I have attached to this message the first 20 lines of the output for
dput(amigo). Unfortunately, i can't send it the overall output.
Thank you.
On Wed, May 16, 2012 at 6:10 PM, David Winsemius [via R] <
ml-node+s789695n4630253...@n4.nabble.com> wrote:
>
> On May 16, 2012, at 1:33 AM, ana24maria w
Hi,
I'm using a command in bioconductor that seems to require a package called
hu6800cdf. I've installed this properly but I still get the same error:
Could not find array definition file ' hu6800cdf.qcdef '. Simpleaffy does
not know the QC parameters for this array type.
See the package vignett
Thanks David for prompt reply. I agree with you. However, I still fail to get
the survfit function to work with newdata. In my previous example I changed the
column names of testX matrix and I still fail.
> colnames(testX)<-names(coxph.model$coefficients)
> sfit<- survfit(coxph.model,newdata=d
Le 17/05/12 03:37, Min Wang a écrit :
Dear Sir or Madam,
I have a question like this. I want to extract the following data from the
square bracket:
1 bandband[0.86]
2 bandband[0.93]
3 noband noban
On May 16, 2012, at 5:08 PM, Damjan Krstajic wrote:
Dear all,
I am confused with the behaviour of survfit with newdata option.
Yes. It has the same behavior as any other newdata/predict from
regression. You need to supply a dataframe with the same names as in
the original formula. Doesn
Dear Sir or Madam,
I have a question like this. I want to extract the following data from the
square bracket:
1bandband[0.86]
2bandband[0.93]
3noband noband [0.95]
4noband
Hello,
I needed this once.
upper.diag <- function(x, byrow=FALSE){
m <- sqrt(1 + 8*length(x))
if(abs(m - floor(m)) < .Machine$double.eps^0.5)
m <- (m - 1)/2
else{
warning("length of 'x' is not a triangular number.")
m <- flo
On Wed, May 16, 2012 at 9:02 PM, Duncan Temple Lang
wrote:
> Hi James.
>
> Yes, you need to identify the namespace in the query, e.g.
>
> getNodeSet(doc, "//x:entry", c(x = "http://www.w3.org/2005/Atom";))
>
> This yeilds 40 matching nodes.
>
> (getNodeSet() is more convenient to use when you do
Dominik,
See this line:
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 30.37 30.37 30.37 30.37 30.37 30.37
The variance of the predictions is zero. caret uses the formula for
R^2 by calculating the correlation between the observed data and the
predictions which uses sd(pred) which
Hi James.
Yes, you need to identify the namespace in the query, e.g.
getNodeSet(doc, "//x:entry", c(x = "http://www.w3.org/2005/Atom";))
This yeilds 40 matching nodes.
(getNodeSet() is more convenient to use when you don't specify a function
to apply to the nodes. Also, you don't need xmlRoo
Hi,
I'm trying to use the XML package to read an RSS feed. To get
started, I was trying to use this post as an example:
http://www.r-bloggers.com/how-to-build-a-dataset-in-r-using-an-rss-feed-or-web-page/
I can replicate the beginning section of the post, but when I try to
use another RSS feed
Thought there might be some Redditors lurking on these mailing lists. I
created a sub-reddit for R (and by extension Bioconductor) discussions,
links, etc.
http://www.reddit.com/r/Rsoftware/
This will be the first and only shameless plug.
-Robert
Robert M. Flight, Ph.D.
University of Louisville
Do leave the posts for anyone else who might google the same question. (I don't
think you really could delete the post anyways, perhaps only on one mirror)
You could probably use some combination or rev() and t() to fill by row, but I
haven't thought through the geometry all the way yet.
Micha
On Tue, 15 May 2012, ilai wrote:
Apologies in advance if I misinterpret " R console insists that I retype
..." but actually makes more sense (than sourcing a script) to use the
group argument (see the last example in ?qqmath) as in 4 groups in each of
30 panels, or allow.multiple=T, outer=T if y
The Matrix package provides good support for many special sorts of
matrices, but here it looks like you probably don't need that
additional machinery for such small case:
makeUpper <- function(vec, diag = FALSE){
n <- (-1 + sqrt(1 + 8*length(vec)))/2
stopifnot(isTRUE(all.equal(n, as.intege
Hi useRs,
I apologize if Ive missed some documentation somewhere, but I cant seem
to find anything related to this question
For a ensemble/data-mining
problem, Im trying to train a neural network on my data set and have it
output predictions (or coefficients) after varying numbers of epoch
Hello,
'sample' is a really bad name for a variable, it's already taken, it's an R
function.
sample>1 && sample<5 # '&&' is not vectorized, it's '&' you want.
