Dear R-users,
I need some assistance.
I am running some interactive variables for categorical variables.
I have dgen(2 levels converted to dummy variables)Â and dtoe(4-levels also
converted to dummy variables). So I have worked with them in two ways:
i created a variable X1 = dgen*dtoeÂ
hello,
i already include the error in blue color word.
i hope it can help you to understand my question.
if not burden you, please give me a guide how to correct the error or maybe you
can correct the coding cause error.
thank you.
> #lda.r
> #
> #Author:Amsha Nahid, Jairus Bowne, G
Hi,
I have generated a skyline plot of a tree in newick format using "ape".
How can I plot the error bars for this graph?
I only have the the tree data.
un<-"8.1:0, 20.1:0):0, 6.1:3):123, (35.1:0, (22.1:0, (43.1:1,
29.1:0):0):0):4, 25.1:6):0, ((42.1:0, 21.1:0):3, (39.1:0, 2.1:
Hi Everyone,
When generating help files for a R package, can I specify the order of the R
functions? Now I can see that the functions are sorted according to their
names. However, for my case, there are two functions are closely related,
but they are separated by some other functions. I think it w
On Mon, 20 Jun 2011, rivercode wrote:
Hi,
When I am editing a command using default R console in Linux, sometimes it
is going into "vi mode"...not too sure how/why this happening. It then
requires me to use vi commands to edit the line, which is very frustrating
when I just want to use the del
On Fri, Jun 10, 2011 at 10:27 PM, Ethan Brown wrote:
> Hi Kieran,
>
> I'm not very familiar with lattice, but here's a workaround that works for
> me. Basically, I just created a new data.frame column that was a factor
> (combo$zf), and forced its levels to be what you're looking for here.
This w
On Wed, Jun 15, 2011 at 6:32 PM, Justin McBride wrote:
> Dennis,
>
> Thanks for your suggestion, but that is not exactly what I was after.
> I was trying to get the legend in the margin on the top right of the
> page and not in the plot frame. Is there a way to do this?
One option is:
xyplot(Yi
cumsum, but you have to add back the information lost in the diff operation by
generating an extra zero at the beginning or end of the sequence and adding the
appropriate offset to the resulting vector.
---
Jeff Newmiller The
Hi,
I have a simple problem where I have two or more predictor variables that
range from 0 to 1 and binary response variable (0 or 1). In the two
variable case, the model to fit with maximum likelihood would simply be:
P(Y=1) = (B1*X1 + B2*X2)/(B1+B2)
or if least squares is to be minimized
On Mon, Jun 20, 2011 at 8:55 PM, Erin Hodgess wrote:
> Too funny!
>
> how about subset?
Sure, that is one option. Each of the following will also work. The
ones wrapped with c() can easily omit more than one at a time.
mtcars[, -which(names(mtcars) == "drat")]
mtcars[, names(mtcars) != "drat"]
Hi users.
I'm new user in R.
I'm workiing with Time series and I would like to know how can I do to undo
the command DIFF(X), for exemple:
If I have the model: m=arima(X, order=c(0,1,1),
seasonal=list(order=c(0,0,1))) (note that have d=1 one difference), to find,
in the same scale, the original num
Hi Erin,
As you've seen from the responses so far there are many ways to do it.
I like to do
xm1[setdiff(names(xm1), c("x1", "x2", ...))]
where xm1 is the data.frame, and "x1", "x2", ... are the columns you
wish to exclude. Just another option.
Best,
Ista
On Mon, Jun 20, 2011 at 11:52 PM, Joshua
Hi Ungku, your error is in your call to lda.default; the object that
you want to plot, lda_result, is thus not created and cannot be
plotted.
The error you're getting with LDA doesn't look to be a coding error,
but a statistical one. I don't know anything about LDA, but simply
googling "variablest
Hi Erin,
See inline.
On Mon, Jun 20, 2011 at 8:45 PM, Erin Hodgess wrote:
> Dear R People:
>
> I have a data frame, xm1, which has 12 rows and 4 columns.
>
> If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as
> it should be.
Okay, so you know how to use the column number to om
Hi Erin,
One option woild be subset(), especially the "select" parameter.
