10))
I get a
I am using 'igraph' package to make some graphs of 'gene-gene interaction'.
I can get a data.frame which has three columns.
gene1 gene2 pvalue
AGT MLR1.2e-04
MLR 11BHSD1 1.71e-05
IFG211BHSD2 2.2e-07
. . .
.
Hi,
I am R beginner and am trying to figure out how to generate a complete list of
species for every point, visit, and year. The code below is close but does not
give me a list of species for every point, visit, and year in my data set.
spplist-unique(sumPtCt$Species)
spplength-length(spplist)
One more question is:
when you plot the gene network, you only get a number on each node, then how
can you match the numbers to the genes?
thank you very much,
Karena
--
View this message in context:
http://r.789695.n4.nabble.com/Questions-about-igraph-package-tp3412734p3412745.html
Sent from
On 28 March 2011 09:00, jouba antr...@hotmail.com wrote:
Your syntax is not very tidy. That makes it hard to check.
x1 -x1, sigmma7, NA
for me this an exogen variable and i am not obliged to specify this
equation
model.se-specify.model()
x1-x2,gamm1,NA
x2-x3,gamm2,NA
x3x4,gamm3,NA
There is a fairly basic interface with gnuplot in the TeachingDemos package,
see ?gp.open
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
On Mar 28, 2011, at 12:31 PM, Luis Felipe Parra wrote:
Thank you. The problem is that I tryed to use the repository
provided in
R-forge and it didn't work.
tryed to use, didn't work ... you did not provide console output
so the source of that failre cannot be identified.
Do you know
Dear jouba,
I think you're using the sem() function in the sem package.
I'm not sure that I understand your question, but I think it is why you need to
specify the variance of the exogenous variable x1 as a parameter. The answer is
that it is a parameter to be estimated from the data, but you
On Mar 28, 2011, at 11:20 AM, Ram H. Sharma wrote:
Dear all
My dataframe has 80,000 variables which I can not everytime load
into R
using *.txt files (read.table option), cost me time and sometime
computer
decomes not responsive. So I need a way to save my dataframe in my
workdirectory
On Mon, Mar 28, 2011 at 3:05 PM, karena dr.jz...@gmail.com wrote:
I am using 'igraph' package to make some graphs of 'gene-gene interaction'.
I can get a data.frame which has three columns.
gene1 gene2 pvalue
AGT MLR 1.2e-04
MLR 11BHSD1 1.71e-05
On Mon, Mar 28, 2011 at 3:12 PM, karena dr.jz...@gmail.com wrote:
One more question is:
when you plot the gene network, you only get a number on each node, then how
can you match the numbers to the genes?
V(g)$name gives the vertex names, in the order of vertex ids. E.g.
here is how to create
-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 a
Try this:
as.data.frame.table(xtabs(Number ~ SPP + Point + Visit, template))
On Mon, Mar 28, 2011 at 3:43 PM, BORGMANN,Kathi kborgm...@audubon.org wrote:
Hi,
I am R beginner and am trying to figure out how to generate a complete list
of species for every point, visit, and year. The code
Hi,
I have a data frame like this:
var1=years
var2=Sex ratio (0value1)
var3=lower 95% confidence interval
var4=upper 95% confidence interval
Is there a way to add these confidence intervals to a plot like this?
plot(years,Sex ratio,type=b)
Thanks in advance for any response
Thank you very much, Gabor! That helps.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Try this:
matplot(DF[,c('SexRatio', 'lower', 'upper')], type ='b', pch = 19, col
= c(2, 1, 1))
On Mon, Mar 28, 2011 at 4:25 PM, Simone Santoro misen...@hotmail.com wrote:
Hi,
I have a data frame like this:
var1=years
var2=Sex ratio (0value1)
var3=lower 95% confidence interval
var4=upper
Good Afternoon,
I am trying to verify if R v2.10.1 can be run with HPUX 11.x or AIX?
Thx!
