On Feb 21, 2012, at 12:08 AM, alexiamelissa wrote:
I have a follow up question to Dr Winsemius' post. You can use the
AIC
criterion against all possible cut off values C to see which
minimizes the
AIC and then that is the ideal cut off in trying to dichotomize a
continuous
variable.
On Feb 21, 2012, at 3:21 AM, David Winsemius wrote:
On Feb 21, 2012, at 12:08 AM, alexiamelissa wrote:
I have a follow up question to Dr Winsemius' post. You can use the
AIC
criterion against all possible cut off values C to see which
minimizes the
AIC and then that is the ideal cut off
David: Brilliant! Thanks very much.
As Berend pointed out, I was not precise in the query, sorry. Please note in
the example that we have a run of three state 2 after 0, and later another run
of two state 2 after 0.
0 2 2 2 0 0 2 2
Whenever the state moves from 0 to 2, I want to compute
Dear Faiz,
in principle, you can include whatever moderators you want in a meta-regression
model. A technical issue here is that the two follow-up variables are probably
strongly correlated, so this may create issues with the model fitting and it
also complicates the interpretation of the
Hi Sebastian,
how about this:
mycurve - function (expr) {
do.call(curve,list(substitute(expr),-100,100))
}
mycurve(x^2)
mycurve(sin(x/20))
mycurve(x+x^2-x^3)
cheers
Am 21.02.2012 06:28, schrieb Sebastian Kranz:
Dear List members,
I really, like the feature that one can call R functions
Dear R Group
in package Deducer, how should i use the
add.cross.strata.test() function to pass the chi.squared.test function as
htest argument?
##Example###
library(Deducer)
dat-data.frame(a=rnorm(100).5,b=rnorm(100)0,c=rnorm(100)(-.5))
tables-contingency.tables(
row.vars=a,
Thanks, that is exactly what I have been looking for!
On Tue, 21 Feb 2012 11:58:01 +0100
Eik Vettorazzi e.vettora...@uke.de wrote:
Hi Sebastian,
how about this:
mycurve - function (expr) {
do.call(curve,list(substitute(expr),-100,100))
}
mycurve(x^2)
mycurve(sin(x/20))
mycurve(x+x^2-x^3)
On Tue, Feb 21, 2012 at 2:25 AM, jpm miao miao...@gmail.com wrote:
Hello,
I try to handle the data using read.csv , zoo and aggregate functions.
The data contains NA values. After aggregating monthly data into quarterly
data, all data become NA. Is it because I don't properly aggregate the
Dear all,
I have a function that for a variable number of inputs plots them to the same
plot
I am doing this quite simply by
plot(seq(from=start, to=stop, length.out=np),
datalist[[1]]$dataset
xlim=c(start, stop), ylim=c(0, 1),
Two comments.
I've not found bootstrapping worthwhile for Cox models. If one has
10-20 events per covariate and no outrageous coefficients (risks of 10
fold) the standard asymptotics are very good. With an multiple events
AG model there is the additional consideration that no one subject is
What do you want to change about the lines? pch (different characters)
might give the desired variety.
Michael
On Tue, Feb 21, 2012 at 7:56 AM, Alaios ala...@yahoo.com wrote:
Dear all,
I have a function that for a variable number of inputs plots them to the same
plot
I am doing this quite
Hello,
How can I display the xlim of the boundaries of all or specific breaks in a
histogram? I generated the attached plot with hist and would like to know which
values of x correspond to the frequency 329 and display these values on the x
axis?
Best,
On 02/20/2012 08:35 PM, ejm0091 wrote:
Hello All,
I am relatively new to R. I would like to know if there is a way to alter
LIMMA defualt options such that the package instead of averaging signal
intensities of probesets selects the probesets with highest baseline
expression/highest signal
?findLineNum in conjunction with ?setBreakpoint and ?traceback or
options(error=recover) seems like a good strategy within R. Not sure
what you can get on the IDE end.