HTH,
Jorge
On Mon, Jun 20, 2011 at 11:45 PM, Erin Hodgess <> wrote:
> Dear R People:
>
> I have a data frame, xm1, which has 12 rows and 4 columns.
>
> If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as
Dear R People:
I have a data frame, xm1, which has 12 rows and 4 columns.
If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as
it should be.
Now, is there a way to use the names of the columns to omit them, please?
Thanks so much in advance!
Sincerely,
Erin
--
Erin Hodgess
A
I a using plink on a large SNP dataset with a .map and .ped file.
I want to get some sort of file say a list of all the SNPs that plink is
saying that I have. ANyideas on how to do this?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
Are there any utilities/packages for showing various performance
metrics of a regression model on some labeled test data? Basic stuff
I can easily write like RMSE, R-squared, etc., but maybe with some
extra utilities for visualization, or reporting the distribution of
prediction confidence/varianc
Hi,
I have generated a skyline plot of a tree in newick format using "ape".
How can I plot the error bars for this graph?
I only have the the tree data.
un<-"8.1:0, 20.1:0):0, 6.1:3):123, (35.1:0, (22.1:0, (43.1:1,
29.1:0):0):0):4, 25.1:6):0, ((42.1:0, 21.1:0):3, (39.1:0, 2.1:0):
Hi Louis,
It seems to me that the easiest way to accomplish what you want is to
just define an extractAIC method for objects of class "mlogit". I am
not familiar with multinomial logistic models, so this may not be
correct (I am especially uncertain about the degrees of freedom I
used), but it sh
Look at the documentation for ?cut, specifically the dig.lab argument:
cut(xm1, 4, dig.lab = 5)
Cheers,
Josh
On Mon, Jun 20, 2011 at 5:46 PM, Erin Hodgess wrote:
> Hello again R People:
>
> I have the following:
>
>> xm1[,1]
> [1] 13053 13068 13068 13053 14853 14853 14850 14850 13053 13053 13
Hello again R People:
I have the following:
> xm1[,1]
[1] 13053 13068 13068 13053 14853 14853 14850 14850 13053 13053 13068 13068
> cut(xm1[,1],4)
[1] (1.31e+04,1.35e+04] (1.31e+04,1.35e+04] (1.31e+04,1.35e+04]
[4] (1.31e+04,1.35e+04] (1.44e+04,1.49e+04] (1.44e+04,1.49e+04]
[7] (1.44e+04,1.49
The way to guarantee a specific number of panels in the histogram, say n, is to
specify n+1 breaks which cover the range of the data. As far as I know this is
the only way.
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Be
Dear R People:
Is there a way to "guarantee" that breaks=n will give you exactly n
breaks, please?
I'm fairly certain that the answer is "no", but thought I'd check.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downt
Hi all,
I have a rather peculiar dataset that I'm not sure how to model
properly. This is data from an instrument that measures the size of
particles but instead of giving a continuous value, it generates a
"histogram" of the counts for a particular bin size. So the data looks
like this:
for (thisOne in structs){
print(get(thisOne)$colA)
}
you need to 'get' (retrieve) the object first.
On Mon, Jun 20, 2011 at 5:18 PM, Noah Silverman wrote:
> Hi,
>
> I have several data structures (xts structures). I then have a list of the
> names of those objects.
>
> I'd like to access t
The advice is always NOT to use Microsoft Word to edit an R file. That stuff
is poisonous. Microsoft word, typical of all Microsoft software, does not do
what you tell it to do but helpfully does what it thinks you meant to ask it to
do but were too dumb to do so.
Even notepad, gawdelpus, wou
On Tue, 21 Jun 2011, james198877 wrote:
> http://r.789695.n4.nabble.com/file/n3612530/PSC.r PSC.r
>
> Hi all,
>
> I just wrote a program in R by editing it in Microsoft Word and then
> pasting into the text editor of R. The above is the file.
>
> And below is what the console complains Why d
http://r.789695.n4.nabble.com/file/n3612530/PSC.r PSC.r
Hi all,
I just wrote a program in R by editing it in Microsoft Word and then pasting
into the text editor of R. The above is the file.
