On Tue, 7 Dec 2010, Santosh Srinivas wrote:
Hello Group,
I have an R-function that works fine for most part but sometime runs into a
long loop! (I'm lazy and short on time to debug right now so want to do
something easy)
For my purpose, it is ok to make few errors is there a way I can put
See also R help thread 'Wait for keystroke or timeout':
http://www.mail-archive.com/r-help@r-project.org/msg97742.html
/H
On Mon, Dec 6, 2010 at 9:11 PM, Michael Bedward
wrote:
> Below is a toy function with one way of doing it. There are bound to
> be better ways :)
>
> function(niter = 10, ti
http://projects.uabgrid.uab.edu/r-group/wiki/CommandLineProcessing
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__
R-help@r-p
Hi Matt,
I'm not certain if this will work with PuTTy, but in general (and on
Konsole/other terminals I've run R in) you can set it with the "width"
argument of options() see:
?options
I believe it is 80 by default, so something like:
options(width = 160)
might be about right.
Cheers,
Josh
Worked perfect, thank you very much!
Matt
On Mon, Dec 6, 2010 at 11:53 PM, Joshua Wiley wrote:
> Hi Matt,
>
> I'm not certain if this will work with PuTTy, but in general (and on
> Konsole/other terminals I've run R in) you can set it with the "width"
> argument of options() see:
>
> ?options
>
Thank you for the help. Much appreciated.
On Mon, Dec 6, 2010 at 9:12 PM, Hadley Wickham wrote:
> It's easiest to see what's going on if you use eval.quoted directly:
>
> eval.quoted(.(cyl), mtcars)
> eval.quoted(.("cyl"), mtcars)
> eval.quoted(.(as.name("cyl")), mtcars)
>
> But you shouldn't ne
Hi,
I'm using R interactively from a PuTTy command line, and when I print a
dataframe (default, but just entering the name of the data frame), the
display is using only half of the available horizontal width in which to
print columns. How to I tell R how many columns it can/should use for
printin
Below is a toy function with one way of doing it. There are bound to
be better ways :)
function(niter = 10, time.out = 3) {
pretend.task <- function() {
Sys.sleep(0.5)
}
start <- proc.time()
for (iter in 1:niter) {
pretend.task()
cur <- proc.time() - start
if (c
Sorry, solved by me RTFM... The directions pertaining to what I needed to do
are found here: http://cran.r-project.org/bin/linux/ubuntu/
Sorry to bother y'all,
Matt
On Mon, Dec 6, 2010 at 10:57 PM, Matthew Pettis wrote:
> Hi,
>
> I'd like to use ggplot2, which I think works best on some R releas
Hi,
I'd like to use ggplot2, which I think works best on some R release >=v2.11
or so. However, with Ubuntu Hardy Heron 8.04 LTS, 'aptitude' packages show
me only a version for 2.6. Can I install more recent version on this Ubuntu
version, and if so, what is the most painless way to do this (is
Given a non-singular fit, the contrast function in the rms package will allow
you to request multi-dimensional contrasts some of which are redundant.
These singular contrasts are automatically ignored. One use for this is to
test for differences in longitudinal trends between two of three treatm
I built my own "outer lapply " as attached code. Is there any R
build-in function can do the same jobs ?
Thanks
oapply<-function(X,Y,FUN, ..., MoreArgs = NULL, SIMPLIFY = TRUE,
USE.NAMES = TRUE){
x=rep(X,each=length(X))
y=rep(Y,times=length(Y))
mapply(FUN,x,y, MoreArgs =
Joonas - as_trix85 hotmail.com> writes:
>
>
> I'm trying to estimate the parameters for GARCH(1,1) process.
> Here's my code:
>
> loglikelihood <-function(theta) {
>
> h=((r[1]-theta[1])^2)
> p=0
>
> for (t in 2:length(r)) {
> h=c(h,theta[2]+theta[3]*((r[t-1]-theta[1])^2)+theta[4]*h[t-1])
>
Hi:
You might find this link useful:
https://github.com/hadley/ggplot2/wiki/%2Bopts%28%29-List
...and apropos your choice of plot,
http://emdbolker.wikidot.com/blog:dynamite
Cheers,
Dennis
On Mon, Dec 6, 2010 at 2:46 PM, Henning Wildhagen wrote:
> Dear R-users,
>
> i encountered some problems
I'm trying to estimate the parameters for GARCH(1,1) process.
