I have submitted a tiny new package to CRAN, lazyData. This has a single
function
requireData
which is designed as a drop-in replacement for base::require. In addition
to doing the same job as base::require, it supplies a LazyData facility for
those packages which have data but do not provide L
A new release of the conf.design package has been submitted to CRAN.
This is largely a "tidying up" release, with no substantial change in
functionality. The one difference some users may notice is that the method
functions
factorize.default
factorize.factor
are no longer exported directly, but
On 02/25/2013 05:07 PM, Anna Zakrisson wrote:
Hi,
I have a data set with two continous variables that I want to plot MEANS (I
am not intrerested in median values) on a double-y graph. I also have 2
factors. I want the factor combinations plotted in different panes.
Dummy dataset:
mydata<- data
This is a basic statistics question and off topic here. Talk to a
statistician (i.e. someone with a good statistics background) or
start reading. You need an extensive statistics tutorial that I
believe is too much for online forums like stats.stackexchange.com.
-- Cheers,
Bert
On Sun, Feb 24,
Hello R user's,
I've read a txt file with the read.table syntax. This file is already in a
form of a contingency table (130 rows, 90 columns) with wich i would like
to do a simple correspondance analysis with the ca() syntax.
Are there a way to do an as.table(my data.frame) transformation ? Or a
Couldn't reproduce your error (but I did get a warning message):
> ads1S <- with(ads1,Surv(INTERVAL_START,EVENT,STATUS,type=c('counting')))
Warning message:
In Surv(INTERVAL_START, EVENT, STATUS, type = c("counting")) :
Stop time must be > start time, NA created
> cox_out <- coxph(ads1S~Meter
On Feb 25, 2013, at 12:35 , Bert Gunter wrote:
> This is a basic statistics question and off topic here. Talk to a
> statistician (i.e. someone with a good statistics background) or
> start reading. You need an extensive statistics tutorial that I
> believe is too much for online forums like sta
Note that ReadImages has been orphaned and is under the BSD license, so
you can become maintainer and re-establish the package on CRAN.
Best,
Uwe Ligges
On 25.02.2013 07:50, PIKAL Petr wrote:
Thanks to all.
For the time being due to urgency I will stay with a copy of R 2.15.x with
which Re
Hello,
If your data.frame is named 'dat', the following might be what you want.
as.table(data.matrix(dat))
Hope this helps,
Rui Barradas
Em 25-02-2013 11:35, franck.berth...@maif.fr escreveu:
Hello R user's,
I've read a txt file with the read.table syntax. This file is already in a
form of
Hi
From: Rasmus Hedegaard [mailto:hedegaard...@hotmail.com]
Sent: Monday, February 25, 2013 2:53 PM
To: PIKAL Petr
Subject: RE: [R] Making the plot window wider and using the predict function
Thank you, Petr - the command
linreg0 <- lm(Hwt ~ Bwt, data = maleData)
works perfectly.
Regarding th
Dear Sarah,
Thanks for your email. I'll describe the problm but without the screenshots
then.
Firstly, I think Ive correctly installed R.
I have installed R for Windows via the R site, CRAN and then UK
University of Bristol or UK Imperial College London. Both times, I have
installed the b
Dear Sir/Madam,
I apologize for any cross-posting. I got a simple question, which I thought
the R list may help me to find an answer. Suppose we have Y_1, Y_2, ., Y_n ~
Poisson (Lambda_i) and Lambda_i ~Gamma(alpha_i, beta_i). Empirical Bayes
Estimator for hyper-parameters of the gamma distr, i
Dear all,
I have got two vectors coding for a stimulus presented in the current trial
(mydat$Stimulus) and a prediction in the same trial (mydat$Prediciton),
respectively.
By applying an if-conditional I want to create a new vector that indicates if
there is a match between both vectors in th
Homework? We don't do homework here.
If not, search (e.g. via google -- "R hierarchical Bayes" -- or some such).
-- Bert
On Mon, Feb 25, 2013 at 1:39 AM, Ali A. Bromideh wrote:
> Dear Sir/Madam,
>
>
>
> I apologize for any cross-posting. I got a simple question, which I thought
> the R list may
Dear all,
I am using the code as below
tdm <- melt(matrixToPlot)
p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
labs(x = "Mz", y = "T", fill = "D") +
geom_raster(alpha=1) +
scale_fill_discrete(h.start=1) +
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Andy Siddaway
> Sent: Monday, February 25, 2013 11:24 AM
> To: Sarah Goslee
> Cc: r-help
> Subject: Re: [R] R software installation problem
>
> Dear Sarah,
>
> Thanks for yo
On Mon, Feb 25, 2013 at 10:24 AM, Andy Siddaway
wrote:
> Dear Sarah,
>
> Thanks for your email. I'll describe the problm but without the screenshots
> then.
>
>
> Firstly, I think I’ve correctly installed R.