And below is what the console complains Why doesn't it recognise 'r'??
I have to mention that at le
Hi,
I have several data structures (xts structures). I then have a list of the
names of those objects.
I'd like to access the object by name.
For example:
foo1 <- as.xts()
foo2 <- as.xts(...)
foo3 <- as.xts(...)
structs <- c("foo1", "foo2", "foo3")
for (thisOne in structs){
prin
Hello all,
I was in direct email contact with Ravi, who was very kind to offer his
assistance. Inbetween correspondence with Ravi, I arrived at a solution.
In case others have the same issue, I am sharing the solution. Thanks Ravi
for your willingness to help.
Regards,
Tim
**
Hi Dennis and all friends,
After upgrading to R 2.11.1, I managed to work this out perfectly. I also added
group information here, since I wanted to color code the profiles according to
the groups they were assigned to.
library("reshape")
library("ggplot2")
sampledata <- matrix(c(1.002, 1.76,
Hi:
(a) What Brian said...
(b) Here's one way to generate a list of model objects from which you
can extract the pieces you may want.
# Generate a fairly minimal, reproducible data set
set.seed(345) # makes results below reproducible
dd <- data.frame(X = rnorm(100), Y = rnorm(100),
Hi Ungku, it's really difficult for us to take a huge block of code and
understand where an error happened. There's several things that can help us
help you:
1) First and foremost, what is the error message or undesired behavior
you're experiencing?
2) Second, please pare down the code to the plac
On 21/06/11 06:52, Runlong Tang wrote:
Hi everyone,
I am writing help files for some R functions in an R package. I am wondering
if we can use the Latex mathematical formula when editing those *.Rd files
and then obtain a pdf manual with beautiful mathematical formula. I tried it
with the defa
On Jun 20, 2011, at 5:16 PM, Alina Sheyman wrote:
I haven't used R in a couple of years, and now am trying something as
simple as importing a csv file and am running into problems right
away.
*
mydata <- read.csv (Wordata1.csv, sep="")
Error in read.table(file = file, header = header, sep =
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
> Sent: Monday, June 20, 2011 2:07 PM
> To: Deepfish
> Cc: r-help@r-project.org
> Subject: Re: [R] General Plotting Commands Help
>
> ... much elided ...
>
> Most
On Jun 20, 2011, at 3:54 PM, Costis Ghionnis wrote:
Hallo everyone! I have a problem about creating a matrix...
Suppose we have a vector y<-c(1,1,1,3,2)
and a zero matrix, m ,with nrows=length(y) and ncol=4.
The matrix would look like this:
0 0 0 0
0 0 0 0
that was it, thank you!!!
On Mon, Jun 20, 2011 at 5:23 PM, Sarah Goslee wrote:
> R is looking for an R object named Wordata1.csv that contains your file
> name.
>
> Instead you want:
> mydata <- read.csv ("Wordata1.csv", sep="")
>
> Sarah
>
> On Mon, Jun 20, 2011 at 5:16 PM, Alina Sheyman wrote:
R is looking for an R object named Wordata1.csv that contains your file name.
Instead you want:
mydata <- read.csv ("Wordata1.csv", sep="")
Sarah
On Mon, Jun 20, 2011 at 5:16 PM, Alina Sheyman wrote:
> I haven't used R in a couple of years, and now am trying something as
> simple as importing
You probably left off the quotes for the file name
mydata <- read.csv ("Wordata1.csv", sep="")
it is looking for an R variable named Wordata1.csv which contains the name
of the file.
Since you are giving it the name of the file, it must be in a
character-valued constant.
On Mon, Jun 20, 2011 at 5
How about:
y <- c(1,1,1,3,2)
m <- matrix(0, nrow=length(y), ncol=4)
m[y==1, ] <- matrix(1:4, nrow=sum(y == 1), ncol=4, byrow=TRUE)
or, depending on your actual problem
y <- c(1,1,1,3,2)
m <- matrix(0, nrow=length(y), ncol=4)
m[y == 1,] <- col(m[y == 1,])
Sarah
On Mon, Jun 20, 2011 at 3:54 PM,
I haven't used R in a couple of years, and now am trying something as
simple as importing a csv file and am running into problems right away.