# Without 'ifelse'
raw.saliva.data$max.cort[index] <- raw.saliva.data$cortisol[index & sample
> 1 & sample < 5]
Negate this last conj
Hi,
Is there any package that deals with triangular matrices?
Say ways of inputting an upper (lower) triangular matrix?
Or convert a vector of length 6 to an upper (lower) triangular matrix (by
row/column)?
Thanks!
-
##
PhD candidate in Statistics
Big R Fan
Big LEGO Fan
Hi,
I get the following error while installing a package. Can someone please
help?
install.packages("memisc")
Warning in install.packages :
argument 'lib' is missing: using 'C:/Users/ravi/Documents/R/R-2.15.0'
Warning in install.packages :
downloaded length 8255 != reported length 200
Error in
Hmmm, I can't reproduce, but I'm not really sure why that would
happen... is there any way you can test this in a --vanilla R session?
(That's the UNIX-y way to start a totally clean session; not sure
exactly how to achieve that on Windows)
Does this happen if you just run
example(TukeHSD)
direc
I recommend that you troubleshoot your own problem using the str function...
for example, str( coeftest(lmodT_WBHO)). The error message is not a "code"...
it is perfectly readable English, and it is telling you that the result of
calling coeftest is not a list with parts that can be pulled out u
It seems like your problem is that R can't find your variable "sample"
and is instead finding its own sample() function which can't be
compared to an integer and is giving your problem.
It seems likely that sample is part of raw.saliva.data? If that's the
case, change sample --> raw.saliva.data$sa
Hello,
I used "correlogram" from "spatial" package to determine correlation scale
for my data but just looking with bare eye it seems that the correlation
scale varies over the domain.
Can someone suggest what would the best way to handle that problem?
Thanks,
Mark
[[alternative HTML versi
There are a couple of options.
First if you want the mean to equal 7, then that means the sum must
equal 21 and therefore you can let optim only play with 2 of the
variables, then set the 3rd to be 21-s1-s2.
If you want the mean to be greater than 7 then just put in a test, if
the mean is less th
Lies Durnez itg.be> writes:
> I want to build a model in R based on animal collection data, that look like
the following
>
> NrVillage DistrictSiteSurvey Species Count
> 1 AX A F Dry B 0
> 2 AY A V Wet A 5
> 3 BX
Dear all,
I am confused with the behaviour of survfit with newdata option.
I am using the latest version R-2-15-0. In the simple example below I am
building a coxph model on 90 patients and trying to predict 10 patients.
Unfortunately the survival curve at the end is for 90 patients. Could som
Hello,
I apologize in advance for not providing sample data, I'm a very new to R
and can't easily generate appropriate sample data quickly. I'm hoping
someone can offer advice without it.
This code below works and does what I want it to do, which is for a given
row in my dataframe, where the vari
On May 16, 2012, at 12:37 , Andras Farkas wrote:
> Dear R Expert
>
> allow me to ask a quick qestion: I have a mean value of 6 and a SD of 3
> describing my distribution. I would like to "convert" this distribution into
> a log normal distribution that would best describe it when resimulated
Duncan,
Thanks for the advice.
It turns out that the web pages are pretty well behaved.
I ended up using
readHTMLTable
str_select
grep
gsub
readLines
When I have time I am going to convert my code to use the html parser and the
more robust getNodeSet method that you mention below.
Thanks fo
On May 16, 2012, at 12:27 PM, aramos wrote:
Thanks, I've already done that!!
But the illustration for how you get the statistics is in the code.
Describe what you want: number of samples, two versus single sided,
two sample versus comparing to theory, which table columns should be
used.
Check out this site:
http://www.gummy-stuff.org/Yahoo-data.htm
It shows how to download a .csv file with the data you might want.
Here is an example URL:
http://finance.yahoo.com/d/quotes.csv?s=XOM+BBDb.TO+JNJ+MSFT&f=snd1l1yrr2
The "r2" in the above URL means P/E ratio.