>
> I have installed R for Windows via the R site, CRAN and then UK –
> University of Bris
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jonas Walter
> Sent: Monday, February 25, 2013 2:38 PM
> To: r-help@r-project.org
> Subject: [R] creating variable that codes for the match/mismatch
> between two other variab
Hi Alex,
See ?theme
Best,
Ista
On Mon, Feb 25, 2013 at 9:44 AM, Alaios wrote:
>
> Dear all,
> I am using the code as below
> tdm <- melt(matrixToPlot)
>p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
> labs(x = "Mz", y = "T", fill = "D") +
>
Yes, it works.
Thank very much you Rui.
Franck Berthuit
France
De :Rui Barradas
A : franck.berth...@maif.fr,
Cc :r-help@r-project.org
Date : 25/02/2013 15:10
Objet : Re: [R] Data frame as table
Hello,
If your data.frame is named 'dat', the following might be what you want.
Hi Petr,
oh, that's really way more easier than the way I did it. Thanks for the hint!
The problem with "no prediction" is that these cases are already coded
within the "Prediction" variable.
0 codes "no prediction required" while 1 and 2 codes for different
predictions. Therefore, there a
Hi
> -Original Message-
> From: Jonas Walter [mailto:jonas.wal...@student.uni-tuebingen.de]
> Sent: Monday, February 25, 2013 4:25 PM
> To: PIKAL Petr
> Cc: r-help@r-project.org
> Subject: RE: [R] creating variable that codes for the match/mismatch
> between two other variables
>
> Hi Pet
hello, all.
one of my students is having an issue with the pie & legend function.
this is her code. (below)
it works just fine for me.
her error is "plot.new has not been called yet". i know this means her pie
chart is coming up blank so the legend will not work.
according to ?graphics thi
Hi,
I created a xyplot with a three-column layout. As suggested by Deepayan I tried
to put titles to each column by using xlab.top. Unfortunately, as my y-axis
scale relation = "free", the column titles are not centered at the three x axes
anymore. Any idea how to center the titles?
#Example:
Hi everyone.
I have two data frames that contain the same variables but with different
number of observation.
I would like to know it was possible to combine the data so I can have
"paired" boxplot on the same figure.
For example,
df1 = data.frame(x = rnorm(100))
df1$type = ifelse(df1$x
Thank you for pointing me to ?dissimilarity.object. I now see that N = Nominal
(factor) and I = Interval scaled (numeric).
Regards.
On Sun, Feb 24, 2013 at 2:36 PM, Peter Ehlers wrote:
> On 2013-02-24 07:57, Joanna Papakonstantinou wrote:
>
>> I am using the iris dataset that contains mixed va
Does the following produce what you want?
> lst <- c(Observed=with(df1, split(x, type)), Simulated=with(df2, split(x,
type)))
> lst <- lst[c(seq(1,length(lst)-1,by=2), seq(2,length(lst),by=2))]
> boxplot(lst, col=rep(c("gray","red"),len=length(lst)))
Bill Dunlap
Spotfire, TIBCO Software
On Mon, Feb 25, 2013 at 2:10 AM, Ulrike Grömping
wrote:
> Gabor,
>
> thanks for your patient answers! I have adjusted the Rtools path to consist
> of both
> the bin and the gcc-4.6.3 sub directory, and that did it. The R path was set
> by R as it was,
> presumably because this is only a 32-bit sys
Am 25.02.2013 18:21, schrieb Gabor Grothendieck:
On Mon, Feb 25, 2013 at 2:10 AM, Ulrike Grömping
wrote:
Gabor,
thanks for your patient answers! I have adjusted the Rtools path to consist
of both
the bin and the gcc-4.6.3 sub directory, and that did it. The R path was set
by R as it was,
presu
On Feb 25, 2013, at 7:37 AM, Nicole Ford wrote:
> hello, all.
>
> one of my students is having an issue with the pie & legend function.
>
> this is her code. (below)
>
> it works just fine for me.
>
> her error is "plot.new has not been called yet". i know this means her pie
> chart is com
Hello again, in the task view I see that "Direct support in R is
starting with release 2.14.0 which includes a new package parallel
...". However I can not get any access to it. When I type
ls('package:parallel'), I get following error:
> ls('package:parallel')
Error in as.environment(pos) :
in the future, please provide R code to re-create some example data :)
read
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-examplefor
more detail..
# create a data table with three unique columns' values..
# treat these values just like letters
x <-
cbind(
On 2013-02-25 10:00, Christofer Bogaso wrote:
Hello again, in the task view I see that "Direct support in R is
starting with release 2.14.0 which includes a new package parallel
...". However I can not get any access to it. When I type
ls('package:parallel'), I get following error:
ls('pa
Hello,
I disagree with the way you've sorted the matrix, like this all A's
become first, then B's, etc, irrespective of the respondents. Each row
is a respondent, and the rows should be kept intact, but with a
different ordering. To this effect, use order():
z <- y[order(y[,1], y[,2], y[,3])
does this work for you?
df1 = data.frame(x = rnorm(100))
df1$type = ifelse(df1$x <= 0 , "type1", "type2")
df1$group<-1
df2 = data.frame(x = rnorm(50,0,2))
df2$type = ifelse(df2$x <= 0 , "type1", "type2")
df2$group<-2
combined.df<-rbind(df1,df2)
boxplot(combined.df$x ~ combined.df$group *co
My point is that you _still_ have not adhered to the Posting Guide request for
sessionInfo() ... I say again. Please read the Posting Guide ... AND PLEASE
STOP posting formatted email.
