Hi
I did not use cfa function, the former poster (Catherine)did and I only advised
almost the same as you did.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of yrosseel
> Sent: Monday, October 15, 2012 8:40 AM
On 10/15/2012 08:28 AM, PIKAL Petr wrote:
Hi
-Original Message- From: r-help-boun...@r-project.org
[mailto:r-help-bounces@r- project.org] On Behalf Of Gunsalus,
Catherine Sent: Friday, October 12, 2012 8:52 PM To:
r-help@r-project.org Subject: [R] Error in rowMeans function
Hello, I a
Hi
>
> While I am generating graphs for the above data I am getting four
> graphs in four pages in a pdf with the following programme.
>
> Page1: Only dots (No lines)
> Page2: Lines observed
> Page3: Lines observed
> Page4: Only dots (No lines)
>
> I want to get rid of page 1 and 4 i.e Only d
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Gunsalus, Catherine
> Sent: Friday, October 12, 2012 8:52 PM
> To: r-help@r-project.org
> Subject: [R] Error in rowMeans function
>
> Hello,
> I am trying to create parcels
On Sun, 14 Oct 2012, Ed wrote:
This was an exceptionally helpful answer, I can only thank you again.
I have plenty of avenues ahead where I was worried before I was
getting trapped in a dead end. If all else fails, the idea of using
anova is brilliant. Thank you!
No prob. Just for the record,
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Noah Silverman
> Sent: Sunday, October 14, 2012 7:14 PM
> To: R-help@r-project.org
> Subject: [R] Date Math
>
> Hello,
>
>
> I have a time series object (xts) that I iterat
Hi,
I have two within-subject factors A and B, and covariate CV, and wanna run
an ANCOVA on the dependent variable DV.
my code is:
ancova=aov(DV~factor(A)*factor(B)+CV+Error(factor(subject)/(factor(A)*factor(B))),data=data)
Is that correct?
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hi Michael,
am sorry for the incomplete reply .
csv file data havinmg like this ,
>getSymbols("IBM")
>weekly_data = to.weekly(week_name)
>dataframe=data.frame(weekly_data)
>outputfile_name="F:\\R-programs\\Outputfile.csv"
>write.table(dataframe, file =outputfile_name,sep = ",",col
Hi,
I have 2 factors A and B, and the covariate CV, and wanna run an
within-subject ANCOVA on the dependent variable DV.
Previously my ANOVA was
anova=aov(DV~factor(A)*factor(B)+Error(factor(subject)/(factor(A)*factor(B))),data=data)
how do I modify this code to included a covariate CV to run
On Sun, 14 Oct 2012, Eiko Fried wrote:
Thank you for the detailed answer, that was really helpful. I did some
excessive reading and calculating in the last hours since your reply,
and have a few (hopefully much more informed) follow up questions.
1) In the vignette("countreg", package = "pscl
Hi Jeff Newmiller,
The script given by you is perfect for my requirement.
The rest of the scripts given by other authors are failing to format the
file header which are giving like eachdatacolumn.1 or eachdatacolumn.2 etc,
whereas I need them as eachdatacolumn_firstactualstepnumber,
eachdata
Hi,
I have two within-subject factors A and B, and covariate CV, and wanna run
an ANCOVA on the dependent variable DV.
my code is:
ancova=aov(DV~factor(A)*factor(B)+CV+Error(factor(subject)/(factor(A)*factor(B))),data=data)
Is that correct?
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I just noticed chinese characters are not printed by using this. Some other
characters gets printed on axis as shown below for the sugested solution.
x <- read.csv(textConnection("名称,类,学生
木材,2,2
表,3,4
笔,4,2
垃圾桶,5,6
杯,6,3"), header = TRUE)
rownames(x) <- x[,1]
x <- x[,-1]
barplot(t(x),
i m installing pkg but get error like this
> install.packages("RMySQL")
Installing package(s) into
‘/home/trendwise/R/x86_64-pc-linux-gnu-library/2.13’
(as ‘lib’ is unspecified)
trying URL 'http://ftp.iitm.ac.in/cran/src/contrib/RMySQL_0.9-3.tar.gz'
Content type 'application/x-gzip' length 165363
You can also get the Tool column repeated by tricking reshape() into letting
you use Tool in the v.name and idvar arguments:
> dta1 <- read.csv( text=
+ "Tool,Step_Number,Data1,Data2
+ A,1,0,1
+ A,2,3,2
+ A,3,2,3
+ B,1,3,2
+ B,2,1,2
+ B,3,3,2
+ ")
# Repeat the first column. R will append .1 to the
ingaschwabe gmail.com> writes:
>
> Dear all,
>
> For a project I need to calculate the conditional AIC of a mixed effects
> model.
