Hi,
I want to do a use case in R language. My problem statement is to upgrade the
passengers from one membership level to another membership level in airlines
based on their characteristics. It is like customer profiling based on their
usage characteristics. Suggest a method that intakes a larg
Hello,
I am using GDAL 2.0 for most of my work, but rgdal depends on GDAL < 2. I
have built and installed GDAL 1.11.2 in /opt/gdal-1.11.2, and rgdal
compiles and installs into R, using the following command:
sudo R CMD INSTALL
--configure-args="--with-gdal-config=/opt/gdal.1.11.2/bin/gdal-config"
Hi Luigi
I should have made up an example to make things easier when I replied today
This should get you going
set.seed(1)
PLATE <-
data.frame(Delta.Rn = rnorm(500),
Cycle = rnorm(500),
Well = rep(1:50, each = 10))
head(PLATE)
xyplot(Delta.Rn ~ Cycle | Well,
da
hi list,
Any updates on this issue? Thank you very much!
Tao
> devtools::install_github("rstudio/packrat")
WARNING: Rtools 3.3 found on the path at c:/Rtools is not compatible with R
3.2.0.
Please download and install Rtools 3.1 from
http://cran.r-project.org/bin/windows/Rtools/, remove t
> Interesting that a 2D matrix of size Nx1 is treated as a different
> animal from a length N vector.
I think we can call this a bug in stl().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Apr 21, 2015 at 6:39 PM, Paul wrote:
> William Dunlap tibco.com> writes:
> > Use the str() functi
William Dunlap tibco.com> writes:
> Use the str() function to see the internal structure of most
> objects. In your case it would show something like:
>
> > Data <- data.frame(theData=round(sin(1:38),1))
> > x <- ts(Data[[1]], frequency=12) # or Data[,1]
> > y <- ts(Data, frequency=12)
> > str(x)
Hi Luigi
The layout.heights is in the wrong place try
par.settings = list(
strip.background=list(col="white"),
axis.text = list(cex = 0.6),
par.xlab.text = list(cex = 0.75),
par.ylab.text = list(cex = 0.75),
superpose.symbol = list(p
Hi Jim,
That solves my problem. Than you.
-- Kumar
ᐧ
Postdoctoral Associate
Fagan Lab, Department of Biology
University of Maryland
On Tue, Apr 21, 2015 at 5:31 PM, Jim Lemon wrote:
> HI Kumar,
> A simple way is:
>
> phimat<-function(x) {
> xcol<-dim(x)[2]
> newx<-matrix(NA,nrow=xcol,ncol=x
It's been fixed in the dev version, and I'm planning on submitting to
CRAN in the near future.
Hadley
On Tue, Apr 21, 2015 at 6:01 PM, Shi, Tao wrote:
> hi list,
>
> Any updates on this issue? Thank you very much!
>
> Tao
>
>
>> devtools::install_github("rstudio/packrat")
> WARNING: Rtools 3.3 f
(1) The manner in which you presented your data was a total mess.
If you ask for help, please have the courtesy to present your data in
such a manner that a potential "helper" can access it without needing to
do a great deal of editing and processing. Like so:
pts <- as.data.frame(matrix(c(4
HI Kumar,
A simple way is:
phimat<-function(x) {
xcol<-dim(x)[2]
newx<-matrix(NA,nrow=xcol,ncol=xcol)
for(i in 1:xcol) {
for(j in 1:xcol) newx[i,j]<-phi(table(x[,i],x[,j]))
}
rownames(newx)<-colnames(newx)<-colnames(x)
return(newx)
}
phimat(df)
Jim
On Wed, Apr 22, 2015 at 6:34 AM, Kumar
I want to calculate phi coefficient for every pair of the columns. Is there
a way to generate a matrix like a correlation matrix? I know cor function
in the case below gives same answer as phi coefficient.
x <- sample(c(0,1), 10, replace=TRUE)
y <- sample(c(0,1), 10, replace=TRUE)
z <- sample(c(0
Dear R-Help:
>From multiple sources comparing methods of tree classification and tree
>regressions on various data sets, it seems that Exhaustive CHAID (distinct
>from CHAID), most commonly generates the most useful tree results and, in
>particular, is more effective than ctree or rpart which a
Have you gone thruway any T tutorials yet? This is a very basic question
that I do not believe would arise if you had done so.