You should be able to
I want to use the standard error values in the summary that is produced using
coeftest, but I am getting an error code- any ideas?
> library(lmtest)
> coeftest(lmodT_WBHO)
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
t1W 5.948190.17072 34.8410 < 2.2e-16 ***
t2W 6.5
Hi, I am seeking help with an error when running the example from R
Documentation for TukeyHSD. The error occurs with any example I run, from
any text book or website. thank you...
> plot(TukeyHSD(fm1, "tension")).
Error in plot(confint(as.glht(x)), ylim = c(0.5, n.contrasts + 0.5), ...) :
err
Hi Troels,
Not sure this is what you want.
> X<-runif(9,0,10)
> FF1<-data.frame(ID=c(1,2,3)[rep(c(1,1,1,2,2,2,3,3,3))],
> PERIOD=c(1,2,3)[rep(c(1,2,3),times=3)],X=X)
> FF1$X[4]<-NA
> FF1
ID PERIOD X
1 1 1 8.27119347
2 1 2 9.64698097
3 1 3 2.74132386
4 2 1
Open source software (what you're driving)
Michael
On Wed, May 16, 2012 at 12:27 PM, aramos wrote:
> Thanks, I've already done that!!
> What is OSS?
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/kolmogorov-Smirnov-critical-values-tp4630245p4630276.html
> Sent from the
Yes here it is. I actually convert them all as strings, initially using
options(stringsAsFactors=F) at the top of my code.
This what the initial dataframe looks like. Please note this is a toy
dataset:
namesXYZAorB
S1BBBBBBA
S2AABBBBA
S3AB
Thanks, I've already done that!!
What is OSS?
--
View this message in context:
http://r.789695.n4.nabble.com/kolmogorov-Smirnov-critical-values-tp4630245p4630276.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mai
I need to fit a t copula with fixed degree of freedom let's say 4. I
do not want to estimate the dof together with correlation matrix
optimally. Instead fix the dof to 4 and only estimate the correlation
matrix in the optimization routine. Is anyone aware of such estimation
method in R.
I think that command will give me the statistics observed value!! Not
quantiles from the k-s distribution!
--
View this message in context:
http://r.789695.n4.nabble.com/kolmogorov-Smirnov-critical-values-tp4630245p4630275.html
Sent from the R help mailing list archive at Nabble.com.
___
All,
Just to get the word out: We are looking for a new Statistical Consultant at
the Division of Statistics and Scientific Computation here at the University
of Texas at Austin. Please pass along to any colleagues who might be
interested...
http://ssc.utexas.edu/people/employment
Thanks, Michae
Sorry for the follow-up, but I dig deeper into the problem.
My text on the R^2 was wrong: In my opinion, and at least to Wikipedia,
R^2 yields a division by zero iff SStot (the total sum of squares) is
zero. SStot is the sum of the sum of the difference between the observed
(not the predicted) val
Thanks Max for your answer.
First, I do not understand your post. Why is it a problem if two of
predictions match? From the formula for calculating R^2 I can see that
there will be a DivByZero iff the total sum of squares is 0. This is
only true if the predictions of all the predicted points from
Hi,
I'm dealing with an optimization problem. I'm using 'optim' to maximize the
output of a function, given some restrictions on the input. I would like to
know if there is a way to impose some restrictions on 'intermediate
variables' of the function. An example..
fx = function (x)
{
s <- 0
for (
Thanks Gabor,
Nifty regexp. I never used strapplyc before and I am sure this will become a
nice addition to my toolkit.
KW
Message: 5
Date: Tue, 15 May 2012 07:55:33 -0400
From: Gabor Grothendieck
To: Keith Weintraub
Cc: r-help@r-project.org
Subject: Re: [R] Scraping a web page.
Message-ID:
On Wed, May 16, 2012 at 06:52:48AM -0700, aramos wrote:
> Hi!
>
> Any one knows how to obtain critical values for the k-s statistic, using R?
Hi.