Here's my code:
loglikelihood <-function(theta) {
h=((r[1]-theta[1])^2)
p=0
for (t in 2:length(r)) {
h=c(h,theta[2]+theta[3]*((r[t-1]-theta[1])^2)+theta[4]*h[t-1])
p=c(p,dnorm(r[t],theta[1],sqrt(h[t]),log=TRUE))
}
-sum(p)
}
The
Hi:
This is called a 'rolling sum', for which there is a very convenient
function called rollapply() in the zoo package. You first need to convert v
to a zoo object (in this case, an indexed vector):
library(zoo)
> rollapply(zoo(v), 2, FUN = sum)
1 2 3 4 5 6 7 8 9
9 14 9 2 4 12 14 11
It's easiest to see what's going on if you use eval.quoted directly:
eval.quoted(.(cyl), mtcars)
eval.quoted(.("cyl"), mtcars)
eval.quoted(.(as.name("cyl")), mtcars)
But you shouldn't need to do any syntactic hackery because the default
method automatically parses the string for you:
eval.quoted
Hello Group,
I have an R-function that works fine for most part but sometime runs into a
long loop! (I'm lazy and short on time to debug right now so want to do
something easy)
For my purpose, it is ok to make few errors is there a way I can put a
timeout on a function and the r-process neede
So i realized putting this
raw.untr <- as.data.frame(read.csv(file='...jsp?y=2008&m=09&d=30',
header = TRUE, as.is = TRUE))
reads the file prefectly, so i was wondering how can i create a set of
dataframes each with observations for one day (say for 3 years with 250 days
of data each year =
Hi Hadley:
I was trying to use ddply using the format . (var1) for splitting.
I thought . ( as.name(grp) ) would do the same thing. But it does not. I was
just trying to know my mistake. I am sorry if it is a basic question.
Thank you and others for your reply.
Best Regards,
S.
On Mon, Dec 6,
In case anyone even remembers that thread, there was a little problem w/
some of the solutions provided because here and there are unary rather
than binary operators or functions. So, I took two of the responses and
fixed them up:
dc<-function(op='+',...) do.call(op,list(...))
or
dcc<-funct
Good evening, folks:
I am trying to run a bootstrap on a CoxPH regression. I have set up a 1,000 x
1,000 matrix of "cases" and am running 1,000 analyses of 1,000 cases each. The
loop keeps breaking because of an error, and I have more or less isolated the
problem to a call of the coxph functi
Hi Benjamin,
If you just had abs(x_i) < c_i, it will reduce to linear inequalities, but
your constraint cannot be reduced to that.
You might try "alabama" or "Rsolnp" packages.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Ge
Dear R-users,
i encountered some problems when trying to adjust the line width of the
axes and stripes in a plot created with ggplot2.
I use the "barley" dataset of the lattice package to illustrate my problem:
library(ggplot2)
library(lattice)
barley[["SD"]] <- 5
limits <- aes(ymax=barley$yie
Dear all,
I get stuck when i try to export the data into SPSS format/file using
"write.foreign()"
Do you know how to do it exactly?
What i have done is
1) First i type the following code in R:
df<-data.frame(id,year,res1)
names(df)<-c("idcode","year","resarrvl")
write.foreign(df,"z:/daten/res.da
On Mon, Dec 6, 2010 at 3:58 AM, Sunny Srivastava
wrote:
> Dear R-Helpers:
>
> I am using trying to use *ddply* to extract min and max of a particular
> column in a data.frame. I am using two different forms of the function:
>
>
> ## var_name_to_split is a string -- something like "var1" which is t
With the result of the 'cut' you can get a direct index for the values:
> x <- runif(20)
> x
[1] 0.26550866 0.37212390 0.57285336 0.90820779 0.20168193 0.89838968
0.94467527 0.66079779 0.62911404
[10] 0.06178627 0.20597457 0.17655675 0.68702285 0.38410372 0.76984142
0.49769924 0.71761851 0.991906
Hi,
Can you give a reproducible example of what you did? My intuition is
that you could do it using par(), but I am not sure what package
metabin() is from (certainly none of the ones that load by default),
and I have even less idea how you created a forest plot (there are
many ways in R). If yo
Hi All,
I have conducted a meta analysis using the metabin function. I want to plot
5 subgroups on the same forest plot. I have managed to do this using the
byvar argument but when i plot the forest plot in R graphics I am unable to
view the very top and very bottom of the image. It is as though
Hi
I think the problem is that the viewer you are using for the PS file
cannot find the ComputerModern fonts that are referenced in the PS file.