> Luckily, I found a reference in the R help forum for a function to be used:
>
> CAIC <- function(model) {
>
> sigma <- attr(VarCorr(model), 'sc')
>
I have spent quite some time building a general framework for unevenly-spaced
time series. The methods do not rely on a transformation to equally-spaced
data, but allow to analyze unevenly-spaced time series in their unaltered
form. Please see http://www.eckner.com/research.html for my notes.
In a
HI,
Just a modification of the column names. I noticed that the "Data." column
starts with "2".
Try this:
dat1<-read.table(text="
Tool Step_Number Data1 Data2 Data3
A 1 0 1 2
A 2 3 1 4
A 3 2 1 3
B 1 3 2 1
B 2 1 2 3
B 3 3 2 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat3<-with(dat1,ag
Hello all-
I got some help on this. While it did not fix the problem with nls, one can
get around the problem by creating a data frame with the NAs excluded. As
follows:
goodidx<-which(! is.na(warming$T10cm))
warming2<-warming[goodidx,]
warm.10<-nls(umoles60~alpha*exp(beta*T10cm),start = start,
Le 14/10/2012 00:00, Duncan Murdoch a écrit :
On 12-10-13 3:20 PM, Christophe Genolini wrote:
Hi the list,
I am about to submit an article describing an R package to the Journal of
Statistical Software but I
encounter a strange behavior of LaTeX: the numbering of the figure is correct
(1. 2.
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Pieter Schoonees
> Sent: Friday, October 12, 2012 6:19 PM
> To: Vining, Kelly; r-help@r-project.org
> Subject: Re: [R] average duplicated rows?
>
> You will have to split() t
Dear R People:
Here is the answer
sudo apt-get install proj
sudo apt-get install gdal-bin
sudo apt-get install libxml2-dev
And all is well to install the R packages.
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: e
Dear R People:
I'm trying to install spsurvey and rgeos in R-2.15.1 on Ubuntu 12.04.
However, I am getting the following:
* installing *source* package ‘rgeos’ ...
** package ‘rgeos’ successfully unpacked and MD5 sums checked
checking geos-config usability... ./configure: line 1342: geos-config:
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the po
Thank you everyone for your response. However, for my actual spreadsheet
where the rows are of unequal length, I was getting blanks (e.g., while
combining multiple rows into a single column). Thus, I added just one line
to the final object to avoid getting such blanks.
blankLess <- dat2 [-which(d
This was an exceptionally helpful answer, I can only thank you again.
I have plenty of avenues ahead where I was worried before I was
getting trapped in a dead end. If all else fails, the idea of using
anova is brilliant. Thank you!
Ed
On 14 October 2012 18:36, Achim Zeileis wrote:
> On Sun, 14
just a side note for your 4th question.
for a small sample, clarke test instead of vuong test might be more
appropriate and the calculation is so simple that even excel can
handle it :-)
On Sun, Oct 14, 2012 at 12:00 PM, Eiko Fried wrote:
> I would like to test in R what regression fits my data
Hi Eiko,
This is not a direct response to your question, but I thought you
might find these pages helpful:
On negative binomial:
http://www.ats.ucla.edu/stat/R/dae/nbreg.htm
and zero inflated poisson:
http://www.ats.ucla.edu/stat/R/dae/zipoisson.htm
In general this page lists a variety of dif
Thank you for the detailed answer, that was really helpful.
I did some excessive reading and calculating in the last hours since your
reply, and have a few (hopefully much more informed) follow up questions.
1) In the vignette("countreg", package = "pscl"), LLH, AIC and BIC values
are listed for
On Sun, Oct 14, 2012 at 8:03 AM, siddu479 wrote:
>
> Hi Gabor,
>
> Thanks for your reply..
> *glob2rx()* function works for my requirement to address global
> pattern(using *) of files like we use in Unix shell.
>
> Sidda
>
Also Sys.glob may be useful here.