Michael
On Tue, Feb 21, 2012 at 2:38 AM, jpm miao miao...@gmail.com wrote:
Hello,
I am using RStudio and have trouble
Both ESS (with ess-tracebug) and StatET (an R plug-in for Eclipse) has
debugger function.
Shige
On Tue, Feb 21, 2012 at 8:16 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
?findLineNum in conjunction with ?setBreakpoint and ?traceback or
options(error=recover) seems like a good
On Tue, Feb 21, 2012 at 12:15 AM, Ben quant ccqu...@gmail.com wrote:
Thanks again for your so far on proto. I have another question.
What is the best way to do stuff based on data prior to calling a
function? I tried the code below without expr (and including commas after
data member
westland westl...@uic.edu writes:
Here is what I have done:
I read in an 1 x 8 table of data, and assign the first four columns to
matrix A and the second four to matrix B
pls -read.table(C:/Users/Chris/Desktop/SEM Book/SEM Stat
Example/Simple Header Data for SEM.csv,
Hi Ista,
I got it.
I try to deselect the all packages and try to run the code again.
Eventually, the console display no digest package.
You are so nice. A lot of things to learn...
Many thanks.
Best,
VD
--
View this message in context:
Hi Alok
Are you certain that you are keeping your connection open? last_insert_rowid()
returns id of last inserted row from the current database connection (which
invoked the function). Is it possible that every time you run the query you
making a new connection?
Hope this helps,
Chris
Dear all,
Thanks all of u for your help. Now I've another similar problem. I want to
plot within the same plot, different lines, each one in a different color
depending on the factor level. I've been able to do it like this, but if i
try with rainbow colors it doesn't work. Can anybody help me
Hi Everyone,
I am working on Reduced model say a ODE and I want to do the Parameter
estimation of the model. Could you please help me in this?
Thanks in advance.
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Sent
HI,
I am trying to run this code
pdf(file=Biology_2012_GOF.pdf, width=8.5, height=4.5)
grid.draw(my.sgpd$Goodness_of_Fit$Biology_2012$Q1_Q2)
dev.off()
I get an error mesage object not found- can someone please help?
Kavita
--
View this message in context:
Guys,
I'm having an error when I use the command:
library(MASS) dados-read.table(inverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)Mensagens de aviso perdidas:1:
In dweibull(x, shape, scale, log) : NaNs produzidos2: In dweibull(x, shape,
scale, log) : NaNs produzidos3: In
On Feb 21, 2012, at 9:21 AM, kmittapalli wrote:
HI,
I am trying to run this code
pdf(file=Biology_2012_GOF.pdf, width=8.5, height=4.5)
grid.draw(my.sgpd$Goodness_of_Fit$Biology_2012$Q1_Q2)
dev.off()
I get an error mesage object not found- can someone please help?
What does this return:
On Feb 21, 2012, at 9:24 AM, Vanúcia Schumacher wrote:
Guys,
I'm having an error when I use the command:
library(MASS) dados-read.table(inverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)Mensagens de aviso
perdidas:1: In dweibull(x, shape, scale, log) : NaNs
Hi All,
I have a left truncated, right censored cox model:
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to know how much of the observed variance (as a number between 0
and 1) is explained by each variable. How could I do that?
Adding terms sequentially and then using
Hi
If you want to have more than 6 line types you had either to use colours
or to follow last part of lty advice from par and go to section Line
types. In that case you can not simúply use lty 1,2,...,n but you have to
prespecify line types in some character vector and choose from that
On Feb 20, 2012, at 7:45 AM, kende jan wrote:
Hello,
I would like to perform triangular test for clinical trial with R.
can you help me please ?
Jan
Hi,
I am not aware of any specific implementations of Whitehead's Triangular
Test[1] in R, though somebody else may be aware of
On Feb 21, 2012, at 16:36 , David Winsemius wrote:
It suggests you managed to send negative or infinite numbers to dweibull, a
distribution which only supports values when given positive, finite numbers.
Look at your data more closely.