If you use something like ghostscript to process the file you can see
this happening ...
[pmur...@stat18 Temp]$ gs plot_example.ps
ESP Ghostscript
Hi,
I'm trying to fit a logit model to a set of data that was collected under a
split plot scheme. The structure of the data is
Whole plot factors: Watering Frequency (2 levels: Hi/Lo) and Fertilizer type (3
factors A/B/C)
Subplot factor: slope type (2 factors up/down)
Response: Proportion of
hello!
I'm trying to make a very simple calculation using R that would called
from bash with two parameters, the names of the input and output files.
the first one is just an 4 by 100 numerical table- the function finds
the index of the maximal elements in each columns and writes it into the
outpu
On Mon, 6 Dec 2010, Roberto Patuelli wrote:
Dear Achim,
Thanks a lot for the superquick reply!
Somehow with your suggestion I can get around the problem, but of course I
run into other problems, such as this:
waldtest(inv.log.leva.base, inv.log.leva, test = "Chisq", vcov = sandwich)
Error
Dear Achim,
Thanks a lot for the superquick reply!
Somehow with your suggestion I can get around the problem, but of course I
run into other problems, such as this:
waldtest(inv.log.leva.base, inv.log.leva, test = "Chisq", vcov = sandwich)
Error in bread. %*% meat. : non-conformable argument
On Mon, 6 Dec 2010, Roberto Patuelli wrote:
Dear All,
I'm trying to use waldtest to test poolability (parameter stability) between
two logistic regressions. Because I need to use robust standard errors (using
sandwich), I cannot use anova. anova has no problems running the test, but
waldtest
No there is no open software to read ESRI "file geodatabases" (the .gdb ones).
If it's not .gdb but .mdb ("personal geodatabase", built on Microsoft
Access database) then you can read it with generic tools (such as
RODBC package) or with an appropriate build of GDAL as part of the
rgdal package.
Hi Jef,
I would extract the information about the smoothers and plot them using
standard plotting routines. After
Mtest <- gam(outcome ~ s(age, by=as.numeric(gender==0)) +
s(age,by=as.numeric(gender==1))+factor(Gender))
run something similar to:
smoothers <- predict(s,type='terms')
and you
Baoqiang Cao gmail.com> writes:
> I have a least square fitting problem with linear inequality
> constraints. pcls seems capable of solving it so I tried it,
> unfortunately, it is stuck with the following error:
> > M <- list()
> > M$y = Dmat[,1]
> > M$X = Cmat
> > M$Ain = as.matrix(Amat)
> > M$
Dear All,
I'm trying to use waldtest to test poolability (parameter stability) between
two logistic regressions. Because I need to use robust standard errors
(using sandwich), I cannot use anova. anova has no problems running the
test, but waldtest does, indipendently of specifying vcov or not
Hi,
It depends on the function used to calculate it---each has their own
way of extracting particular values. Here's an example for t.test:
x <- rnorm(10, 100, .1); y <- x - 100
t.test(x, y)$p.value
You could find this yourself by looking at the documentation for t.test()
?t.test
The heading
In this case I only get three values back. My intention was to get every
value between 10% and 90% i.e. let's say i have a matrix
1 2 3 4
5 6 7 8
9 10 11 12
i only want the values above 10% and under 90%.
Maybe i should try it with a loop.
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View this message in context:
http://r.789695.n4.
Dear,
I have the data in the following form:
>head(matrices_m)
Location Variable Value Week
1 Africa Africa21 4 weeks
2 Asia Africa 0 4 weeks
3 Canada Africa17 4 weeks
4China Africa29 4 weeks
5 Europe AfricaNA 4 weeks
6Japan Africa68 4 week
Hello-
Does anyone know if R can read in ESRI file geodatabase feature
classes, directly from the file geodatabase? I currently have to
export the data out of Arc, to a csv file, then read the csv file. I'd
like to bypass this step and read directly from the file geodatabase.
