Hi r-users,
Â
I really need help in fitting the t-copula. I try to reproduce the example
given by Jun Yan in âEnjoy the joy of copulaâ but Iâm not sure how to
correct the error based on the error message. I tried so many ways but still
could not get it working.
Â
loglik.marg <- funct
I have just installed the latest versions of R and Tinn-R (running Windows XP
prof.)
R 2.11.0
Tinn-R version 2.3.5.2
Everything seems fine, except for the following:
I usually do this: Open Tinn-R and click on the R icon to open R - this splits
the screen into two parts horiozontally, with Tinn-R
Hello all!
I am relatively new at R and was having a few issues with the following
graph:
barplot(c(5724.688,7290.7875), ylab=expression(paste(Delta, "AUC 45-200 min
(μg/ml * 155 min)")), xlim=c(0,2), width=0.5,
border="black",font.lab=2,cex.lab=1.7, names.arg=c("Saline (n=8)", "Exendin
(9-39)
Ok.
Thank you very much for setting me straight :)
On Wed, Jun 9, 2010 at 7:22 PM, Joris Meys wrote:
> On Wed, Jun 9, 2010 at 5:19 PM, Or Duek wrote:
> > Hi,
> > I would like to compare to regression models - each model has a different
> > dependent variable.
> > The first model uses a number
On Wed, Jun 9, 2010 at 10:03 PM, Marius Hofert wrote:
> How do I remove the labels (numbers)?
>
> Is splom(~iris[,1:4], axis.line.tck = 0, axis.text.cex =0) the way to go?
Just use 'pscales = 0'. There is even an example of this is in
help(splom). You should really get into the habit of reading
d
Bill and Erik, thank you very much for your help. In addition to
solving my problem, both solutions contain other good things I didn't
know about.
Regards,
Mark Seeto
On Thu, Jun 10, 2010 at 2:44 PM, Erik Iverson wrote:
> Hello,
>
>> How does one specify a formula to lm inside a function (with v
Dear R users,
I posted a couple of questions and got no response, so I am giving it
another shot.
I ran an experiment with a TWO-WAY within subject design. A sample dataset
is in http://www-scf.usc.edu/~hex/data.txt
I already ran ANOVA by using the following formula:
aov(RT~Factor1*Factor2 + Er
How do I remove the labels (numbers)?
Is splom(~iris[,1:4], axis.line.tck = 0, axis.text.cex =0) the way to go?
Thanks,
Marius
On 2010-06-10, at 06:42 , Deepayan Sarkar wrote:
> On Wed, Jun 9, 2010 at 1:32 PM, Marius Hofert wrote:
>> Dear ExpeRts,
>>
>> why does
>>
>> splom(~iris[,1:4],sc
This is a lazy way, and a slightly extravagant way if your memory is limited
and you are dealing with large numbers of rows.
NCols <- 5
NRows <- 7
myMat <- matrix(runif(NCols*NRows), ncol=NCols)
d <- dist(myMat)
dm <- as.matrix(d)
diag(dm) <- Inf
ij <- which(dm == min(dm), arr.ind = TRUE)[1,]
Here is one way.
f <- function(d) {
y <- names(d)[1]
xs <- names(d)[-1]
nx <- length(xs)
xs <- sort(sample(xs, sample(1:nx, 1)))
form <- as.formula(paste(y, "~", paste(xs, collapse="+")))
Call <- substitute(lm(FORM, data = d), list(FORM = form))
eval(Call)
}
d <- within(data.frame(y
Hello,
How does one specify a formula to lm inside a function (with variable
names not known in advance) and have the formula appear explicitly in
the output?
For example,
f <- function(d) {
in.model <- sample(c(0,1), ncol(d)-1, replace=T)
current.model <- lm(paste(names(d)[1], "~",
paste(
On Wed, Jun 9, 2010 at 1:32 PM, Marius Hofert wrote:
> Dear ExpeRts,
>
> why does
>
> splom(~iris[,1:4],scales = list(alternating = c(0,0), tck = c(0,0)))
>
> not remove the ticks and labels (xyplot does)?