__
Dear Jeff Newmiller,
I apologize if then I did not clarify what I am looking properly. I used
the combination you suggested in the query but I did not see any of the
resultings adressing what I asked in my title: Rusboost implemented over
AdaBoost.M2. Searching for the words 'rusboost', 'smoteboos
HI,
If you need the "Tool" column repeated,
Try this:
dat1<-read.table(text="
Tool Step_Number Data1 Data2 Data3
A 1 0 1 2
A 2 3 1 4
A 3 2 1 3
B 1 3 2 1
B 2 1 2 3
B 3 3 2 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat3<-with(dat1,aggregate(dat1[,3:5],list(Tool=Tool),function(x) x))
d
Hello,
Try the following.
dta1 <- read.csv( text=
"Tool,Step_Number,Data1,Data2
A,1,0,1
A,2,3,2
A,3,2,3
B,1,3,2
B,2,1,2
B,3,3,2
")
sp <- split(dta1[-2], dta1$Tool)
dta2 <- do.call(rbind, lapply(sp, function(x) as.vector(unlist(t(x)
dta2 <- data.frame(dta2, stringsAsFactors = FALSE)
idx <- r
Jeff,
My understanding is that the lag command will lag an entire time series. That
isn't what I'm looking for.
I just want, for example, today, and 5 entries back.
for exmple:
iter <- '2011-05-18'
observations[iter] # works fine, returns the row at that date.
index(observations[iter[)
First, here is your message as it appears on R-help.
On 10/14/2012 05:00 AM, r-help-requ...@r-project.org wrote:
> I?m trying to set up proportional hazard model that is stratified with
> respect to covariate 1 and has an interaction between covariate 1 and
> another variable, covariate 2. Both va
?embed
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer
The following shows how to use trace() to track and store information
about evaluations of the objective function:
https://stat.ethz.ch/pipermail/r-help/2010-September/252790.html
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-proje
Have you tried other values for trace? Higher values will likely provide more
information (depending on the method - BFGS here) - e.g. use trace = 2 or 3 etc
instead of TRUE (which is similar to using 1).
Optim calls an internal function (implemented in C I presume) so putting a
print statemen
Asking questions and giving answers about R is what this mailing list is about.
It is not really a theory list. If you understand the theory you can search
among the libraries yourself. But when the first search I make brings up what
to me looks like the answer to your question, then you either
Hi,
You can download the package here
http://cran.r-project.org/web/packages/XML/index.html
Regards,
Anindya
On Sun, Oct 14, 2012 at 10:14 PM, balaji sarangarajan
wrote:
>
>
>
> Hello,
>
>
>
> I have installed R version 2.15.1 and I am trying to work with
> rattle package. I have got an error
Jeff,
My understanding is that the lag command will lag an entire time series. That
isn't what I'm looking for.
I just want, for example, today, and 5 entries back.
for exmple:
iter <- '2011-05-18'
observations[iter] # works fine, returns the row at that date.
index(observations[iter[)
Hi Pieter, thank you for your response.
I am indeed using the "optim" function. I had already tried the trace option
under the control argument - however it only provides me with iteration no.
10, 20, 30 etc. I would like to be able to generate the results for every
iteration.
Here is an example
Dear Jeff Newmiller,
Thanks for the suggestion I did not know the website but the problem is
more dry than simple multiclass classification. In fact, there are even few
papers addressing the problem. I was in hope that if not yet available in
CRAN someone from the R community could point out any '
Pardon, *I hope it is not inappropriate to ask such questions here.
Carlos Andrade
http://carlosandrade.co
2012/10/14 Carlos Andrade
> Dear Jeff Newmiller,
>
> Thanks for the suggestion I did not know the website but the problem is
> more dry than simple multiclass classification. In fact, th
Hello,
I have installed R version 2.15.1 and I am trying to work with
rattle package. I have got an error
stating as below;
Error in loadTooltips() : could not find function
"xmlTreeParse"
In addition: Warning messages:
1: package XML is not available (for R version 2.15.1)
2: In
Hi,
If you don't want the column "Tool" to be repeated, try this:
dat1<-read.table(text="
Tool Step_Number Data1 Data2
A 1 0 1
A 2 3 1
A 3 2 1
B 1 3 2
B 2 1 2
B 3 3 2
",sep="",header=TRUE,stringsAsFactors=FALSE)
with(dat1,aggregate(cbind(Data1,Data2),list(Tool=Tool),function(x) x))
To
On Sun, 14 Oct 2012, Ed wrote:
First up, thanks hugely for your response. I've been beating my head
against this!
On 14 October 2012 16:51, Achim Zeileis wrote:
I'm not sure what you mean by "integral vector". If you want to apply the
approach to hundreds of thousands of observations, I gues
May I suggest you use a more appropriate tool for your search:
RSiteSearch("multiclass classification")
On Sun, 14 Oct 2012, Carlos Andrade wrote:
Hi,
I have been searching everywhere for an implementation of those algorithms,
but I have only observed them in Matlab and on the literature.