Thanks for your help
I was wondering if there is a way to get an exact p-value at times where R
gives me just a range . for example
t.test(x,y)
p-value < 2.2e-16
thanks,
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Sent from the R help mailing l
>> if(FALSE) {
>> stuff your don't want executed
>> }
>>
>
> Switching a block of code off/on with editing a single character
> may be done using 0/1 instead of FALSE/TRUE.
Or even F/T
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer
Hi,
On Fri, Dec 3, 2010 at 10:23 AM, manuel.martin
wrote:
> Dear all,
> I am currently looking for a book about support vector machines for
> regression and classification and am a bit lost since they are plenty of
> books dealing with this subject. I am not totally new to the field and
> would l
Btw, forgot to mention I am using the standard Matrix package and I am
running version 2.10.1 of R.
On Mon, Dec 6, 2010 at 11:04 AM, scott white wrote:
> I have a very sparse square matrix which is < 20K rows & columns and I am
> trying to row standardize the matrix for the rows that have non-mis
thanks a lot, duncan for that super-fast reply and help!
i could install rgl via synaptic and afterwards ca in R directly. and it works!
:-)
have a nice evening, kat
Von: Duncan Murdoch [murdoch.dun...@gmail.com]
Gesendet: Montag, 6. Dezember 2010 19:55
An
I have a very sparse square matrix which is < 20K rows & columns and I am
trying to row standardize the matrix for the rows that have non-missing
value as follows:
row_sums <- rowSums(M,na.rm=TRUE)
nonzero_idxs <- which(row_sums>0)
nonzero_M <- M[nonzero_idxs,]/row_sums[nonzero_idxs]
M[nonzero_idx
On 06/12/2010 1:45 PM, Manderscheid Katharina wrote:
hi,
for some unknown reason i cannot install the package ca in R running in a
ubuntu mint system. i keep getting the following error message:
You need the development package for OpenGL or MesaGL to install rgl. I
don't know Ubuntu so I do
hi,
for some unknown reason i cannot install the package ca in R running in a
ubuntu mint system. i keep getting the following error message:
configure: error: missing required header GL/gl.h
ERROR: configuration failed for package ‘rgl’
* removing ‘/home/kat/R/i486-pc-linux-gnu-library/2.12/rgl’
Ista,
Thanks for this method, it works! Unfortunately 'R' still spits
everything out to std out, but the variable as you describe below grabs only
the final output variable, similar to how functions or scripts work
internally within 'R'. Thanks!
Kjetil,
Thanks also for your input, but
Have you tried setting singular.ok=TRUE in the call to lm? This will start
with the full set of contrasts, but only fit those that it is able to.
Otherwise you can set specific contrasts or subsets using the C (note case) or
contrasts functions.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Dat
You can use quantile and cut:
> x <- runif(1000)
> x.cut <- cut(x, c(-Inf, quantile(x, prob=c(.1, .9)), Inf),
> labels=c('lo','mid','hi'))
> table(x.cut)
x.cut
lo mid hi
100 800 100
>
On Mon, Dec 6, 2010 at 12:53 PM, Marty_H wrote:
>
> I have got a matrix with 100 values.
> Is there a easy w
Hi Steven,
thank you again so much for your help and for the transformation of the data.
In fact I knew that there are these empty lines in the data. I couldn't take
them off with the program I have. I tried to solve this problems with using the
"levels".
You are right, the statistical tests a
Benjamin B. googlemail.com> writes:
> Hello list reader,
>
> I am trying to form some constraints for an optimization I am working on.
> I think I have understand the use of the constraints in matrix form. I use
> them like:
>
> [...]
>
> Now I would like to formulate a constraint like
>
> Su
peter dalgaard gmail.com> writes:
>
> On Dec 6, 2010, at 15:15 , Jonathan P Daily wrote:
>
> > Correct me if I'm wrong, but isn't the minimal x value in your example
> > the same regardless of what positive coefficient you apply to x? If that
> > is the case, you would expect the same min(x) for
Andreas Jensen gmail.com> writes:
> Hello.