I do not know, whether there is a function for this. However, the following
randomized approach allows to extract a table of statistic/p.value pairs
f
Take a look at this SO question:
http://stackoverflow.com/questions/10575005/output-a-boolean-from-an-rscript-into-a-bash-variable
None of the solutions are "Boolean specific" so you should be good
with them (the key is printing and capturing)
Michael
On Wed, May 16, 2012 at 2:36 PM, Jannis wro
Dear R community,
is there any way to invoke R in batch mode, do some calculations and get
the values of some R variables back into the (bash)shell ? I only
managed to get some output saved into a text file with:
R --slave --args 2 2 test2.R
test.R contains:
a <- as.numeric(commandArgs()[
Hello All,
This will probably be easy for some but isn't for me. Currently am working on a
text mining exercise. Want to be able to predict whether cancer patients got
KRAS testing, and, if so, whether the test yielded a result of wild
type/negative or mutant/positive. I've begun with a "bag-of
Thanks a lot - beautiful
Troels
Den 16-05-2012 19:29, David Winsemius skrev:
On May 16, 2012, at 11:56 AM, Troels Ring wrote:
Dear friends - I hope you will forgive me another simple question,
illustrated by
ID <- c(1,1,1,2,2,3,3,3)
PERIOD <- c(1,2,3,2,3,1,2,3)
X <- runif(8,0,10))
Extrane
On May 16, 2012, at 11:56 AM, Troels Ring wrote:
Dear friends - I hope you will forgive me another simple question,
illustrated by
ID <- c(1,1,1,2,2,3,3,3)
PERIOD <- c(1,2,3,2,3,1,2,3)
X <- runif(8,0,10))
Extraneous paren removed:
FF <- data.frame(ID=ID,PERIOD=PERIOD,X=X)
I need to the
Dear Researchers,
I am looking a way to disable the Error massage in read.table() as warn =
TRUE in readLines(), when the lines are empty
Error in read.table(con, header = F, sep = " ", nrow = n) :
no lines available in input
thanks for all suggestions
Gianni
[[alternative HTML versio
If you want a confidence based in new x values you can do. I have this post
with steps to do this. It's written in Portuguese but the R code is useful.
http://ridiculas.wordpress.com/2011/05/19/bandas-de-confianca-para-modelo-de-regressao-nao-linear/
Bests.
Walmes.
==
x[is.na(z)] <- NA
This might send you a nasty bug if x and z are different lengths
though -- just a head's up.
Michael
On Wed, May 16, 2012 at 12:55 PM, Mintewab Bezabih
wrote:
> Dear R users,
>
> I was wondering how I can replace the values of a vector with the values
> from in another vecto
Dear R users,
I was wondering how I can replace the values of a vector with the values from
in another vector in the same row
For example, how can I replace the value of x below with NA when the value of Z
in the same row is NA?
x <-1:20
z<- c(11, 15, 17, 2, 18, 6, 7, NA, 12, 10,21, 25, 27,
Dear R users,
I was wondering how I can replace the values of a vector with the values from
in another vector in the same row
For example, how can I replace the value of x below with NA when the value of Z
in the same row is NA?
x <-1:20
z<- c(11, 15, 17, 2, 18, 6, 7, NA, 12, 10,21, 25, 27, 1
On May 16, 2012, at 6:37 AM, Andras Farkas wrote:
Dear R Expert
allow me to ask a quick qestion: I have a mean value of 6 and a SD
of 3 describing my distribution. I would like to "convert" this
distribution into a log normal distribution that would best describe
it when resimulated usin
Dear friends - I hope you will forgive me another simple question,
illustrated by
ID <- c(1,1,1,2,2,3,3,3)
PERIOD <- c(1,2,3,2,3,1,2,3)
X <- runif(8,0,10))
FF <- data.frame(ID=ID,PERIOD=PERIOD,X=X)
I need to the fourth value of X as NA, and ID and PERIOD is updated to
1,1,1,2,2,2,3,3,3 and 1,
Hi Yang,
Did you try a different CRAN mirror?
Best,
Ista
On Wed, May 16, 2012 at 10:21 AM, Yang, Ming wrote:
> Has one try to install the ggplot2 package recently? I tried to install
> it on my new system and had trouble:
>
>
>
>> install.packages("ggplot2")
>
> Installing package(s) into 'C:/P
It looks like there might be a mirror problem -- use
chooseCRANmirror() to select a different mirror.
Best,
Michael
On Wed, May 16, 2012 at 10:21 AM, Yang, Ming wrote:
> Has one try to install the ggplot2 package recently? I tried to install
> it on my new system and had trouble:
>
>
>
>> instal
Looks like your mirror was in an inconstant state. Seems to be fixed by
a finished rysnc in the meantime ...