It does actually, though not in the way you think. Compare with
splom(~iris[,1:4], scales
Hi there,
I am sure there is a better way to do it, but here is a suggestion:
res <- matrix(NA, ncol = 2, nrow = 5)
for(i in 1:5) res[i, ] <- which(as.matrix(d) == sort(d)[i], arr.ind =
TRUE)[1,]
res
HTH,
Jorge
On Wed, Jun 9, 2010 at 11:30 PM, Jeff08 <> wrote:
>
> Dear R Gurus,
>
> As you pro
On Wed, Jun 9, 2010 at 6:49 PM, Xin Ge wrote:
> Hi All,
>
> I need a small help plotting median lines on lattice boxplots.
>
> # Data
> a <- rep(c("A","B"), each=10)
> b <- rep(c("a","a+b","b","b+a"), each=5)
> c <- c(1,9,5,2,7,7,8,8,8,5,4,5,3,2,5,6,7,8,9,1)
> x <- data.frame(a, b, c)
> med.A <- m
Dear R Gurus,
As you probably know, dist calculates the distance between every two rows of
data. What I am interested in is the actual two rows that have the least
distance between them, rather than the numerical value of the distance
itself.
For example, If the minimum distance in the following
Hello,
How does one specify a formula to lm inside a function (with variable
names not known in advance) and have the formula appear explicitly in
the output?
For example,
f <- function(d) {
in.model <- sample(c(0,1), ncol(d)-1, replace=T)
current.model <- lm(paste(names(d)[1], "~",
paste(na
Hello experts.
Sorry this is such a novice question, but I have been trying, fruitlessly to
get my x axis to angle at 45 degrees. I have tried the text() call with srt
and adj and just cannot get my labels to tilt. Does anyone have any other
suggestions? The following is my command for the gra
Dear R Forum members,
I am wondering of anyone has used R to pull data from an Ingres SQL database?
My initial thoughts would be to use the RJDBC package and Ingres's JDBC driver.
Before I get this thing together, has anyone used this combination and can
verify that it works, or can forewarn m
John Antonakakis wrote:
is there a good way to generate reports in R? I found the following code
using split.screen but this only works with plots and i would like to
generate reports with tables and charts. thx
The answer is definitively yes, and you can spend a lot of time learning about
how
is there a good way to generate reports in R? I found the following code using
split.screen but this only works with plots and i would like to generate
reports with tables and charts. thx
if (interactive()) {
par(bg = "white") # default is likely to be transparent
split.screen(c(
One possibility is
aggregate(iris[,-5],list(iris[,5]),mean)
Group.1 Sepal.Length Sepal.Width Petal.Length Petal.Width
1 setosa5.006 3.4281.462 0.246
2 versicolor5.936 2.7704.260 1.326
3 virginica6.588 2.9745
Dear All,
I have the data some thing like this, I am showing here three days data
only:
> dummy.data <- read.table(file='dummy.txt',sep='', header=TRUE)
> dummy.data
StDate Domaindesc Logins
1 05/01/10xxx 10
2 05/01/10xxx 45
3 05/01/10xxx 2
4 05/01/10
Here is an alternative
with(iris, rowsum(iris[, -5], Species)/table(Species))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Peter Langfelder
Sent: Thursday, 10 June 2010 12:27 PM
To: SH.Chou
Cc: r-help@r-project.org
Subject: R
apply(iris[, -5], 2, tapply, iris$Species, mean)
On Wed, Jun 9, 2010 at 3:43 PM, SH.Chou wrote:
> Hi there:
> I have a question about generating mean value of a data.frame. Take
> iris data for example, if I have a data.frame looking like the following:
> -
>Sepal.Len
I have not found anything about this except the following from the DBI
documentation :
Bind variables: the interface is heavily biased towards queries, as opposed
> to general
> purpose database development. In particular we made no attempt to define
> bind
> variables; this is a mechanism by wh
thats beautiful
> apply(m[, 3:14], 2,
+ function(x) tapply(x, m[,2], function(x) sum(!is.na(x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
1800332332333 3 3 3
1801332332333 3
Joris: Very, very nice. I have spent the day learning about indices. Thank
you very much.
Mr. Natural
--
View this message in context:
http://r.789695.n4.nabble.com/correcting-a-few-data-in-an-unreshaped-data-frame-tp2248219p2249599.html
Sent from the R help mailing list archive at Nabble.com.
Hi there:
I have a question about generating mean value of a data.frame. Take
iris data for example, if I have a data.frame looking like the following:
-
Sepal.Length Sepal.Width Petal.Length Petal.WidthSpecies
15.1 3.5
Hello R-Team,
I am trying to construct a Copula from a multivariate Gamma distribution with
its marginals gamma-distributed.