I
I am not sure exactly what errors you are getting, but here are some comments.
It is good practice to initialize a vector to the total length that you will
require instead of changing the dimension each time you substitute a value.
I.e. use something like
deter <- rep(NA, length(true.dose) )
Hi,
On Sun, Oct 14, 2012 at 11:23 AM, Shant Ch wrote:
> n=10
> x1<-(1:n)/n
> y1<-rnorm(n,x1^2,1)
> m=20
> x2<-(1:m)/m
>
> The value of y2 will be rnorm (m, x2^2,1) if none of the elements of x2 is
> same as x1, but for every same elements in x1 and x2, the value of y2 will be
> same as y1. I kn
There are a few ways. The xts package has a "lag" function. So does
"zoo". Pay careful attention to the conventions used for specifying
relative time in these various packages. You can also infill your missing
data to create a regularly-spaced time series. There is no shortage of
web informa
Hello all,
I would like to transform the paths of MST {ape} in SpatialLines objects.
But I don't understand the mechanical procedure in SpatialLines-class of
'sp' package.
Please, somebody can help me?
Thanks in advanced.
Cleber N.Borges
### my (half) trial
library(sp);
library(ape) # for
Optimization is not something "R" does... this is something implemented by
specific functions that may be in the base install or in add-on packages,
but you have not indicated how you are currently approaching your problem.
If you are using the optim() function, then I refer you to the fine
ma
If you are writing the loop yourself, you can use print() statements in the
loop. If you are using something like optim(), have a look at the documentation
and use the trace option under the control argument. Otherwise, put a print
statement in the function you are passing to optim. If this fail
Not clear to me what your successive rows indicate in your output.
However, you may get something similar to what you want using the reshape2
library:
library(reshape2)
dta1 <- read.csv( text=
"Tool,Step_Number,Data1,Data2
A,1,0,1
A,2,3,2
A,3,2,3
B,1,3,2
B,2,1,2
B,3,3,2
")
# put into long fo
Hi Marco,
a small example will be helpful to get better to the point.
But I suggest you either to address this question to
R geo mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-geo
and eventually take a look to GeoXp library.
Cheers
Anna
Anna Freni Sterrantino
Department of Statisti
Hello,
I have a time series object (xts) that I iterate over in a loop. Works fine.
My challenge is that I want to be able to reference other entries in the series
by math. i.e. For today's observation, what were the last 5 observations? If
indexed numerically, it is trivial, but I can fi
First up, thanks hugely for your response. I've been beating my head
against this!
On 14 October 2012 16:51, Achim Zeileis wrote:
> I'm not sure what you mean by "integral vector". If you want to apply the
> approach to hundreds of thousands of observations, I gues that these are
> categorical (m
Hello all,
I have a .csv file like below.
Tool,Step_Number,Data1,Data2... etc up to 100 columns.
A,1,0,1
A,2,3,1
A,3,2,1
.
.
B,1,3,2
B,2,1,2
B,3,3,2
.
.
.. so on upto 50 rows
where the column "*Tool*" has distinct steps in second column
"*Step_Number*",but both have same entries in Step_Numbe
1000x Thanks for all of your helps! All of it workd perfectly! :)
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__
R-help@r-pr
Hello,
I have trouble with saving and showing candlestick graph.
I want to draw daily 1-minute candle chart per each day and save it
repeatedly.
The data I saved and is used in my code is as the following format:
-
Index,Open,High,Low,Close # hea
Hi,
I have been searching everywhere for an implementation of those algorithms,
but I have only observed them in Matlab and on the literature.
I noticed a package called 'ada' in CRAN but it is not for multi class. I
would be happy with just Adaboost.m2, Smoteboost over adaboost.m2 or any
other c
When I tried to run r console in eclipse,got this error:
--
Please make sure that R package 'rj' (1.1 or compatible) is
installed...
So I tried to intall it in R console like this:
---
n=10
x1<-(1:n)/n
y1<-rnorm(n,x1^2,1)
m=20
x2<-(1:m)/m
The value of y2 will be rnorm (m, x2^2,1) if none of the elements of x2 is same
as x1, but for every same elements in x1 and x2, the value of y2 will be same
as y1. I know the following is correct, but for large vectors, this won't work
effi
Hi Gabor,
Thanks for your reply..
*glob2rx()* function works for my requirement to address global
pattern(using *) of files like we use in Unix shell.