The answer is that it is a logical that controls whether means and sd's or
medians and mads are calculated and returned. But I don't think this answer
will be comprehensib
> On Apr 21, 2015, at 1:05 PM, Luciano La Sala
> wrote:
>
> Dear everyone,
>
> The following function, taken from Quick-R, gets measures of central tendency
> and spread for a numeric vector x.
>
> I can't figure out what the argument npar means in each instance.
> Any tips will be most appr
Dear everyone,
The following function, taken from Quick-R, gets measures of central
tendency and spread for a numeric vector x.
I can't figure out what the argument npar means in each instance.
Any tips will be most appreciated.
mysummary <- function(x, npar=TRUE, print=TRUE) {
if (!npar) {
Thanks! The package still cannot be installed and I've found an alternative
way which is using package "limma"
On Tue, Apr 21, 2015 at 10:20 AM, Marc Schwartz
wrote:
>
> > On Apr 21, 2015, at 12:01 PM, Ye Lin wrote:
> >
> > Hi All, after installing the new version of R (3.2.0), I cannot find
>
> On Apr 21, 2015, at 12:01 PM, Ye Lin wrote:
>
> Hi All, after installing the new version of R (3.2.0), I cannot find
> package "colbycol", is there anyway to use it with the new version?
>
> I want to use function cbc.read.table, which is in package "colbycol". If
> this package is no longer
Hi All, after installing the new version of R (3.2.0), I cannot find
package "colbycol", is there anyway to use it with the new version?
I want to use function cbc.read.table, which is in package "colbycol". If
this package is no longer available in the new version, is there anyway
around it?
Tha
Hi R users,
I want to calculate Thiessen weights to compute areal rainfall from number
of point measurements. I am using R and thanks to some previous question in
the same topic, I got to know that I can usedeldir. But the problem is my
boundary polygon is not a rectangle; it's an irregular polygo
Good Morning,
I am not asking for a method by how to implement, if possible, the method in R.
I am pretty confident of my statistical approach, I do not want to do it by
hand :-). Although, I do have the formulas to do it by hand. The groups are
measured once. The animal is treated short or lo
The statistical analysis I want to conduct is a 2x2 factorial analysis with a
single control group. There are 3 animals per group.The low n is accepted in my
area of research. My factors are Treatment (No Treatment and Treatment) and
Duration (Acute and 3 weeks). The No Treatment level represent
Dear Carlijn
You might shed some light on what is going on by using
profile.rma.mv
Michael
On 21/04/2015 09:42, Carlijn Wibbelink wrote:
Thank you for your reaction, it worked.
However, I'm wondering if this is the right way to test whether there is
significant variation on one of the two leve
On 04/21/2015 05:00 AM, r-help-requ...@r-project.org wrote:
Dear All,
I am in some difficulty with predicting 'expected time of survival' for each
observation for a glmnet cox family with LASSO.
I have two dataset 5 * 450 (obs * Var) and 8000 * 450 (obs * var), I
considered first one as t
This is correct (for getting likelihood ratio tests). Manually setting a
component to 0 is also the same as just leaving out the corresponding random
effect. So, you could also do:
model2 <- rma.mv(y, v, random = list(~ 1 | y, ~ 1 | ID), data=dat)
model3 <- rma.mv(y, v, random = ~ 1 | y, data=da
Thank you for your reaction, it worked.
However, I'm wondering if this is the right way to test whether there is
significant variation on one of the two levels. The results of the anova tests
do not correspond to the results of the Z-test in metaSEM. (In metaSEM only one
of the variances is sign
Dear all,I am trying to fit a heavy tailed distribution and I have tried
working with the mix function of the mixdist package.It looks like that this
package allows fitting two distributions (or move) of the same family and not
combining different distributions (so mixing a geometric with a nor
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