The multivariate Gamma should be able to contain a correlation coeficient or
matrix.
I have studied the book "Continuous Multivariate Distributions vol.I Models and
app
Xin,
Xin Ge wrote:
Hi All,
I need a small help plotting median lines on lattice boxplots.
# Data
a <- rep(c("A","B"), each=10)
b <- rep(c("a","a+b","b","b+a"), each=5)
c <- c(1,9,5,2,7,7,8,8,8,5,4,5,3,2,5,6,7,8,9,1)
x <- data.frame(a, b, c)
med.A <- median(subset(x$c, x$a=="A"))
med.B <- media
Hi All,
I need a small help plotting median lines on lattice boxplots.
# Data
a <- rep(c("A","B"), each=10)
b <- rep(c("a","a+b","b","b+a"), each=5)
c <- c(1,9,5,2,7,7,8,8,8,5,4,5,3,2,5,6,7,8,9,1)
x <- data.frame(a, b, c)
med.A <- median(subset(x$c, x$a=="A"))
med.B <- median(subset(x$c, x$a=="B"
Hello,
steven mosher wrote:
# create a matrix with some random NAs in it
m<-matrix(NA,nrow=15,ncol=14)
m[,3:14]<-52
m[13,9]<-NA
m[4:7,8]<-NA
m[1:2,5]<-NA
m[,2]<-rep(1800:1804, by=3)
y<-order(m[,2])
m<-m[y,]
m[,1]<-rep(1:3,by=5)
# what we want is a result that looks like this
1800 3 3
# create a matrix with some random NAs in it
> m<-matrix(NA,nrow=15,ncol=14)
> m[,3:14]<-52
> m[13,9]<-NA
> m[4:7,8]<-NA
> m[1:2,5]<-NA
> m[,2]<-rep(1800:1804, by=3)
> y<-order(m[,2])
> m<-m[y,]
> m[,1]<-rep(1:3,by=5)
> m
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,
On Jun 9, 2010, at 8:38 PM, TND (Ing. Marcos Ortiz Valmaseda) wrote:
> Regards to all R-Help list
> I ´m searching a R list on Spanish
> Do you know any?
>
> Regards and thanks a lot
Go here:
https://stat.ethz.ch/mailman/listinfo/r-help-es
HTH,
Marc Schwartz
__
Regards to all R-Help list
I ´m searching a R list on Spanish
Do you know any?
Regards and thanks a lot
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-g
Dear Duncan
I am not the only one who thinks RGoogleDocs is fantastic. See below. I
think it would save many people a lot of time and/or disappointment if
the omegahat site were updated to show/offer 0.4-1 instead of 0.4-0. Am I
correct in believing that I cannot change it, only you can? If so, th
Hi D.,
One option would be to use the boot library:
require(boot)
x <-c(1, 0, 0.153846154, 0, 0.142857143)
mycv <- function(x, id){
xs <- x[id]
sd(xs)/mean(xs)
}
my.boot <- boot(x, mycv, R = 999)
my.boot
plot(my.boot)
boot.ci(my.boot)
Also, take a look at [1], [2] and [3].
HTH,
Jorge
[1]
Hi Paul,
On 6/9/10 1:12 AM, Paul Murrell wrote:
> grid.polygon() can do multiple polygons in a single call, but rather
> than using NA's to separate sub-polygons, it uses an 'id' argument (or
> an 'id.lengths' argument) to identify sub-polygons within the vectors of
> x- and y-values (see the exam
Maybe this can be display-dependent. I just measured
the two diagonals on the plot and they're equal to
within less than one mm (possibly closer but that
was the accuracy of my measuring device (pencil and
strip of paper)).
-Peter Ehlers
On 2010-06-09 16:57, g...@ucalgary.ca wrote:
For me, the
Dear R Help,
I have a general question - I know this is the R list, but I hope
someone can help me out a little as I've always found the help here to
be absolutely fantastic.