Zeros are a common cause, too. For a 1 (the shape
You do not have your data organized in a way that persp() can use it. Here
is a simple example. The function will not work with x, y, z vectors. X and
Y are vectors in order with no repeated values, Z is a matrix containing one
value for every combination of x and y:
x - 1:10
y - 1:15
z -
Thanks Ilai this helped.
Cheers,
Nino
Il 16/02/2012 23:15, ilai ha scritto:
# All days in years 2006 to 2009 by month in 48 (12x4) files.
days- seq(as.Date(2006/1/1), as.Date(2009/12/31),by=day) # one
long vector
out- paste(rep(format(days,'%d%m%y'),each=2),c('aaa','bbb'),sep='_')
# reformat
GUYS,
I NEED HELP WITH ERROR:
library(MASS)
dados-read.table(mediaRGinverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)
Erro em fitdistr(dados[[1]], densfun = weibull) :
Weibull values must be 0
WHY RETURN THIS ERROR? WHAT CAN I DO?
BEST REGARDS
Vanúcia SchumacherCurso de graduação em meteorologia - UFPELBolsista do
Programa de Educação Tutorial - PET
From: vanucia-schumac...@hotmail.com
To: r-help@r-project.org
Subject: help error: In dweibull(x, shape, scale, log) : NaNs produzidos
Date: Tue, 21 Feb 2012 14:24:02 +
i am trying to run stepwise regression for two models(lower,upper),
family=gamma
and i get the same error despite the models i use.
Error in UseMethod(extractAIC) :
no applicable method for 'extractAIC' applied to an object of class
formula
In addition: Warning message:
In nobs.default(object,
should i keep the first line of the code like-
pdf(file=Biology_2012_GOF.
pdf, width=8.5, height=4.5)
str( my.sgpd$Goodness_of_Fit$
Biology_2012$Q1_Q2)
Or should I just have your code?
str( my.sgpd$Goodness_of_Fit$
Biology_2012$Q1_Q2)
thanks again.
On Tue, Feb 21, 2012 at 10:32 AM, David
I have some data set which has some values -999.000 I would like to remove
whole row of this kind of values.
e.g
a-matrix(c(1,2,3,4,4,5,6,6,-999.99,5,9,-999.00),nrow=4)
a-
[,1][,2] [,3]
[1,]14 -999.99
[2,]255.00
[3,]369.00
[4,]46 -999.00
Hello,
carol white wrote
Hello,
How can I display the xlim of the boundaries of all or specific breaks in
a histogram? I generated the attached plot with hist and would like to
know which values of x correspond to the frequency 329 and display these
values on the x axis?
Best,
carol
Hello,
Try
a-matrix(c(1,2,3,4,4,5,6,6,-999.99,5,9,-999.00),nrow=4)
ix - apply(a, 1, function(x) sum(trunc(x) != -999) == ncol(a))
a[ix, ]
Hope this helps
Rui Barradas
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On 21.02.2012 16:48, Vanúcia Schumacher wrote:
GUYS,
I NEED HELP WITH ERROR:
library(MASS)
dados-read.table(mediaRGinverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)
Erro em fitdistr(dados[[1]],
densfun = weibull) :
Weibull values must be 0
WHY RETURN THIS ERROR?
Dear all,
I am using R to work on huge numbers of telemetry data divided by day.
Each file (an xlsx file) contains 2 rows, the first one for sst readings
and the second one for chl readings, and 72360 columns, each
corresponding to the centre of a cell in my study area. The columns have
no
On Tue, Feb 21, 2012 at 07:52:20AM -0800, uday wrote:
I have some data set which has some values -999.000 I would like to remove
whole row of this kind of values.
e.g
a-matrix(c(1,2,3,4,4,5,6,6,-999.99,5,9,-999.00),nrow=4)
a-
[,1][,2] [,3]
[1,]14 -999.99
[2,]2
On Feb 21, 2012, at 11:43 AM, peter dalgaard wrote:
On Feb 21, 2012, at 16:36 , David Winsemius wrote:
It suggests you managed to send negative or infinite numbers to
dweibull, a distribution which only supports values when given
positive, finite numbers. Look at your data more closely.
labbig lkaimaka at windowslive.com writes:
i am trying to run stepwise regression for two models(lower,upper),
family=gamma
and i get the same error despite the models i use.