Sidda
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Dear R-user,
I'm having some difficulty with working PFIM 3.2, a package for
implementing population PK/PD in R. I wish to evaluate the determinant of
Fisher information matrix each time with successive dose from a pre defined
sequence of doses and want to store those values in a vector. It's
impor
Dear All,
thanks to Berend, my question posted yesturday was solved succesfully
here: http://r.789695.n4.nabble.com/hep-on-arithmetic-covariance-conversion-to-log-covariance-td4646068.html .
I posted the question with the assumption of using the results with
rlnorm.rplus() from compositions. U
Hi,
I am new to R, and am working on some optimization problems - I was
wondering if there was a way that I could show all the iterations in R -i.e.
showing for each iteration, what the iteration is, how much the function is,
what the norm of the gradient would be...
Any help is greatly appreciated
On Sun, 14 Oct 2012, Eiko Fried wrote:
I would like to test in R what regression fits my data best. My dependent
variable is a count, and has a lot of zeros.
And I would need some help to determine what model and family to use
(poisson or quasipoisson, or zero-inflated poisson regression), and
I would like to test in R what regression fits my data best. My dependent
variable is a count, and has a lot of zeros.
And I would need some help to determine what model and family to use
(poisson or quasipoisson, or zero-inflated poisson regression), and how to
test the assumptions.
1) Poisson R
Ed:
I'm experiencing some problems using the party package (specifically
mob) for prediction. I have a real scalar y I want to predict from a
real valued vector x and an integral vector z. mob seemed the ideal
choice from the documentation.
I'm not sure what you mean by "integral vector". If
On Fri, Oct 12, 2012 at 1:26 AM, Jay Rice wrote:
> Below I have written out some simplified data from my dataset. My goal is
> to interpolate Price based on timestamp. Therefore the closer a Price is in
> time to another price, the more like that price it will be. I want the
> interpolations for e
Hello,
Try the pattern "A_B.*_C.*\\.csv".
Hope this helps,
Rui Barradas
Em 14-10-2012 08:13, siddu479 escreveu:
Hi Experts,
This might be silly question that I am asking, but no way as I am new to
R.
I want to list the files in a directory using regular expression like
A_B*_C*.csv etc
Dear all,
I have a dataset representing several geographical points (locations), each
one having a specific value after performing a PCA.
Now I'm trying to understand how to separately represent on a geographic
map (of Europe) the pattern of PC1,2, ecc.with colors (e.g.heatmap)
I have to add that t
Dear R Helpers, I need to find the root of following equationy=x^2+3*x-1 by
substitution a random number from less to more in the certain segment.I had
tried using this codef <- function (x) x^2+x-12str(xmin <- uniroot(f, c(2,
4), tol = 0.0001)) but $ f.root: num -7.85e-07not 0, and so I'm not
On Sun, Oct 14, 2012 at 3:13 AM, siddu479 wrote:
> Hi Experts,
>
> This might be silly question that I am asking, but no way as I am new to
> R.
> I want to list the files in a directory using regular expression like
> A_B*_C*.csv etc.
> How to make this possible in R ?
> I tried like this list
Hi Experts,
This might be silly question that I am asking, but no way as I am new to
R.
I want to list the files in a directory using regular expression like
A_B*_C*.csv etc.
How to make this possible in R ?
I tried like this list.files(dir=".", pattern="A_B*_C*.csv") but this gives
no output
On Sat, 13 Oct 2012, Rui Barradas wrote:
Hello,
Try package tseries, function white.test().
That's not White's test for heteroskedasticity but White's network test
for nonlinearity, i.e., a different test.
Z
Hope this helps,
Rui Barradas
Em 13-10-2012 21:40, Afrae Hassouni escreveu:
Hel
On Sat, 13 Oct 2012, Afrae Hassouni wrote:
Hello,
Is there a way to perform a White test (testing heteroscedasticity)
under R?
White's test is a special case of the Breusch-Pagan test using a
particular choice of auxiliary regressors. The Breusch-Pagan test is
available in bptest() from "l
options(error = recover)
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
hi all:
Following is simple example,
for(i in 1:1000){
a<-function(i){
#some function that has an error
}
}
What I want to know is the way to find the error point in for-loop, What "i"
makes the error.
is there any way to solve it instead of debugging and finding an error
manually?
Thanks,
Hello, I'm trying to take a PPS systematic sample of a data set, and I've
gotten stuck at the last point. I have selection numbers, and just need to
pick out the units whose cumulative size are the least upper bounds of
these numbers
This is as close as I've gotten:
selected<-array(dim=c(97,4))
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