--
David.
On Feb 25, 2013, at 10:20 AM, Nicole Ford wrote:
> I did look at ??pie ??graphics, as per my repl
Hi manuel
Have a look at http://tolstoy.newcastle.edu.au/R/e2/help/07/05/17666.html
Just plugging in your xlabs everything is OK for Deepayan's example
xyplot(1:9 ~ 1:9 | gl(3, 1, 9),
layout = c(3, 1),
xlab = myXlabGrob('Trial number',
'Subject number',
Hi all~
I was wondering if there is an R package out there that can do multivariate
interpolation. I conducted a simulation across an even grid and would like to
now use this information to interpolate values for a single dependent variable
using data measured from 10 or so variables. I've look
Hi all-
New to R, and I've managed to learn a bit about Lavaan package for SEM, and
pscl package for analyzing count data.
I would like to find a way to combine these analyses as I am hoping to run a
model with a multi-trait, multi-reporter factor (2 reporters by 4 traits
apiece, with one o
I did look at ??pie ??graphics, as per my reply. which netted nothing of
value.
thanks.
~Nicole Ford
Ph.D. student
Graduate Assistant/ Instructor
University of South Florida
Government and International Affairs
office: SOC 012M
e: nmhi...@mail.usf.edu
http://gia.usf.edu/student/nford/
On
I am having difficulty getting the dynamic tree cut package to work.
Given the data table "myddtable"
LengthPlaceColorAge5HRed224ABlue205WGreen243GRed222GBlue236WGreen255ARed194H
Blue23
I created a similarity matrix using DAISY and Gower metric and specified
Place and Color columns as characters (
Or if David's answer seems like too much work you could use the `mvrnorm`
function in the MASS package to generate 2 vectors with the given
correlation and sample size and feed those vectors to the `cor.test`
function.
Or Pearson's test can be computed in 1 line of R code without needing any
speci
On 2013-02-25 07:47, Manuel Kunz wrote:
Hi,
I created a xyplot with a three-column layout. As suggested by Deepayan I tried to put
titles to each column by using xlab.top. Unfortunately, as my y-axis scale relation =
"free", the column titles are not centered at the three x axes anymore. Any
I use to work whit stata dataframe, so, when I use R I type read.dta
Until today I do that without any problem, after type:
mydata<-read.dta("C:/dropbox/dataframe.dta")
attach(mydata)
Everything works great... but today, when I typed:
mydata<-read.dta("C:/dropbox/dataframe.dta")
attach(mydata)
I'm learning R and am converting some code from SPSS into R. My background
is in SAS/SPSS so the vectorization is new to me and I'm trying to learn
how to NOT use loops...or use them sparingly. I'm wondering what the
most efficient to tackle a problem I'm working on is. Below is an example
piece
Here's a direct translation:
Variable <- 0
Variable <- ifelse(item1 == 1, Variable +1, Variable)
Variable <- ifelse(item2 == 1, Variable +1, Variable)
Variable <- ifelse(item3 == 1, Variable +1, Variable)
Variable <- ifelse(item4 == 1, Variable +1, Variable)
Here's another way to do it:
Vari
Dear R-Help group:
I have been tinkering with how I want my personal standard library
functions to look like. They are not designed to be professional and
heavyweight, but lightweight. There are probably dozens of little bugs,
because I don't know or have not properly taken care of a variety of
Note that in
Variable <- 0 + (item1 == 1) + (item2 == 1) + (item3 == 1) + (item4 == 1)
the '0 +' is not needed, since adding logicals produces integers.
Your second solution
Variable <- sum((item1 == 1), (item2 == 1) , (item3 == 1) , (item4 == 1),
na.rm=TRUE)
gives the wrong answer, since su
If you do this sort of thing a lot you may find the psych package helpful:
# make example data
x <- 1:3
dat <- data.frame(expand.grid(Item1=x, Item2=x, Item3=x, Item4=x))
# make scoring ky
key <- c(Item1=1, Item2=2, Item3=1, Item4=1)
# load psych library
library(psych)
# score
(scores <- score.
I sometimes need to return multiple items from a loop. Is it possible to have
the <<- operator work the same for mclapply as for lapply ?
> extra <- list()
> squares <- mclapply(1:10, function(x){extra[[x]] <<- x; x^2;})
> extra
list()
> squares <- lapply(1:10, function(x){extra[[x]] <<- x; x^2
by the time your rude reply came ( you are often rude to people so i shouldn't
have been surprised but somehow was) , i had already found my answer, by doing
it MYSELF on her computer and found had not followed some simple instructions.
be well.
~Nicole Ford
Ph.D. student
Graduate Assistant/ In
Why do you equate using <<- with returning multiple items from a loop? There
are valid reasons to use <<-, but the people who want to use it practically
never have them.
Just return a list of the items you want to return from within the function.
squares <- mclapply(1:10, function(x){result <-
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