I have run a psychological study where participants are given multiple
stimuli and their responses to those stimuli are me
Dear R-Helpers,
I am trying to bootstrap the coefficient of variation on a suite of
vectors, here I provide an example using one of the vectors in my
study. When I ran this script with the vector x <-c(0.625,
0.071428571, 0.1, 0.125, 0), it returned CV(boot) [the second
one], and s
Hello R help
I have a dataframe, with 71 samples (rows) and 30 variables. I got linear
models for some of the variables, and I want to join fitted and residuals of
these models to the data frame. Sometimes, these vectors have the same length
of the dependant variable, but in a few cases, NA va
For me, the angles circled are not exactly right.
See the pdf file: plot.pdf.
But it is OK.
Thanks Peter for your directions
-james
> On 2010-06-09 14:17, g...@ucalgary.ca wrote:
>> Thank.
>> Better. Seems that angles are close to but not equal to pi/2.
>> It may be because the plot box is not a sq
On 2010-06-09 14:17, g...@ucalgary.ca wrote:
Thank.
Better. Seems that angles are close to but not equal to pi/2.
It may be because the plot box is not a square: the length of
x-axis is not the same as the length of y-axis.
Even curves y = x and y = 1-x look like not orthogonal but
they should si
Hi everyone,
I'm using the metafor package to meta-analyze a set of proportions. This is
working really well for the raw proportions, but is there a way to
back-transform the arcsine transformed proportions in the rma or forest
functions with the atransf option? The estimates and CIs for the tr
It works! How fantastic a capability, and how embarrassing that I missed
that version change! Thanks very much!
-Harlan
On Wed, Jun 9, 2010 at 5:52 PM, Farrel Buchinsky wrote:
> Harlan Harris harris.name> writes:
>
> >
> > Hello,
> >
> > I'm trying to figure out how to use the RGoogleDocs pa
Harlan Harris harris.name> writes:
>
> Hello,
>
> I'm trying to figure out how to use the RGoogleDocs package from OmegaHat,
> and am having a bit of trouble. I emailed Duncan Temple Lang directly, but
> didn't receive a response, so I thought I'd try here to see if anyone else
> can help.
>
>
Samuel Okoye wrote:
I can't find his email and I have asked the same question to
bioconduc...@stat.math.ethz.ch
Regards,
Samuel
So you need the handy function maintainer() added to 2.11.0:
> require(GeneralizedHyperbolic)
> maintainer("GeneralizedHyperbolic")
[1] "David Scott "
David Sc
If your data came in using read.csv then it is most likely a data frame rather
than a matrix. Try as.matrix to convert it to a matrix and then use barplot.
If that does not work then give us a sample of your data and the exact commands
(copy/paste) that use used along with any errors/warnings.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Brigid Mooney
> Sent: Wednesday, June 09, 2010 1:49 PM
> To: r-help@r-project.org
> Subject: [R] Problem Matching Exact Values
>
> Sorry for the basic question - bur I ran into s
Hello dear R-help mailing list,
I wish to perform a non-parametric repeated measures anova.
If what I read online is true, this could be achieved using a mixed Ordinal
Regression model (a.k.a: Proportional Odds Model).
I found two packages that seems relevant, but couldn't find any vignette on
the
Good morning,
I've been dabbling in R, so my knowledge has quite a few holes in it. I'm
hoping that this has a simple answer and just falls into one of those holes.
I have a table of percentages that I want to display as a barchart. Groups
1-4 in columns and Variables 1-5 in rows, with the perce
Try this:
f$E <- diag(as.matrix(f[f$D]))
On Wed, Jun 9, 2010 at 11:03 AM, Malcolm Fairbrother <
m.fairbrot...@bristol.ac.uk> wrote:
> Dear all,
>
> I have a data frame f, with four variables:
>
> f <- data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
> f
> A B C D
> 1 0 0 1 3
Sorry for the basic question - bur I ran into something I haven't
noticed before and would appreciate a little more perspective on my
problem.
I am using R to determine if various thresholds are hit (or surpassed)
in a data set. If a threshold is surpassed, I have had no problems
identifying it.
To get the counts, assuming your data frame is called factors and it only
contains the 17 factors, you can do
n = nrow(factors)
aux = rep(1, n);
tab = tapply(aux, as.list(factors), sum);
example:
factors = matrix(sample(c(1:3), 3000, replace = TRUE), 1000, 3)
lfactors = as.list(data.fran = nrow(
Dear ExpeRts,
why does
splom(~iris[,1:4],scales = list(alternating = c(0,0), tck = c(0,0)))
not remove the ticks and labels (xyplot does)?