>
> I'm trying to solve a quadratic programming problem of the form min
> ||Hx - y||^2 s.t. x >= 0 and x <= t using solve.QP in the quadprog
> package but I'm having problems with Dmat not being positive definite,
> which is kinda okay since I expect it
I have got a matrix with 100 values.
Is there a easy way to cut the upper 10% (every value which is above 90%)
and the lower 10% of my matrix-values away and create a new matrix with all
the rest.
Thx for your help
Marty
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hi: i don't know that book but there's another pretty new book on svms by
lutz hamel. It's pretty gentle but very nice, especially if you have ZERO
background on svms.
there's also another one titled "an introduction to svms" by cristianini and
shawne-taylor that's also quite good but much more de
Let us look at the objective function:
f(x) = x^2 + Si * x, where Si is the sum of the i-th column. This function
has a stationary point at x = -Si/2, and the second derivative is 2, so it
is a minimum.
Now, the column sums of your data matrix are all positive. So, your minimum
has to be negati
My favorite book on SVM is Learning with Kernels by Scholkopf and Smola. You
might also want to consider a relevance vector machine, which is a more recent
development. RVM is Bayesian-based and usually produces a sparser
representation than a SVM. Check out Mike Tipping's web site at
http://www
Hi Marianne,
You have to be very careful with subsetting when doing something like that -
that's where you went wrong with your original construct. This version works:
> v<-c(3,6,8,1,1,3,9,5,6,3)
> a <- numeric(length(v)-1)
> for (i in 2:length(v)) {a[i-1] <- v[i-1] + v[i]}
> a
[1] 9 14 9 2 4
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Nordlund, Dan (DSHS/RDA)
> Sent: Monday, December 06, 2010 9:21 AM
> To: r-help@r-project.org
> Subject: Re: [R] How can I refer to actual (n) and previous (n-1)
> elements in a v
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Marianne Stephan
> Sent: Monday, December 06, 2010 9:13 AM
> To: r-help@r-project.org
> Subject: [R] How can I refer to actual (n) and previous (n-1) elements
> in a vector?
>
>
Hello,
How can I apply a function on a vector that refers to actual (n) and previous
elements in the vector (e.g. n-1)?
For example:
I would like to calculate the sum of (n-1) + n for each element of a vector and
get a vector as a result.
Besides others I tried this:
v<-c(3,6,8,1,1,3,9,
Hi,
I have a least square fitting problem with linear inequality
constraints. pcls seems capable of solving it so I tried it,
unfortunately, it is stuck with the following error:
> M <- list()
> M$y = Dmat[,1]
> M$X = Cmat
> M$Ain = as.matrix(Amat)
> M$bin = rep(0, dim(Amat)[1])
> M$p=qr.solve(as.
On Mon, Dec 6, 2010 at 5:57 AM, Gabor Grothendieck
wrote:
> On Mon, Dec 6, 2010 at 6:52 AM, Ted Zeng (曾振兴) wrote:
>>
>> Dear all,
>>
>> I am using the as.xts function to transfer a data frame to the xts
>>
>> The following is the code and result:
>>
>> a<-read.csv("price.csv")
>> a$Date<-as.POSIX
When I try to read data from excel, I save it in csv-format and read the files
via:
fileData <-read.csv(fkFileName, header = fkHasHeader, stringsAsFactors=FALSE,
sep=fkSep)
where fkSep is depending on your localization of excel (I think).
For some columns that don’t get the correct data format
Hello list reader,
I am trying to form some constraints for an optimization I am working on.
I think I have understand the use of the constraints in matrix form. I use
them like:
constr_mat<- -diag(2)
constr_vec<- rep(-0.05,2)
constr_mat<- rbind(constr_mat, diag(2))
constr_vec<- c(constr_vec, re
I suspect that R is reading the data just fine, but is not displaying it as you
want to see it. Read the R Windows FAQ 3.3.
"Dan Winetsky" wrote:
>To whom it may concern,
>
>I have a database (an SPSS .sav file) with some vectors containing
>strings
>of words in Cyrillic. My console, however,
I'd like to make a less than full rank design using dummy variables
for factors. Here is some example data:
when <- data.frame(time = c("afternoon", "night", "afternoon",
"morning", "morning", "morning",
"morning", "afternoon", "afternoon"),
Hi List,
I'm using npRmpi to run some density equality tests and place the
output into a matrix. I've put together some crude functions for the
purpose, but I'm receiving the following error when npdeneqtest()
reached the bootstrap;
FATAL ERROR: Memory allocation failure (type DBL_VECTOR). Progra
Dear All,
I am having trouble getting my data into R as i need it! I am used to using
read.delim("") to open .txt files to do work on. The function i am using
requires a matrix like the one below.