Khan
804-284-7201
/pre
The information contained in this e-mail is confidential...{{dropped:27}}
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Thank you David and Dennis
On Mon, Mar 28, 2011 at 3:43 PM, David Winsemius dwinsem...@comcast.netwrote:
On Mar 28, 2011, at 11:20 AM, Ram H. Sharma wrote:
Dear all
My dataframe has 80,000 variables which I can not everytime load into R
using *.txt files (read.table option), cost me
Dear list,
I'm running windows xp with R 2.12.0. I'm trying to load a excel
spreadsheet into R using the xlsx package. I posted my code below with
the error I get.
res - read.xlsx(Copy of test_excel_input_data.xlsx, 6)
Error in .jcall(RJavaTools, Ljava/lang/Object;, invokeMethod, cl, :
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
Dear R helpers
Suppose I have a data.frame as given below -
my_dat = data.frame(class = c(XYZ, XYZ, XYZ, XYZ, XYZ,ABC, ABC,
ABC, ABC, ABC ), var1 = c(20, 14, 89, 81, 17, 44, 36, 41, 11, 36), var2
= c(1001, 250, 456, 740, 380, 641, 111, 209, 830, 920))
my_dat
class var1 var2
1 XYZ 20
On 3/25/2011 3:13 AM, Denis Kazakiewicz wrote:
Hello
I simply want to plot two variables against one 'year' variable in
qplot.
Is any way of doing this without reshaping data in long format and using
facet function afterwards?
Yes, its is possible. But it is harder and more convoluted than
Try this:
my_dat[order(my_dat$class, -my_dat$var1, decreasing = TRUE),]
On Mon, Mar 28, 2011 at 5:55 PM, Vincy Pyne vincy_p...@yahoo.ca wrote:
Dear R helpers
Suppose I have a data.frame as given below -
my_dat = data.frame(class = c(XYZ, XYZ, XYZ, XYZ, XYZ,ABC, ABC,
ABC, ABC, ABC ), var1
Dear R helpers
I am resending my mail as the output I desire was not properly visible and I
apologize for the same.
Suppose I have a data.frame as given below -
my_dat = data.frame(class = c(XYZ, XYZ, XYZ, XYZ, XYZ,ABC, ABC,
ABC, ABC, ABC ), var1 = c(20, 14, 89, 81, 17, 44, 36, 41, 11, 36),
Dear sir,
Thanks for the great solution.
Regards
Vincy
--- On Mon, 3/28/11, Henrique Dallazuanna www...@gmail.com wrote:
From: Henrique Dallazuanna www...@gmail.com
Subject: Re: [R] Ordering data.frame based on class
To: Vincy Pyne vincy_p...@yahoo.ca
Cc: r-help@r-project.org
Received:
Mingwei Min mm809 at cam.ac.uk writes:
I am trying to calculating derivatives for vectors using R. I have a
numerical vector, which is in time series manner and contains NAs. I
want to know the changing rate throughout the whole time range. I am
wondering if there is function in R can do
The 16th Summer Institute in Statistical Genetics will be held June
13-July 1, 2011, in Seattle, at the University of Washington.
The course offerings include a 2.5 day module in Advanced R
Programming, June 27-29 with Thomas Lumley and Ken Rice as
instructors.
This module covers use of R for
Shahriar, Khandaker (CONT Khandaker.Shahriar at capitalone.com writes:
Good Afternoon,
I am trying to verify if R v2.10.1 can be run with HPUX 11.x or AIX?
Thx!
Khan
804-284-7201
You might get a quick answer, but probably your best bet
is simply to try compiling:
I have a fair bit of experience with S-Plus and have been asked to port
some of my S-Plus bootstrapping functions to R, to which I am relatively
new, Needless to say, I've run into some problems. In particular, I need
to perform bootstrap resampling of the colMeans function using a moving
I will be collecting data where one of the fields may be of length
zero to some variable number of elements, along with other items of
fixed size.
As an illustration if mydata.dat is:
V1 V2 V3 V4 V5
1, 2.3, Bob, {1.7,2.3,3.4}, 4.5
2, 3.4, Carol, {}, 3.4
3, 2.2, Ted,
Try this -- I added commas in the first line (header) so it looks like
a CSV file:
x - textConnection('V1, V2 , V3 ,V4, V5
+ 1, 2.3, Bob, {1.7,2.3,3.4}, 4.5
+ 2, 3.4, Carol, {}, 3.4
+ 3, 2.2, Ted, {1.0,2.5}, 3.5')
# read in the lines and then replace {} with quotes
Duncan I have been trying to work out the solution you gave me but I haven't
really got to sort it out. I tried first the option with install packages
and got this:
filename - file.choose()
filename
[1] C:\\Users\\Hp\\Documents\\R\\win-library\\2.12\\Rsymphony_0.1-12.tar.gz
I have the following situation.
H12 is an array of dimension (n,k,m) and hessian_lambda is a numeric
of length m.
I need to multiply each matrix H12[,,1], H12[,,2], ..., H12[,m] by
hessian_lambda[1], hessian_lambda[2], ..., hessian_lambda[m],
respectively, and then add the resulting (n,k)
On 2011-03-28 13:08, Henrique Dallazuanna wrote:
Try this:
matplot(DF[,c('SexRatio', 'lower', 'upper')], type ='b', pch = 19, col
= c(2, 1, 1))
Or you could have a look at plotCI() in the gplots package.