Cheers,
Marius
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do re
You could try something like this (untested):
> tmpfun <- function(i) {
+ tmp <- paste( "^k.", i, sep="")
+ apply( subset(mydf, select=grep(tmp, names(mydf) ) ), 1, min )
+ }
> out1 <- lapply(1:10, tmpfun)
> names(out1) <- paste( "min.k.", 1:10, sep="" )
> mydf2 <- cbind(mydf, out1)
--
Gre
Hi,
The maPalette() function (in the marray package) does not always
return the number of colors requested. More specifically, if a middle
color is given and the number of colors requested is odd, then
maPalette returns either one more or one less than the requested
number of colors:
maPalette(lo
If you cannot set the aspect ratio to 1 for the entire plot (Peter's solution),
then you may consider using either subplot or my.symbols (TeachingDemos
package) to create an asp=1 region within the plot that you can use to plot the
rotated rectangle.
The other option is to do a little more matr
What does coordinates() give for a polygon? centeroid?
I suppose I could look up the code, but it would be immensely helpful
if the help page could specify it.
regards
Nikhil
On Jun 9, 2010, at 3:50 PM, Rodrigo Aluizio wrote:
> I'm not sure if this is what you want. But the function
> coordi
Thank.
Better. Seems that angles are close to but not equal to pi/2.
It may be because the plot box is not a square: the length of
x-axis is not the same as the length of y-axis.
Even curves y = x and y = 1-x look like not orthogonal but
they should since multiplication of their slopes is -1.
-jame
Hi Josh,
One way would be:
res <- apply(b, 1, function(Names) t.test(a[, Names[1]], a[, Names[2]]))
do.call(rbind, lapply(res, function(l) c(l$statistic, l$parameter, p =
l$p.value)))
# t df p
# test.1 1.775490 17.35589 0.09335398
# test.2 -1.489210 15.82584
Try this:
subset(x, select = grep("cru", names(x)))
On Wed, Jun 9, 2010 at 2:41 PM, Josh B wrote:
> Hello R listserve,
>
> I would appreciate someone's help with this problem. Consider the following
> toy dataset:
>
> x <- read.table(textConnection("worldclim.1 worldclim.2 cru.1 cru.2
> indv.1
Hi Josh,
Here is a suggestion:
colnames(x)[1] <- 'mynewname'
colnames(x)
# [1] "mynewname" "apples""bananas" "cherries"
HTH,
Jorge
On Wed, Jun 9, 2010 at 1:45 PM, Josh B <> wrote:
> Hi all,
>
> I have a very simple problem that I cannot seem to find the answer to.
> Consider the follow
I'm not sure if this is what you want. But the function coordinates() in sp
package gives you the coordinates of SpatialObjects.
Regards.
Rodrigo.
2010/6/9 Nikhil Kaza
> You need to execute gpclibPermit() to enable gpclib.
>
> library(maptools) should have issued a warning to that effect.
>
>
Try:
names(x)[1] <- 'pears'
On Wed, Jun 9, 2010 at 2:45 PM, Josh B wrote:
> Hi all,
>
> I have a very simple problem that I cannot seem to find the answer to.
> Consider the following toy dataset:
>
> x <- read.table(textConnection("V1 apples bananas cherries
> indv.1 7 8 4 3
> indv.2 7 7 4 9")
I have dataframe with 17factors variables (for example every factor have
3levels)
I have maybe 5000 observation.
And i need to do table where is in every raw 1 of possible combination of
this factors and the numbur how many time is this combination in my dataset.
I wrote one code, but this is ver
Hello Listserve,
Here is another question to keep you on your toes. Please consider the
following toy dataset:
a <- read.table(textConnection("fred sam joe alex
measure.1 10 4 10 1
measure.2 10 4 2 8
measure.3 3 1 8 3
measure.4 5 1 3 3
measure.5 8 6 8 3
measure.6 9 5 1 0
measure.7 4 6 10 1
measu
Can someone tell me how to make up (eg) a library's html help files
by hand? I think I ought to be able to use RCMD Rdconv for this but
(R-2.10.0, MS-win) when I type (in a dos session) "rdcmd rdconv
--help" I get a message to the effect that a perl script rdconv can't
be opened.