My data is from excel and then saved as a txt file. I have tried the usual
read.delim("") approa
Hi,
I use Oracle sql developer to connect to a server. I would like to connect
to the server directly from R by using odbcConnect("DB name") function. I am
trying to make first the connection below Control Pannel->Datasource (ODBC).
Which software should I select here? Which one is suitable for O
I suppose I should have been more clear. I saw that her interval did not
include the actual minimum, but I was asking if (and if, why) she was
expecting the minimum x value to be different for each run. If the y value
were returned the same on each run that would be puzzling.
As for the returne
I have the following problem.
After having created an xts importing data from a csv with:
data <- read.csv(mib.csv, header = TRUE, dec = ".", sep="\t")
dates <- as.POSIXct(strptime(data[,1],format="%m/%d/%Y"))
mib <- xts(data[,c(2:6)],order.by=dates))
I work out weekly log returns:
p <- apply.we
I am using the function apply.weekly() from the xts package and I
realized that for the program the week starts Tuesday.
So for instance, with a series like this:
Close
2006-07-03 3.00
2006-07-04 36738.00
2006-07-05 36207.00
2006-07-06 36534.00
2006-07-07 36434.00
2006-07-10 36588.00
On Dec 6, 2010, at 15:15 , Jonathan P Daily wrote:
> Correct me if I'm wrong, but isn't the minimal x value in your example the
> same regardless of what positive coefficient you apply to x? If that is
> the case, you would expect the same min(x) for each iteration.
>
> i.e. in the interval [0
Correct me if I'm wrong, but isn't the minimal x value in your example the
same regardless of what positive coefficient you apply to x? If that is
the case, you would expect the same min(x) for each iteration.
i.e. in the interval [0,1] the minimum x value of x^2 + x is the same as
x^2 + 10
Thanks Chris Campbell,
I didn't though about that.
Cheers,
Maarten
On Mon, 2010-12-06 at 10:08 +, Chris Campbell wrote:
> data$subjectID <- paste(data$groups, data$subjects) # create a
> character
> label
>
> xyplot(responses~time|subjectID, groups = groups, data = data,
> aspect="xy")
Hi,
I usually use optimize function for ML Estimation. Now I´ve got a data frame
with many sets, but I can´t save estimates each time I run the code for each
data set (I´m using a for loop with my loglikelihood function and works ok but
when I apply another for loop to:
optimize(my.loglikeliho
Hi!
I'm a new programmer in R
Last week i wrote many function in C and i used R CMD SHLIB to make dll.
Since friday i have this error:
cygwin warning:
MS-DOS style path detected: C:\PROGRA~1\R\R-211~1.1/bin/SHLIB.sh
Preferred POSIX equivalent is:
/cygdrive/c/PROGRA~1/R/R-211~1.1/bin/SHLIB.s
I am also finding the link between TINN - R (2.3.7.0) and R (2.12.0 2010
- 10 - 15) to be problematic.
source(.trPaths[5], echo=TRUE, max.deparse.length=150)
Error in source(.trPaths[5], echo = TRUE, max.deparse.length = 150) :
object '.trPaths' not found
Steve Friedman Ph. D.
Ecologist /
Hi
I get this error with Rserve:
eval failedeval failed, request status: R parser: syntax erroreval failed,
request status: R parser: input
incompleteorg.rosuda.REngine.Rserve.RserveException: eval failed
at org.rosuda.REngine.Rserve.RConnection.eval(RConnection.java:233)
at se.
Ye
Thank you so much, finally i have something to look at. Just got to
interpret the bloomin' thing now!
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OK, try...
model3=gam(Symptoms~as.factor(txerad)+as.factor(maritalStatus),family=binomial,data=data2)
or
model3=gam(data2$Symptoms~as.factor(data2$txerad)+as.factor(data2$maritalSt
atus),family=binomial)
... both types of construction work with the current mgcv:gam (and with glm).