(There's also a plotCI() in the plotrix package.)
Peter Ehlers
On Mon, Mar 28, 2011
Hello R users,
If I generate a random tree with n=10 tips as rtree(n=10) say, is there a
way to have the distances between all tips put into a n by n matrix?
Sincerely,
Brian
[[alternative HTML version deleted]]
__
R-help@r-project.org
Hi:
Is this what you're after?
m - array(1:27, c(3, 3, 3))
xx - 1:3
Reduce('+', lapply(xx, function(k) xx[k] * m[, , k]))
m
, , 1
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
, , 2
[,1] [,2] [,3]
[1,] 10 13 16
[2,] 11 14 17
[3,] 12 15
Brian Pellerin brianpatrickpellerin at gmail.com writes:
If I generate a random tree with n=10 tips as rtree(n=10) say, is there a
way to have the distances between all tips put into a n by n matrix?
You should probably send this to the r-sig-phylo (phylogenetics)
special interest group
?pdf
On Mon, Mar 28, 2011 at 12:57 PM, chakri_amateur chakri2...@yahoo.co.inwrote:
Hello Everyone,
I am using R for creating heatmap from a square matrix. Below is my script
to do so
my_map - read.csv(Desktop/input.csv, sep=,)
my_matrix - data.matrix(my_map)
my_heatmap -
Dennis,
Yes, Reduce works, but unfortunately it does not speed up the
evaluation time (actually, it makes it a little bit slower).
Thank you anyway.
On Tue, Mar 29, 2011 at 2:21 AM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Is this what you're after?
m - array(1:27, c(3, 3, 3))
xx -
Hi,
I was wondering if the following model can be fitted by nlme/lme4. The
difficult part is that the error term is combined with other parameters.
The model that we need to fit is as below and the R codes to generate data is
provided as well. Many thanks!!!
# Y_{ijk} = (E_0 +
Dear Christoph
Are there formulas in any cells of sheet 6?
One thing you might try is to open up the spreadsheet in Excel and create a new
sheet,
call it Sheet 7.
In Sheet 6, select all the cells with data (or just click the
upper left icon at the border of the spreadsheet that selects the
Benno,
That helps but it only makes the color bar symmetrical. I want to be able to
compare 2 different heatmaps so that 0.7 (for example) is always the same tone
of green and not shifted slightly. Is this possible?
Paul
On 28 Mar 2011, at 10:42, Benno Pütz wrote:
On 27.Mrz.2011, at
Hi Kathi,
Are you looking for something like this?
Alist - letters[1:4]
Blist - 1:5
Clist - LETTERS[24:26]
Alist
[1] a b c d
Blist
[1] 1 2 3 4 5
Clist
[1] X Y Z
expand.grid(Alist, Blist, Clist)
Var1 Var2 Var3
1 a1X
2 b1X
3 c1X
4 d1X
5
On 2011-03-28 09:33, Pam Allen wrote:
Thank you Jim and David for your help.
The 'levels' call is not a misdirection, in my actual dataset it is
necessary because the flows aren't symmetrical. So while your solution is
quite elegant David, it doesn't apply to my actual data, just the example.
Here's a start...
require(maps)
foo - map(state, new york)
lines(x = range(foo$x, na.rm = TRUE), y = range(foo$y, na.rm = TRUE))
You can figure out from this how to specify the coordinates that you
want for dividing up the map, put them in a file etc.
Steven McKinney
On 2011-03-28 10:25, Steve Greiner wrote:
Hello,
I've seen various threads on people reporting:
step factor 0.000488281 reduced below `minFactor' of 0.000976563
While I know how to set the minFactor, what I'd like to have happen is for nls
to return to me, the last or closest fitted
Doesn't use apply but maybe the following will work for you?
temp1- data.frame(ID=c(Herb,Shrub),stat=c(4,5),pvalue=c(.03,.04))
temp2- data.frame(ID=c(Herb,Shrub,
Tree),stat=c(12,15,13),pvalue=c(.2,0.4,.3))
L-list(a=temp1,b=temp2)
d.f = do.call(rbind,L)
d.f$tableName = substring(rownames(x),1,1)
?cophenetic
Cheers,
Simon.
On 29/03/11 10:27, Ben Bolker wrote:
Brian Pellerinbrianpatrickpellerinat gmail.com writes:
If I generate a random tree with n=10 tips as rtree(n=10) say, is there a
way to have the distances between all tips put into a n by n matrix?
You should
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