Can I do t
Hello R listserve,
I would appreciate someone's help with this problem. Consider the following toy
dataset:
x <- read.table(textConnection("worldclim.1 worldclim.2 cru.1 cru.2
indv.1 7 8 32 658
indv.2 7 7 39 422"), header = TRUE)
How could I create a subset of the data based on the column prefi
Hi all,
I have a very simple problem that I cannot seem to find the answer to. Consider
the following toy dataset:
x <- read.table(textConnection("V1 apples bananas cherries
indv.1 7 8 4 3
indv.2 7 7 4 9"), header = TRUE)
How would I change the column name of ONLY the first column, not the othe
How about this:
f <- data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
N <- nrow(f)
mat <- cbind(1:N,f$D)
f$E <- f[mat]
f
A B C D E
1 0 0 1 3 1
2 0 1 1 1 0
3 1 0 0 2 0
4 1 1 1 3 1
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project
Ar,
I think I've found a bug in the behavior of the stats::cor function when
NAs are present, but in case I'm missing something, could you look over
this example and let me know what you think:
> a = c(1,3,NA,1,2)
> b = c(1,2,1,1,4)
> cor(a,b,method="spearman", use="complete.obs")
[1] 0.8164
Your transformation assumes that the x- and y-axes are on the
same scale. Add 'asp = 1' to your plot() call to set the
appropriate aspect ratio.
-Peter Ehlers
On 2010-06-09 10:13, g...@ucalgary.ca wrote:
Rectangle R centered at (x,y) with width 2w and height 2h is given by
x1=x-w
y1=y-h
x2=x
Dear all,
I'm trying to fit a variogram model using variofit function in geoR
package. There is an option to fix the nugget but is there anyway to force
the variogram sill to equal some defined value? I'm working with
standardized longitudinal data so the process variance is 1 and I'd like
I've really apreciated your advice! Thanks!
On Wed, Jun 9, 2010 at 14:05, Peter Ehlers wrote:
> Soapbox:
> Well, if you're just starting out with R it would be
> a VERY good idea to learn right away that T is not TRUE
> and F is not FALSE, at least not always. Sooner or
> later you WILL have pro
Why?
See ?seq
seq(0,1,.01)
--- On Wed, 6/9/10, suman dhara wrote:
> From: suman dhara
> Subject: [R] for loop incremented by 0.01
> To: r-h...@stat.math.ethz.ch
> Received: Wednesday, June 9, 2010, 12:30 PM
> Sir,
> I want to use a for loop which will be incremented by 0.01
> in 0 to 1.
> P
On 2010-06-09 5:11, Kaveh Vakili wrote:
Hi list,
in the Rglpk_solve_LP function (::Rglpk),
on line 26, the function calls a function
as.glp_bounds() that i cannot access.
i'm trying to alter the Rglpk_solve_LP function
to add a line to retrieve column/row dual values.
everytime i change the s
Uwe,
Very nice, thank you.
Rich
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide comm
It looks in the environment of your formula (which in this case is the
global environment) and then in data. Thus here are two solutions:
f2 <- function(c) {
environment(fo) <- environment()
nls(fo, data=d, start=list(a=1, b=1))
}
f3 <- function(c) nls(fo, data=c(d, c = c), start
Hi R users!
Is it possible to cut the edges from a surface graph without to cut the axes
using persp function?
I am using the following code:
op <- par(bg = "white")
persp(phi1, phi2, z,main="Bullwhip generated with AR(2) demand when L=1",
xlab ="phi1" , ylab ="phi2", zlab ="Bullwhip", t
Jie TANG gmail.com> writes:
>
> hi ,R user folks .
>
> Nowadays I read a paper which draw a probability ellipse circle figure
> shown in the appendix.
> I wonder how to draw this figure by R ?
> the x-axis and y-axis both express the error but in different direction .
Your question is quit
doris gomez yahoo.fr> writes:
>
> Dear lmer users,
You should probably redirect this sort of question to
r-sig-mixed-mod...@r-project.org , rather than the generic
R help list.
>
> The experiment includes 15 groups of (3 males and 1 female).
> The female is characterized by its quality Q1
Soapbox:
Well, if you're just starting out with R it would be
a VERY good idea to learn right away that T is not TRUE
and F is not FALSE, at least not always. Sooner or
later you WILL have problems. So do yourself a favour
and get into the habit of using TRUE/FALSE instead of T/F.