*
mydata <- read.csv (Wordata1.csv, sep="")
Error in read.table(file = file, header = header, sep = sep, quote = quote,
:
object 'Wordata1.csv' not found
Hi, I have a dataset (see attached) with 2 variables "Y" is binary, "x" is a
continuous variable. I want to calculate area under the curve (AUC) for the ROC
curve, but I got different AUC values using ROC() from Epi package vs.
rcorr.cens() from rms package:
test<-read.table("test.txt",sep='\t'
On Jun 20, 2011, at 4:52 PM, Deepfish wrote:
I am rather new to R and struggling with the syntax. I am trying to
do simple
plots and I can't get anything but errors. Specifically, I would
like to
know how to plot 3D scatterplots and using polar coordinates.
So far I've tried installing the
--- Begin Message ---
Ivan, try this
d1 <- matrix(rnorm(10),50,50)
d2 <- matrix(rnorm(10),50,50)
regs <- lapply(1:ncol(d1), function(i) lm(d1[,i] ~ d2[,i])
HTH, Filipe
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Em nome de ivan
Enviada
Hallo everyone! I have a problem about creating a matrix...
Suppose we have a vector y<-c(1,1,1,3,2)
and a zero matrix, m ,with nrows=length(y) and ncol=4.
The matrix would look like this:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0
I am trying to perform a backwards stepwise variable selection with an mlogit
model. The usual functions, step(), drop1(), and dropterm() do not work for
mlogit models.
Update() works but I am only able to use it manually, i.e. I have to type in
each variable I wish to remove by hand on a se
Hi everyone,
I am writing help files for some R functions in an R package. I am wondering
if we can use the Latex mathematical formula when editing those *.Rd files
and then obtain a pdf manual with beautiful mathematical formula. I tried it
with the default R CMD check command but could not ge
I am rather new to R and struggling with the syntax. I am trying to do simple
plots and I can't get anything but errors. Specifically, I would like to
know how to plot 3D scatterplots and using polar coordinates.
So far I've tried installing the scatterplot3D package and have copied and
pasted exa
Hi,
When I am editing a command using default R console in Linux, sometimes it
is going into "vi mode"...not too sure how/why this happening. It then
requires me to use vi commands to edit the line, which is very frustrating
when I just want to use the delete key instead of "x" to delete a charac
On 6/20/2011 12:23 PM, ivan wrote:
Hi,
I have two datasets, x and y. Simplified x and y denote:
X
Y
A B C A B C . . . . . . . . . . . . . . . . . .
I want to implement all possible models such as lm(X$A~Y$A), lm(X$B~Y$B),
lm(X$C~Y$C)... I have tried the following:
fun<- function(x,y){
Your suggestion would have it return after fitting
the first model, which is not what the OP wants.
The basic problem is that
models <- list(lm(...))
replaces the old value of models with a new length-1
list. You want to add the new fitted model to the
list of fitted models. E.g.,
models
You probably want to use the merge() function.
imaginary example for data frames named "KS" and "US," where the common
identifier variable is named "ID"
merged.data<-merge(KS,US,by.x="ID",by.y="ID",all.x=F,all.y=F)
Note that this will retain only observations for which there is a common ID
in b
To be more accurate and helpful, try this:
fun<- function(x,y){
for(i in 1:length(colnames(x))){
for(j in 1:length(colnames(y))){
if(colnames(x)[i]==colnames(y)[j]){
models=list(lm(ts(x[i])~ts(y[j])))
return(models)
Hi,
can you put "return(models)" within the inner braces and report what it
does. That might do the trick, since it should return the 'models' for every
combination of i and j.
HTH,
Daniel
hazzard wrote:
>
> Hi,
>
> I have two datasets, x and y. Simplified x and y denote:
>
> X
>
> Y
>
>
On 6/20/2011 11:25 AM, Vickie S wrote:
Hi Dennis,
It looks like something is wrong about configuration of ggplot dependency with
plyr.Since i saw some threads about this particular error message.
I tried several times by installing different versions of plyr but it did not
work.
sa
Hi,
I have two datasets, x and y. Simplified x and y denote:
X
Y
A B C A B C . . . . . . . . . . . . . . . . . .