Error in UseMethod(extractAIC) :
no applicable method for 'extractAIC' applied to an object of class
formula
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of uday
Sent: Tuesday, February 21, 2012 7:52 AM
To: r-help@r-project.org
Subject: [R] removing particular row from matrix
I have some data set which has some values -999.000 I
On 21.02.2012 18:50, Vanúcia Schumacher wrote:
Yes, my data dados[[1]] have values=0, but what can I do to fix thi problem/?
I replace values equal to 0 for 0.1??
If they are really 0, they are not from a weibull. If you think they
are, there is something wrong with the data. If not,
Aniruddha,
Thank you, this is much more clear (and makes sense). Unfortunately I am
out of my element here so I don't think I'll be much help, but see inline
On Tue, Feb 21, 2012 at 4:14 AM, Aniruddha Mukherjee
aniruddha.mukher...@tcs.com wrote:
-- I ran the socketConnection command and the
On Feb 21, 2012, at 10:39 AM, kmittapalli wrote:
should i keep the first line of the code like-
pdf(file=Biology_2012_GOF.
pdf, width=8.5, height=4.5)
str( my.sgpd$Goodness_of_Fit$
Biology_2012$Q1_Q2)
Or should I just have your code?
str( my.sgpd$Goodness_of_Fit$
Biology_2012$Q1_Q2)
You are
On 21.02.2012 19:57, zheng wei wrote:
Thank you both for helping. Still could not figure out.
I was contacting different supporting IT departments in my university but did
not get any help.
For the moment, I just want to what does the instruction of the package means.
You could find this
Hi Everyone,
I am a graduate student who will be using R to do my analysis. I need to do
a spatial analysis, and the first step is to calculate the geographic
distance between my study sites. I am hoping to use earth.dist because it
allows for multiple pairwise distances to be calculated at one
Vanúcia SchumacherCurso de graduação em meteorologia - UFPELBolsista do
Programa de Educação Tutorial - PET
From: vanucia-schumac...@hotmail.com
To: r-help@r-project.org
Date: Tue, 21 Feb 2012 15:48:44 +
Subject: [R] HELP ERROR Weibull values must be 0
GUYS,
I NEED HELP
Hi,
then, If I replace the midpoint 0.0025 average values equal to 0there won't be
prejudice in the results?Would then be the best thing to do?
regards,
Subject: Re: [R] help error: In dweibull(x, shape, scale, log) : NaNs
produzidos
From: pda...@gmail.com
Date: Tue, 21 Feb 2012
Yes, my data dados[[1]] have values =0, but what can I do to fix thi
problem/? I replace values equal to 0 for 0.1??
thank you
Date: Tue, 21 Feb 2012 18:44:34 +0100
From: lig...@statistik.tu-dortmund.de
To: vanucia-schumac...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R]
Here is my code
slidingwindowplotATGC = function(windowsize, inputseq)
{
starts = seq(1, length(inputseq)-windowsize, by = windowsize)
n = length(starts)
chunkGs = numeric(n)
chunkAs = numeric(n)
chunkTs = numeric(n)
chunkCs = numeric(n)
for (i in 1:n) {
Thank you both for helping. Still could not figure out.
I was contacting different supporting IT departments in my university but did
not get any help.
For the moment, I just want to what does the instruction of the package means.
You could find this instruction on page
Dear Bjørn-Helge Mevik:
Thank you so much (!!) for your speedy help. This works perfectly, no
fuss. My understanding of data.frames is still fuzzy, so I knew it was
some flaw in my understanding.