(I know that Pe
Thanks Erik
I can't figure out how to use the various x_apply functions in this setting,
nor post datasets to reproduce. But anyhow: the table structure is something
like this:
id (integer), handedness(R,L,A), gender(M,F), cat1(patient, control).
cat2(stroke, MS, dement, control), accuracy(int
Thanks to all replies.
I was able to get it running with the R.oo package.
I hope this reply makes it to the proper thread.
Michael
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Can your data.frame be properly coerced to a matrix like your example?
If so,
apply(f, 1, function(x) x[eval(x)["D"]])
Malcolm Fairbrother wrote:
Dear all,
I have a data frame f, with four variables:
f <- data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
f
A B C D
1 0 0 1
not tested
for(i in seq(0, 1, by = 0.01)) {
print(i)
}
Read FAQ 7.31 though.
suman dhara wrote:
Sir,
I want to use a for loop which will be incremented by 0.01 in 0 to 1.
Please help.
Regards,
Suman Dhara
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Hello,
This post is for Dr. Thomas Lumley or anybody familiar with the survey
package. I am estimating a proportional hazards model with weights using
svycoxph. Are there functions already built in the survey package that allow
me to do validation and calibration of the model?
Thanks
--
View t
Dr. Hadley Wickham - Data Visualisation in R: Harnessing the power of ggplot2
to produce elegant data graphics
London: 1st - 2nd November 2010
Mango Solutions is delighted to offer a one-off 2 day training course with Dr.
Hadley Wickham, R Project Data Visualisation Guru and creator of ggplot 2.
Thanks for all your help. I found adding the 'stringsAsFactors' condition
solved the problem.
On 1 June 2010 17:09, Joris Meys wrote:
> Hi Jessica,
>
> this tells me that your text is saved as a factor.
> Try :
> names <- read.csv(file="Names.csv",stringsAsFactors=F)
>
>
> Cheers
> Joris
>
> O
Dear all,
I have a data frame f, with four variables:
f <- data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
f
A B C D
1 0 0 1 3
2 0 1 1 1
3 1 0 0 2
4 1 1 1 3
I want to create a new variable (f$E), such that each of its elements is drawn
from either f$A, f$B, or f$C, accordin
dear R wizards: environment-related programming question:
myformula = y ~ a*x^2+b*x+c
d= data.frame( x=rnorm(20), y=rnorm(20) )
cat("\nunconstrained works: \n");
nls( myformula, data=d, start=list(a=1, b=1, c=1), trace= TRUE)
cat("\nconstrained works: \n");
b=1; nls( myformula, data=d, start=li
Sir,
I want to use a for loop which will be incremented by 0.01 in 0 to 1.
Please help.
Regards,
Suman Dhara
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PLEASE do read t
Dear listers,
Does anyone have any experience running marginal structural models in r or can
point me in the direction of any good tutorials on this?
Regards,
//M
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PLEAS
Hello,
This post is for Dr. Thomas Lumley or anybody familiar with the survey
package. I am estimating a proportional hazards model with weights using
svycoxph. Are there functions already built in the survey package that allow
me to do validation and calibration of the model?
Thanks
--
View th
On Wed, Jun 9, 2010 at 5:19 PM, Or Duek wrote:
> Hi,
> I would like to compare to regression models - each model has a different
> dependent variable.
> The first model uses a number that represents the learning curve for reward.
> The second model uses a number that represents the learning curve
Jon Erik Ween wrote:
Hi!
Would anyone know how to generate a list of variable names from a
data frame by the class of the variable?
a start...
df <- data.frame(f1 = factor(1:10),
f2 = factor(1:10),
n1 = 1:10,
n2 = 1:10)
sapply(df, class)
Hi all,
I wrote the following function to generate data following a mixture-ar1
model.
The model is described as below:
theta is a m-vector with each entries identically and
independent Bernoulli trials with
success probability pi1.
x is a m-vector with entries follow a ar1 mod
Rectangle R centered at (x,y) with width 2w and height 2h is given by
x1=x-w
y1=y-h
x2=x+w
y2=y-h
x3=x+w
y3=y+h
x4=x-w
y4=y+h
polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4))
Rotating a point (u,v) at (0,0) by theta degree is given by matrix
[cos(theta),-sin(theta)
sin(theta),cos(theta)]
so we have a new
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