Uwe ligges
On 16.05.2012 16:21, Yang, Ming wrote:
Has one try to install the ggplot2 package recently? I tried to install
it on my new system and had trouble:
install.packages(
On 16.05.2012 15:52, aramos wrote:
Hi!
Any one knows how to obtain critical values for the k-s statistic, using R?
ks.test(.)$statistic
Uwe ligges
Thanks,
Alex
--
View this message in context:
http://r.789695.n4.nabble.com/kolmogorov-Smirnov-critical-values-tp4630245.html
Sent from
On 16.05.2012 15:51, Bharat Warule wrote:
Hello R user,
I have four data sets in dir "D:/Bharat Warule/Rdata_file" which are
output_data_prod_1.rda, output_data_prod_2.rda, output_data_prod_3.rda,
output_data_prod_4.rda.
Each data set is huge size like number of rows 343297 and columns are nea
On 15.05.2012 16:30, tudor wrote:
Dear useRs:
Is there a way I could predict the terminal node associated with a new data
entry in an rpart environment? In the example below, if I had a new data
entry with an AM of 5, I would like to link it to the terminal node 2. My
searches led to http://to
Fine for me, and I cannot investigate anything since there is not even a
single piece of reproducible code given.
Uwe Ligges
On 15.05.2012 23:20, Daniel Carr wrote:
I have doubled buffered animations that I show in class.
They used to work but now flash.
The default windows() option is buffe
Has one try to install the ggplot2 package recently? I tried to install
it on my new system and had trouble:
> install.packages("ggplot2")
Installing package(s) into 'C:/Program Files/R/R-2.14.2/library'
(as 'lib' is unspecified)
also installing the dependency 'scales'
trying URL
'http:/
On Wed, May 16, 2012 at 9:52 AM, aramos wrote:
> Hi!
>
> Any one knows how to obtain critical values for the k-s statistic, using R?
>
Take a look at ?ks.test and the code of ks.test to see how R does it.
OSS is super helpful for these sorts of things.
Michael
> Thanks,
> Alex
>
> --
> View thi
Hello,
Your data.frame is composed exclusively of factors, but try this
(I've changed the name to 'sampl', because 'sample' is an R function.)
# logical index vectors
iA <- sampl$AorB == "A"
iB <- sampl$AorB == "B"
new.sampl <- data.frame(
apply(sampl, 2, function(x){
i
Dear all,
I want to build a model in R based on animal collection data, that look like
the following
Nr Village DistrictSiteSurvey Species Count
1 AX A F Dry B 0
2 AY A V Wet A 5
3 BX B F W
On 15.05.2012 23:47, rl269 wrote:
Hello,
I am having trouble asking R to read individual numeric vectors for a box
plot of the residuals of a linear regression. It is performing arithmetic
addition on the 16 individual variables that I want individual box plots
for.
I have 16 race*treatment
Dear Sirs.
I am working with the R packages Hmisc and PredictABEL to make NRI estimates
from my Cox models with and without a specific biomarker.
According to Pencina et al (Statistics in Medicine 2010, DOI: 0.1002/sim.4085
), a continuous/non-categorical NRI (NRI>0) is to be used when there ar
Hello R user,
I have four data sets in dir "D:/Bharat Warule/Rdata_file" which are
output_data_prod_1.rda, output_data_prod_2.rda, output_data_prod_3.rda,
output_data_prod_4.rda.
Each data set is huge size like number of rows 343297 and columns are near
to 50.
For example:
x1 <- data.frame(x11=
Hi!
Any one knows how to obtain critical values for the k-s statistic, using R?
Thanks,
Alex
--
View this message in context:
http://r.789695.n4.nabble.com/kolmogorov-Smirnov-critical-values-tp4630245.html
Sent from the R help mailing list archive at Nabble.com.
___
On 15.05.2012 19:23, Zhiqiu Hu wrote:
r-help@r-project.org
Dear friends,
I want to make the following change of R setting on a windows 7 desktop.
$R_check_force_suggests = "FALSE"
You can change it globally in the operating systems defaults for
environment variable, or for the current ses
More information is needed to be sure, but it is most likely that some
of the resampled rpart models produce the same prediction for the
hold-out samples (likely the result of no viable split being found).