Your constructi
Ok, i just tried this:
library(mgcv)
model3=gam(data2$Symptoms~as.factor(data2$txerad)+as.factor(data2$maritalStatus),family=binomial,data=data2)
And im still getting this error!
Error in dl[[i]] : subscript out of bounds
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On Mon, Dec 6, 2010 at 6:52 AM, Ted Zeng (曾振兴) wrote:
>
> Dear all,
>
> I am using the as.xts function to transfer a data frame to the xts
>
> The following is the code and result:
>
> a<-read.csv("price.csv")
> a$Date<-as.POSIXct(a$Date)
>
> str(a)
> 'data.frame': 15637 obs. of 2 variables:
>
Probably the problem is with trying to smooth maritalStatus, which is a factor
variable. Smooths are generally functions of metric variables (i.e. variables
that take numerical values, where the ordering is meaningful). You can have
random effect smooths and markov random field smooths in more r
Dear all,
I am using the as.xts function to transfer a data frame to the xts
The following is the code and result:
a<-read.csv("price.csv")
a$Date<-as.POSIXct(a$Date)
str(a)
'data.frame': 15637 obs. of 2 variables:
$ Date : POSIXct, format: "2010-01-04 09:45:01" "2010-01-04 09:45:02"
"2
On 05/12/2010 10:13 PM, pythonomics wrote:
So I am working on an economic model and I need to change the parameters of
the runif statement as time goes on.
Ex.
X<-runif(1:50,0,5)
X<-runif(51:100,100,150)
X<-runif(100:T, 1,2)
Not sure how to go about entering this in to R properly.
It's not
I have a question regarding understanding the output from a nested linnear
mixed effect model.
My model looks like the following
lme(Poaceae~Site+Ditch_block+Tree_cut+as.factor(Dist_start)+(as.factor(Dist_start)*Tree_cut)+((as.factor(Dist_ditch))/Ditch_block),random=~1|Mainplot/Transect/Obsplot)
I
I have a question regarding understanding the output from a nested linnear
mixed effect model.
My model looks like the following
lme(Poaceae~Site+Ditch_block+Tree_cut+as.factor(Dist_start)+(as.factor(Dist_start)*Tree_cut)+((as.factor(Dist_ditch))/Ditch_block),random=~1|Mainplot/Transect/Obsplot)
Thank you for your reply!
It says:
This is mgcv 1.5-5 . For overview type `help("mgcv-package")'.
Its version: R 2.9.2
And summary(data2) gives:
id age bmi siblingsgender
Min. : 1.0 Min. :32.19 Min. :16.41 Min. :0.000 Fem
On Mon, Dec 6, 2010 at 11:30 AM, Peter Ehlers wrote:
> If you have a *tab*-delimited file, then why are your using
> read.csv??
>
> Try this:
> 1. read your data with read.table() or read.delim()
>
If you're very new to R, try Rcmdr. Data > Import > Text file.
Liviu
_
Hi,
I typically install new versions of R on windows using the downloadable
executable file rather than the full tar.
I need to now document the success of the installation in addition to my
preferred procedure of running an old dataset against the new build.
I found quickly that this is all
Here is another approach to try:
> require(data.table)
> var <- "g10"
> df <- data.table(df)
> str(df)
Classes ‘data.table’ and 'data.frame': 6 obs. of 5 variables:
$ g10: int 1 1 1 10 10 10
$ l1 : num 0.41 0.607 0.64 -1.478 -1.482 ...
$ d1 : num 0.918 0.959 0.773 0.474 0.591 ...
$ l13: n
On 2010-12-06 01:58, Sunny Srivastava wrote:
Dear R-Helpers:
I am using trying to use *ddply* to extract min and max of a particular
column in a data.frame. I am using two different forms of the function:
## var_name_to_split is a string -- something like "var1" which is the name
of a column i
Try excluding the first column.
cor(gse20437[, 2:4])
chintan85 wrote:
>
>
> Tab delimited file looks like this
>
> Id v1 v2v3
> df 56 9045
> gh 87 9878
> ty 897867
>
> I used this code
>
>
> [code]
>
> gse20437 <- read.csv("C:/Users//Desktop/data/GSE20437_matrix
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