I enjoyed your very informative writeups on the package in JSS and R-News;
they told me a lot
Someone has given me a data set that produces a failure with all the
hallmarks of a memory overrun in a C routine. I've tried running it
under an instrumented version of R with valgrind and don't catch
anything. I do get one error message out though:
==642== Warning: set address range perms:
seq(1,4,1)
[1] 1 2 3 4
seq(1,4,-1)
Error in seq.default(1, 4, -1) : wrong sign in 'by' argument
On Tue, Feb 21, 2012 at 1:53 PM, bioinformatics uccstud...@live.ie wrote:
Here is my code
slidingwindowplotATGC = function(windowsize, inputseq)
{
starts = seq(1,
Thats all well and good but the number im using is a positive so that
shouldn't be happening.
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Sent from the R help mailing list archive at Nabble.com.
Thanks Prof. Terry for the response.
To answer the second comment, the weights I am considering are inverse
probability of treatment weights (IPTW). To perform bootstrapping, my
initial thought would be to
- select patients (sampling from unique id with replacement) and
- then include all the
Don't delete the context of the message. If you are sure that windowsize is
positive, are you also sure that it is less than length(inputseq)?
seq(1, -10, 1)
Error in seq.default(1, -10, 1) : wrong sign in 'by' argument
--
David L Carlson
Associate
I haven't quite worked it out for the Wiebull but that will certainly (down-)
bias your variance estimates for, e.g., the normal.
I think the right thing to do is to use the correct distribution rather than
forcing an incorrect distribution to fit incorrectly.
Michael
On Feb 21, 2012, at
Please do the following:
install.packages(fortunes)
require(fortunes)
fortune(brain surgery)
cheers,
Rolf Turner
On 22/02/12 07:42, cmartin wrote:
Hi Everyone,
I am a graduate student who will be using R to do my analysis. I need to do
a spatial analysis, and the
Hi R users,
I'm trying to build an R package and most of the work have been done in
Fortran programming.
I looked for help for a while and seems like R can only call Fortran
subroutines. Now my fortran code has modules, subroutines and main program.
Is there anyway that R can call such fortran
On Tue, Feb 21, 2012 at 10:53:44AM -0800, bioinformatics wrote:
Here is my code
slidingwindowplotATGC = function(windowsize, inputseq)
{
starts = seq(1, length(inputseq)-windowsize, by = windowsize)
n = length(starts)
chunkGs = numeric(n)
chunkAs = numeric(n)
Hi, I have a microarray dataset from Agilent chips. The data were really log
ratio between test samples and a universal reference RNA. Because of the nature
of log ratios, coefficient of variation (CV) doesn't really apply to this kind
of data due to the fact that mean of log ratio is very
On Tue, Feb 21, 2012 at 1:44 PM, array chip arrayprof...@yahoo.com wrote:
Hi, I have a microarray dataset from Agilent chips. The data were really log
ratio between test samples and a universal reference RNA. Because of the
nature of log ratios, coefficient of variation (CV) doesn't really
Tura,
meus dados possuem mais de 700 elementos, fica muito inviavel passa-los aqui
por summarya questão e que nesses dados eu tenho valores ausentes e valores 0,
li algumas sugestões em que deve-se mensurar os valores em 0 para 0.1,sendo
assim ele roda tranquilo , mas retorna um aviso de
I am wondering if i got this thing right
I have a function of two variables ( lambda and nu) and three
hyperparameters(a,b,c)
library(compoisson)
# function for finding inverse of k
invk - function(x,a=1,b=1,c=1){
if (b/c=log(factorial(floor(a/c)))+(a/c-floor(a/c))*log(floor(a/c)+1))
Hi,
I want to impute a data set multiple times with Amelia, but the data set is
large so it takes a long time. As a result, I'm trying to run the multiple
imputation with parallel processors in Windows, but am having trouble. Here
is a quick example:
##
library(foreach)
library(doSNOW)
Hi list,
I want to draw a bar plot with color indicating one grouping and different
shading on top of the color indicating another grouping. How should I proceed?
Thanks!
...Tao
__
R-help@r-project.org mailing list
Inline below.