I want to implement all possible models such as lm(X$A~Y$A), lm(X$B~Y$B),
lm(X$C~Y$C)... I have tried the following:
fun<- function(x,y){
for(i in 1:length(colnames
On Mon, Jun 20, 2011 at 1:31 PM, albeam wrote:
> Hi everyone,
>
> Thank you for the help, I apologize for not "providing commented, minimal,
> self-contained, reproducible code." I was looking for some pointers about
> how to do this in general, but it would have been helpful for me to post a
> sp
On Jun 20, 2011, at 2:38 PM, Idris Raja wrote:
I have R library's installed in several places on my windows machine.
They are in:
1) C:\Program Files\R\R-2.12.2\library
2) C:\Users\MyUser\Documents\R\win-library\2.12
I can access a library from either location with RStudio, but I
can't acc
Dear All,
I'm trying to use dynlm to fit a time series.
I have 3 seasonal terms. Here is an example of the problem.
This is my time variable, hourly data:
timeSeries <- seq(as.POSIXct("2011-01-01 00:00:00"), as.POSIXct("2011-12-31
23:00:00"), by="hour")
My response
is:
y <- rnorm(length(
Hi everyone,
Thank you for the help, I apologize for not "providing commented, minimal,
self-contained, reproducible code." I was looking for some pointers about
how to do this in general, but it would have been helpful for me to post a
specific example. Anyway, after the feedback this is the solu
I have R library's installed in several places on my windows machine.
They are in:
1) C:\Program Files\R\R-2.12.2\library
2) C:\Users\MyUser\Documents\R\win-library\2.12
I can access a library from either location with RStudio, but I can't access
a library from 2) when using R from the command l
As a follow up to my own question, the problem was with my version of vim. I
needed vim 7.3, was using 7.2.
I updated and it works as expected. Just make sure your version of python
matches the version of pywin you install if on windows.
On Fri, Jun 17, 2011 at 6:51 PM, Idris Raja wrote:
> I am
On Mon, Jun 20, 2011 at 9:22 PM, David Winsemius wrote:
>
> On Jun 20, 2011, at 10:49 AM, Bert Gunter wrote:
>
>> Hi all:
>>
>> Thanks, David. This is a good example of knowledgeable "R forensic
>> investigation." I leave it to Frank whether it meets his criteria.
>>
>> However, I would argue that
Hi Dennis,
It looks like something is wrong about configuration of ggplot dependency with
plyr.Since i saw some threads about this particular error message.
I tried several times by installing different versions of plyr but it did not
work.
sampledata= matrix(c(1.002, 1.76, 0.67, 0.99
Thanks for the clarifications.
On Mon, Jun 20, 2011 at 12:40 PM, Bert Gunter wrote:
> Sorry, I was unclear. The comment after the second should be:
>
> z[ ,a] # the column of z whose name is the value of the object a
>
> -- Bert
>
> On Mon, Jun 20, 2011 at 9:38 AM, Bert Gunter wrote:
> > Ben:
Hi all, I need to know two quick things regarding RPut function from RExcel.
1/ How to assign colnames and rownames in the command, i.e., something like
'=RPut("example"; A10:F50; rownames=TRUE; colnames=TRUE)'.
2/ When I use RPut and I have NAs in Excel selection, the entire matrix is
coerced
Interesting!
I get nice convergence in both 32 and 64 bit systems on 2.13.0. I agree the
older versions
are a bit of a distraction. The inconsistent behaviour on current R is a
concern.
Maybe Philip, Uwe, and I (and others who might be interested) should take this
off line
and see what is going
Sorry, I was unclear. The comment after the second should be:
z[ ,a] # the column of z whose name is the value of the object a
-- Bert
On Mon, Jun 20, 2011 at 9:38 AM, Bert Gunter wrote:
> Ben:
>
> 1. One doesn't ask questions like this. Syntax is syntax.