Almost every incarnation of R^2 requires the variance of the
prediction. This particular fai
On 16.05.2012 02:13, Gundala Viswanath wrote:
Given the attached plot,
Nothing came through.
how can I locate the center text with "Mean and SD" so that it can be
placed exactly under "---emp".?
The current code I have is this:
L = list(bquote(Em.Mean ==.(new_avg)),bquote(Em.SD==.(new
On 16.05.2012 08:11, umai88 wrote:
I got this code below and i want to repeat the loop for 100 times..
And what is the problem? What are you aiming at?
Uwe Ligges
x<-rnorm(60)
mat1<-matrix(x,nrow=15,ncol=4)
trim<-numeric(ncol(mat1))
win<-numeric(ncol(mat1))
ssd<-numeric(ncol(mat1))
fo
On 16.05.2012 12:37, Andras Farkas wrote:
Dear R Expert
allow me to ask a quick qestion: I have a mean value of 6 and a SD of 3 describing my
distribution. I would like to "convert" this distribution into a log normal
distribution that would best describe it when resimulated using log normal
On May 16, 2012, at 1:33 AM, ana24maria wrote:
Thank you very much.
After using dput and the easy way (> result <-
eblup.mse.f.wrap(domain.data
= amigo, lme.obj = fit.lme)),
i have got the following error:
Error in `[.data.frame`(sample.data, , variabs) :
undefined columns selected
What
Thanks! Now it is clear.
Francisco
On Wed, 16 May 2012 07:32:56 -0400, Gabor Grothendieck wrote
> On Tue, May 15, 2012 at 11:20 PM, Gabor Grothendieck
> wrote:
> > On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
> > wrote:
> >> Hi all
> >>
> >> I have fitted a model usinf nls function to
I can't replicate your problem. I created a spreadsheet in Excel 2007
consisting of three columns. Numbers from 1 - 15, rand(), and the sum of
the first two columns. Using all the defaults with read.xlsx() (package:
xlsx), I get the values of each column and using keepFormulas=TRUE, I get
the form
Fascinating... dput() has never given me anything that looks like
that I would have expected something much more like
z <- structure(c(123.53, 123.78, 124.24, 124.2, 124.07, 123.91, 123.44,
123.0616, 123.06, 123.13, 123.745, 123.96, 123.99, 123.99, 124.3,
124.38, 124.67, 125.19, 124.9, 125.27,
try this:
dtm <- DocumentTermMatrix(examplecorpus, control = list(wordLengths=c(1,100)))
On Wed, May 16, 2012 at 6:22 AM, C.H. wrote:
> Dear All,
>
> The following code illustrate the problem.
>
> [R code]
> require(tm)
> exampledoc <- c("R is good", "R is really good")
> examplecorpus <- Corp
? cor
e.g.,
x <- data.frame(rnorm(5), rnorm(5), rnorm(5), rnorm(5), rnorm(5))
cor(x)
Best,
Michael
On Wed, May 16, 2012 at 6:52 AM, Andrea Sica wrote:
> Dear all,
>
> I have created a subset from my dataset, which contains 6 variables.
> I need to make the correlation among all of them, possi
Can you show us what you want the final data.frame to look like? You've
created five variables stored as factors and you seem to be trying to change
those to numeric values? Is that correct?
Since AB and BA are always set to 1, you could just replace those values
globally rather than mess with th
Rui-
Just a quick question. I understand your comment on using ANOVA, but
doesn't this only test for similarities of the mean. We are trying to see if
a the same model can fit for two or three months, therefore have the similar
slope and intercept. The ANOVA would only do one part of this correct
Hello,
Your bug is obvious, each pass through the loop you read twice and write
only once. The file pointer keeps moving forward...
Use something like
while (length(pv <- readLines(con, n=n)) > 0 ) { # note that this line
changed.
i <- i + 1
write.table(pv, file = paste(fileNames.t
Dear list members:
I am trying to calculate power for an ANCOVA analysis.
I have found different solutions such as power.t.test and power.anova.test
but they seem to refer to the ANOVA part of the ANCOVA.