On Tue, Feb 21, 2012 at 2:07 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
On Tue, Feb 21, 2012 at 1:44 PM, array chip arrayprof...@yahoo.com wrote:
Hi, I have a microarray dataset from Agilent chips. The data were really log
ratio between test samples and a universal
Good advice. But perhaps ?mad or some other perhaps robust plain old
measure of spread?
The problem is not (lack of) robustness to outliers, the problem is to
find genes whose expression variation is small compared to (mean)
expression. Trouble is, Agilent throws the mean expression
Hi all,
SETUP:
I have pairwise data on 22 chromosomes. Data matrix X for a given
chromosome looks like this:
1 13 58 1.12
6 142 56 1.11
18 307 64 3.13
22 320 58 0.72
Where column 1 is person ID 1, column 2 is person ID 2, column 3 can
be ignored, and column 4 is how much chromosomal sharing
On Feb 21, 2012, at 22:44 , array chip wrote:
Hi, I have a microarray dataset from Agilent chips. The data were really log
ratio between test samples and a universal reference RNA. Because of the
nature of log ratios, coefficient of variation (CV) doesn't really apply to
this kind of data
Hi,
Is there a short way of doing this?
I have the following table in R:
12.0 0.5 0.6 0.2 0 0
12.3 1.2 0.8 0 0 0
13.1 0 1.2 0 0 0
10.1 0 0 0 1.3 0
10.2 1.3
There's almost always a better way than a loop (although
sometimes it isn't worth the effort to figure it out). This time
it's straightforward:
apply(c5[, 2:4], 1, function(x)sum(x 0))
[1] 3 2 1 0 3
Sarah
On Tue, Feb 21, 2012 at 6:04 PM, Valerie Moore vmoore2...@yahoo.com wrote:
Hi,
Is
Valerie,
In additio to Sarah's suggestion, you could also use
rowSums(c5[, 2:4] 0)
HTH,
Jorge.-
On Tue, Feb 21, 2012 at 7:45 PM, Sarah Goslee wrote:
There's almost always a better way than a loop (although
sometimes it isn't worth the effort to figure it out). This time
it's
Your problem isn't in earth.dist but one step earlier in the call to
data(). As the warning message says, it can't find your data...most of
the time, data() is only used for built-in data sets; if you are
bringing your own data to R you need to get it in another way: this
might help
Hello,
I have built a zoo object. I try to see it in a tabular form. Is there
any way to view the date and the data in the excel format? If I use
as.matrix, I can only see the data but not the date.
x2zooq CPI CPIZF UNRUNRSA
1978 Q1 NaN NaN 1.87
Thank you very much! I'll follow-up with more questions as I dabble...if I
have any.
Thank you
ben
On Tue, Feb 21, 2012 at 7:01 AM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
On Tue, Feb 21, 2012 at 12:15 AM, Ben quant ccqu...@gmail.com wrote:
Thanks again for your so far on proto.
What is the excel format in R? From what you put below, it sure
looks like the date is visible...perhaps you want time()?
Michael
On Tue, Feb 21, 2012 at 9:01 PM, jpm miao miao...@gmail.com wrote:
Hello,
I have built a zoo object. I try to see it in a tabular form. Is there
any way to view
Please cc the list on all correspondence.
The easiest way to get data in/out of Excel will likely be to use read.zoo
and write.zoo: see the accompanying vignette for details. For more
information, see the R Import/Export manual
M
On Tue, Feb 21, 2012 at 9:35 PM, jpm miao miao...@gmail.com
hello,
I am using johnson distributions to make distributions. I want to recover
statistics from these distributions, but all I can see how to do is plot
them.
How would I get, say, the gini on a johson distribution i create?
thanks,
Ryan
--
Ryan Murphy
2012
B.A. Economics and Mathematics
This seems like a natural fit for ggplot2 graphics, but I'm not aware
of shading as a widely implemented graphical element: this might have
some useful information --
http://had.co.nz/ggplot2/geom_histogram.html
Can you find an example of shading in R? (perhaps in the R Graphics
Gallery) If you
Hi all,
I got this time for my code,
proc.time()-pt
user systemelapsed
132541.743 0.004 132533.526
As you can see, there is huge difference btw elapsed time and system time.