>
> 2. This has nothing to do with p
The quotes around 'Major.Gleason' and 'Minor.Gleason' are required for
accessing data frame columns by name. You could alternately refer to
the columns by number if you're sure you know which column is which:
> output = paste(df[ ,1], df[ ,2], sep = '+')
It's just a requirement for accessing thin
Ben:
1. One doesn't ask questions like this. Syntax is syntax.
2. This has nothing to do with paste; it's the syntax of "[" ,
subscripting/extraction
3. But it does make sense:
a <- "b"
z <- data.frame(a=1:3, b=4:6)
z[ ,"a"] # the "a" column of z
z[ ,a] # the column of z with the value of th
Thanks! Very glad you pointed me to the paste function, it looks very
helpful.
I have a quick follow-up after reading through the online tutorial on the
"paste" function:
Why do we need quotation marks around "Major Gleason" and "Minor Gleason"
in: output = paste(df [,'Major.Gleason'], df[ ,'Mi
Actually, I 'm using "mgcv" package.
But I learnt basic linear regression in my school, it is really hard for me
to handle these problem!
by the way, i need to work harder!
--
View this message in context:
http://r.789695.n4.nabble.com/About-GAM-in-R-Need-YOUR-HELP-tp3610522p3611796.html
Sent f
On Jun 20, 2011, at 11:47 AM, Luke Miller wrote:
If we assume that your data are in a data frame (which doesn't allow
spaces in column names, hence the periods in the call below):
df = data.frame(Major.Gleason = c(4,5,2,3), Minor.Gleason =
c(3,2,4,3))
You can paste together the contents of
On Jun 20, 2011, at 10:49 AM, Bert Gunter wrote:
Hi all:
Thanks, David. This is a good example of knowledgeable "R forensic
investigation." I leave it to Frank whether it meets his criteria.
However, I would argue that that this is bad practice and quite
unwise. In general, these details are
If we assume that your data are in a data frame (which doesn't allow
spaces in column names, hence the periods in the call below):
>df = data.frame(Major.Gleason = c(4,5,2,3), Minor.Gleason = c(3,2,4,3))
You can paste together the contents of the two columns with a plus
sign in between using the
Thanks very much Bert and David. I was looking for an approach that would
use objects that are active while the panel function was being executed.
Perhaps which.packet is the best way to go, e.g. if using y ~ x | a*b, fetch
levels(b)[which.packet()[2]] from the environment in which b is defined.
Hi --
I had a pretty quick R question since unfortunately I have not been able to
find an answer on Google. It shouldn't take much more than a minute to
answer.
I'm trying to add up the major gleason grade and minor gleason grade for an
analysis of patients with prostate cancer. One column has
Dear list,
I am using the plot.coinertia function from the ade4 package. Can
someone tell me if it is possible to change the label size in the upper,
right plot? My RV value is so high that the arrows become shorter than
the labels.
Thank you for your help.
Oana Tomescu
--
Dipl.-Ing. Oana T
Hi all:
Thanks, David. This is a good example of knowledgeable "R forensic
investigation." I leave it to Frank whether it meets his criteria.
However, I would argue that that this is bad practice and quite
unwise. In general, these details are implementation dependent and
could change. Yes, they
This does not seem like a core R issue... at best it would be relevant on the R
Debian special interest group mailing list
(https://stat.ethz.ch/mailman/listinfo/r-sig-debian). However, using your
system software installation tools to make sure you have installed relevant
font packages for X se
On Mon, Jun 20, 2011 at 9:46 AM, Gabor Grothendieck
wrote:
> On Mon, Jun 20, 2011 at 9:08 AM, albeam wrote:
>> Please allow me to clarify my original question. What I really need to be
>> able to do it is to take arbitrary functions and evaluate them for arbitrary
>> parameter values. I'm doing t
Hi:
Typically a profile plot is a plot over time for each of n subjects
(aka a 'spaghetti plot'). Here's a revision of your code to yield a
profile plot in ggplot2, although it's not clear for what the grouping
variable is meant.
sampledata= matrix(c(1.002, 1.76, 0.67, 0.99, 1.38, 1.0, 0.5, 0.78,
Hi everyone,
I've searched the internet and lots of stats books high and low for this one,
but nothing seems to be quite what I want. I've got continuous data on four
different state activities recorded in seconds, however each continuous session
is not equally long, so the data are best expres
Hi,
Thank you for the replies.