My model is of the form:
lm (y ~ factor + x1 + x2 + x2*myfactor)
where myfactor is a fact
You can achieve that with a combination of as.formula and paste.
library(nlme)
data(petrol, package = "MASS")
lme(as.formula(paste(Y.VAR, "~EP")), random= ~1|No, data=petrol)
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
F
Hai, I have change it to these, but error and I couldn't fix it. Do you have
any idea why?
file <- system.file("D:\\FYP\\image\\Cropped Images\\user61",
"forgerUser61.xlsx", package = "xlsx")
wb <- loadWorkbook("forgerUser61.xlsx")
sheets <- getSheets(wb)
sheet <- sheets[["all"]]
res <- readRows(s
Dear R Expert
allow me to ask a quick qestion: I have a mean value of 6 and a SD of 3
describing my distribution. I would like to "convert" this distribution into a
log normal distribution that would best describe it when resimulated using log
normal distribution. Currently I am using another
Hi,
Probably you could check this:
>?quantile
Particularly the 'type' option.
Best Regards,
Pascal
- Mail original -
De : Retep32
À : r-help@r-project.org
Cc :
Envoyé le : Mercredi 16 mai 2012 16h22
Objet : [R] Wrong Q3 + Mean.
Hi.
> a
[1] 13 13 14 14 15 15 16 20 21 26
> summary(a)
Michael Weylandt wrote
>
> Can you provide a reproducible example?
>
Of course, Michael.
Consider the following time series:
11/2/2011 14:30 123.53
11/2/2011 15:00 123.78
11/2/2011 15:30 124.24
11/2/2011 16:00 124.2
11/2/2011 16:30 124.07
11/2/2011 17:00 123.91
11/2/2011 17:30 123.44
11/2/20
Hi,
many thanks for your answer. if i set tz="GMT" it does the job. Great!
thanks
cheers
Benedikt
Am 16.05.2012 12:20, schrieb jholtman [via R]:
> Is the a daylight saving time problem? Check your timezone and see
> when it occurred; these times might not be legal.
>
> Sent from my iPad
>
> O
On 05/16/2012 01:18 PM, Santosh wrote:
Hello R/Splus users..
I am posting in R discussion group in hope of wider response compared to
what I received from Splus user groups
Was wondering if there is any function available in Splus 8.2 that is
equivalent to "reshape" of R?
Below is a sample
On Tue, May 15, 2012 at 11:20 PM, Gabor Grothendieck
wrote:
> On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
> wrote:
>> Hi all
>>
>> I have fitted a model usinf nls function to these data:
>>
>>> x
>> [1] 1 0 0 4 3 5 12 10 12 100 100 100
>>
>>> y
>> [1] 1.281055090 1.5
Dear All,
The following code illustrate the problem.
[R code]
require(tm)
exampledoc <- c("R is good", "R is really good")
examplecorpus <- Corpus(VectorSource(exampledoc), encoding = "UTF-8")
dtm <- DocumentTermMatrix(examplecorpus, control = list(minWordLength = 1))
as.matrix(dtm)
[/R code]
Th
At 08:22 16/05/2012, Retep32 wrote:
Hi.
> a
[1] 13 13 14 14 15 15 16 20 21 26
> summary(a)
Min. 1st Qu. MedianMean 3rd Qu.Max.
13.014.015.016.719.026.0
> mean(a)
[1] 16.7
> quantile(a)
0% 25% 50% 75% 100%
13 14 15 19 26
Clearly, this is not rig
Dear all,
I have created a subset from my dataset, which contains 6 variables.
I need to make the correlation among all of them, possibly, without
making it one by one. Is there any command that can permits me to
do it directly for all of them in the same time?
Thank you so much in advance.
Andr
Hello,
the following lines
m <- matrix(c(1,1,9,1,2,6,1,3,7,2,1,4,2,2,5,2,3,1,3,1,2,3,2,-1,3,3,-2), 9,
3, byrow = TRUE, dimnames=list(NULL, cbind('A','B','Y')))
md <- as.data.frame(m)
md$A <- as.factor(md$A)
md$B <- as.factor(md$B)
mm <- model.matrix(Y~A+B+A:B, data=md)
produce
> mm
(Interce
Is the a daylight saving time problem? Check your timezone and see when it
occurred; these times might not be legal.
Sent from my iPad
On May 16, 2012, at 3:27, Benedikt Gehr wrote:
> Hi
>
> I have a rather large data frame (>7000 rows with 28 columns) which I want
> to sort by date. Below I
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