Does that mean lots of I/O? Or some bad coding?
Thanks,
Libo
[[alternative HTML version
The time relationships aren't strictly linear between any of the three
measures, but *very generally* I've interpreted them as something
like:
User: stuff you do (i.e., doing all the commands)
System: stuff at the OS level (memory allocations and whatnot)
Elapsed: Clock time
None is a great
Hi Chris,
Apologies for getting back late. In my case I am creating a new
connection everytime I insert a new trade. I thought last_insert_rowid()
would insert max(rowid) for given table? I can try to insert bunch of
rows in a data frame, but whenever I create a new connection, I will get
Thanks.
I was just reminded by the tech support in my university that cplex is an
independent software owned by ILOG, which in turn is now owned by IBM. I
suceeded in installing the software cplex under the directory of C:/Program
Files/IBM/ILOG/CPLEX_Studio_Academic124/cplex
I guess Rcplex
Thanks. Shall I sum the user time and system time, which roughly equals to
the elapsed time?
I also tried to improve the code by using 'cmpfun(myfunction)' in
'compiler' package, however, it doesn't help too much.
Libo
On Tue, Feb 21, 2012 at 8:11 PM, R. Michael Weylandt
Hi all,
I am looking for good packages in R for data-fitting including cross
validation... so that I don't have to write my own cross-validation
programs...
For your information, I have also posted more detailed description of my
problem including a few plots here:
Thank you Peter.
John
From: Peter Langfelder peter.langfel...@gmail.com
To: Bert Gunter gunter.ber...@gene.com
Sent: Tuesday, February 21, 2012 2:51 PM
Subject: Re: [R] CV for log normal data
Good advice. But perhaps ?mad or some other perhaps robust
Thanks Weidong for your help.I had earlier tried Step AIC also but no use.
Trying other options as suggested by the R-group.
Subha
[[alternative HTML version deleted]]
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Thanks Steve.
From: Steve Lianoglou mailinglist.honey...@gmail.com
To: David Winsemius dwinsem...@comcast.net
ject.org
Sent: Friday, February 17, 2012 9:27 PM
Subject: Re: [R] stepwise selection for conditional logistic regression
Also, when you're
Suppose i have a working correlation matrix R, and i want to generate a
longitudinal poisson data with n clusters. the cluster size is assumed to
be t and the marginal mu for response variable is lambda.
thanks for help!
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View this message in context:
Hi David
My data set has about 20 significant variables and step function with logistic
regression is working fine( in R-commander). I tried to get conditional
logistic by introducing the stratum variable and clogit. The clogit is not
converging but is giving the summary of the model. When
hi
i checked the package corcounts, but it seems to be used for generating
high dimensional correlated count data. is it also proper for generating
data with N=50 for instance?
thanks
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Hi Michael,
Thanks for your reply!
I'm not sure if I was clear. I was really thinking about shading lines.
I have just came up with a solution on my own:
barplot(1:10, col=rep(0:1, each=5))
barplot(1:10, angle=20, density=c(0,20), col=2, add=T)
Here is my initial failed attempt:
On 02/21/2012 11:56 PM, Alaios wrote:
Dear all,
I have a function that for a variable number of inputs plots them to the same
plot
I am doing this quite simply by
plot(seq(from=start, to=stop, length.out=np), datalist[[1]]$dataset
xlim=c(start, stop), ylim=c(0, 1), type=l)
Hi everyone,
I have 2 data sets and I like to carry out a test to find out if they
come from the same distribution.
Any suggestions ? thanks in advance.
M.O
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On 02/22/2012 09:24 AM, Shi, Tao wrote:
Hi list,
I want to draw a bar plot with color indicating one grouping and different
shading on top of the color indicating another grouping. How should I proceed?
Hi Tao,
You can build a barplot with the rectFill function (plotrix), but it is
not a
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