Yes I would say it does resemble a randomized block design in that each
person completed 4 out of a possible 16 versions/conditions of the vignette.
There was approximately 20 responses in each condition.
The commands I have been using are;
modela <-lmer(DV ~ IV*I
Frank Harrell wrote:
>
> I know about the current.row, current.column, and panel.number functions
> that are useful within panel functions written for lattice. Are there
> easy ways to obtain the names of the conditioning variables (those
> appearing after |) and their values for the current pan
Chris,
This doesn't cover all possible cases, but does work for your example.
It should be enough for you to tweak for your actual data.
diffsum <- function(x) {
# first identify the decreasing values
# and the difference between increasing values
xdif <- x[2:length(x)] -
On Mon, 2011-06-20 at 13:38 +0200, Dominik P.H. Kalisch wrote:
> Hi,
>
> I would like to cluster a dataset with the ward algorithm.
I'm assuming that this refers to the agglomerative partitioning method
[1]. That is, the number of clusters is selected according to the data
partition that is seque
On Jun 20, 2011, at 15:08 , albeam wrote:
> Please allow me to clarify my original question. What I really need to be
> able to do it is to take arbitrary functions and evaluate them for arbitrary
> parameter values. I'm doing the optimization myself, so I need to be able to
> take a user's funct
On Mon, Jun 20, 2011 at 9:08 AM, albeam wrote:
> Please allow me to clarify my original question. What I really need to be
> able to do it is to take arbitrary functions and evaluate them for arbitrary
> parameter values. I'm doing the optimization myself, so I need to be able to
> take a user's f
Dear Uwe,
> If you think the relevant section is not sufficient, it's
> your turn to provide a short and precise suggestion for a
> change, preferable a diff against the R-devel sources. That
> way you may be able to convince an R Core member to work on it.
I would be more than happy to rearra
Hi:
I was curious to see how to do this. I generated two versions of the
same function - one for 10-fold predictions when the number of
observations is an exact multiple of 10, returning a matrix, and
another that lets the user define the number of folds and works with
lists. The function also ret
See the help page!
type: character string, one of ‘"Xlib"’ or ‘"quartz"’ (some Mac OS
X builds) or ‘"cairo"’. The latter will only be available if
the system was compiled with support for cairo - otherwise
‘"Xlib"’ will be used. The default is set by
I wish for help for the following issue with GNU R.
I have on my Ubuntu-Linux-10.04-LTS built R-2.13.0 from source these days.
After I launched the R interactive console, the plot() etc. graphics functions
reported the following errors:
-- could not find any X11 fonts
-- locale not supported
I wish for help for the following issue with GNU R.
I have on my Ubuntu-Linux-10.04-LTS built R-2.13.0 from source these days.
After I launched the R interactive console, the plot() etc. graphics functions
reported the following errors:
-- could not find any X11 fonts
-- locale not supported
Please allow me to clarify my original question. What I really need to be
able to do it is to take arbitrary functions and evaluate them for arbitrary
parameter values. I'm doing the optimization myself, so I need to be able to
take a user's function and evaluate them at the current parameter value
> Date: Mon, 20 Jun 2011 11:25:54 +0200
> From: lig...@statistik.tu-dortmund.de
> To: p.e.b...@dunelm.org.uk
> CC: r-help@r-project.org; pb...@astro.uni-bonn.de
> Subject: Re: [R] different results from nls in 2.10.1 and 2.11.1
>
> Since one is a 32-bit and the other one a 64-bit, and therefore t
I wish for help for the following issue with GNU R.
I have on my Ubuntu-Linux-10.04-LTS built R-2.13.0 from source these days.
After I launched the R interactive console, the plot() etc. graphics functions
reported the following errors:
-- could not find any X11 fonts
-- locale not supported
Hi thanks for the help.
Here is the error message i got. I could not figure out what goes wrong here.
==
require(ggplots2)
sampledata=data.frame(c(1.002, 1.76, 0.67, 0.99), c(1.38,1.0,0.5, 0.78),
c(1.003,0.57,0.99, 0.58),c(0.76,1.23, 1.45, .78), c(1.43, 1.34, 1.0, 0.9))
rownames(sampledat
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