On Wed, 14 Nov 2007, Christian Gold wrote:
> Thanks for your reply. I have tried the reproducible example you have
> provided (thanks!) in 3 different versions of R, and get what you get in R
> version 2.5.1, but in 2.3.1 (which I normally use) and 2.1.0 (which I still
> have installed) I get f
I don't understand the output from a pararell regression ,
I don't know how to extract components of the result .
can anyone send me files explaining the pararell regression detailly ?
I will be very grateful to you
--
View this message in context:
http://www.nabble.com/Is-there-any-document-ex
Thanks for your reply. I have tried the reproducible example you have
provided (thanks!) in 3 different versions of R, and get what you get in
R version 2.5.1, but in 2.3.1 (which I normally use) and 2.1.0 (which I
still have installed) I get factors.
I guess this means I should switch to the ne
try this
x<-c(1,2,3,2,3,1,1,3,2)
A<-matrix(x,3,3,byrow=T)
B<-matrix(c(1,2,2,1),2,2)
A1<-matrix(rep(A,each=2),6,3)
B1<-rbind(B,B,B)
data.frame(A1,B1)
X1 X2 X3 X1.1 X2.1
1 1 2 312
2 1 2 321
3 2 3 112
4 2 3 121
5 1 3 212
6 1 3 221
Hi elijah
Thank you again very much, the case I am having is simple data like
the following:
NodetotalOutgoing link1 link2
1 2 2 3
2 2 4 5
3 2 5 6
4 2 7 8
5 2 8 9
6 2 9 10
7
On Nov 13, 2007 5:53 AM, Shoaaib Mehmood <[EMAIL PROTECTED]> wrote:
> i want to make survival plots for a coxph object using survfit
> function. mod.phm is an object of coxph class which calculated results
> using columns X and Y from the DataFrame. Both X and Y are
> categorical. I want survival
Try cronbach in the psy package
Regarding your other question, see classic web reference below.
http://www.ats.ucla.edu/STAT/SPSS/library/negalpha.htm
Also, note Harold Doran's explanation from a similar R listserv post back in
January
http://tolstoy.newcastle.edu.au/R/e2/help/07/01/9125.html
alpha.scale (psych)
cronbach (multilevel)
On Nov 14, 2007 7:33 AM, raymond chiruka <[EMAIL PROTECTED]> wrote:
> hie
> 1...i'm trying to carryout a relibility testusing cronbach's alpha what
> fuctin do i use.
>
> 2.. this is more of a statistical question.if the alpha value for all
> th
On Nov 13, 2007 10:28 PM, Daniel Malter <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I want to illustrate when individuals had an event over time. I am using
> dotplot from the lattice library. The plot itself works well, but as I have
> many individuals, the labels on the y-axis for these individuals over
Dear Brian,
Thanks for your prompt response.
I had tried both 'relevel' and 'reorder'
preg_fyear5a.msm<-msm(outcome~ipi, subject=id, data, qmatrix=crudemat,
exacttimes=TRUE, covariates=~factor(relevel(year5a),1))
preg_fyear5a.msm<-msm(outcome~ipi, subject=id, data, qmatrix=crudemat,
exacttimes=T
On Nov 13, 2007 7:23 PM, James D Forester <[EMAIL PROTECTED]> wrote:
>
>
> hadley wrote:
> >
> >
> > That's from a previous version, now you need to do:
> >
> > p + scale_x_continuous("Predictor") + scale_y_continuous("Response")
> >
> > (or scale_x_discrete if discrete)
> >
> >
>
> Thanks Hadley -
hadley wrote:
>
>
> That's from a previous version, now you need to do:
>
> p + scale_x_continuous("Predictor") + scale_y_continuous("Response")
>
> (or scale_x_discrete if discrete)
>
>
Thanks Hadley -- that worked fine. However, I just noticed that when I try
to add an abline to this plo
Pardon me for intruding, but I had recently to explain something
analogous to this to a distinguished physician who had hurt himself
playing with survival models and was bleeding graphs and analyses all
over the place...
Dylan Beaudette a écrit :
> On Tuesday 13 November 2007, Prof Brian Ripley wr
On Tuesday 13 November 2007, Peter Dalgaard wrote:
> Prof Brian Ripley wrote:
> > On Tue, 13 Nov 2007, Dylan Beaudette wrote:
> >> Hi,
> >>
> >> I have setup a simple logistic regression model with the glm() function,
> >> with the follow formula:
> >>
> >> y ~ a + b
> >>
> >> where:
> >> 'a' is a
Hi Amit,
Please read carefully the Mailing List Posting Guide (available at
http://www.r-project.org/posting-guide.html). In particular, this section:
"For new subjects, compose a new message and include the
'[EMAIL PROTECTED]' (or '[EMAIL PROTECTED]') address
specifically. (Replying to an ex
Here's a version using reshape and ggplot:
mydata <- read.table("http://ca.geocities.com/jrkrideau/R/heartdata.txt";,
sep="\t", header=FALSE)
mydata[,1] <- as.Date(mydata[,1],"%m/%d/%y")
names(mydata) <- c("dates", "sy","dys","pulse", "weight")
molten <- melt(mydata, m = c("sy", "dys"))
qplot(da
On Nov 13, 2007 4:30 PM, James D Forester <[EMAIL PROTECTED]> wrote:
>
> Hi all,
>
> For various reasons, I need to use ggplot instead of qplot for a complex
> figure. Everything is working fine, except I cannot figure out how to rename
> the axis labels in ggplot. I have pasted a simple example be
Prof Brian Ripley wrote:
> On Tue, 13 Nov 2007, Dylan Beaudette wrote:
>
>
>> Hi,
>>
>> I have setup a simple logistic regression model with the glm() function, with
>> the follow formula:
>>
>> y ~ a + b
>>
>> where:
>> 'a' is a continuous variable stratified by
>> the levels of 'b'
>>
>>
>> Lo
Hi ,
I am trying to modify the same example script to calculate AOV.
Below is the script file (aov.R) I am trying to execute:
aov.R
--
data1<-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40
,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40)
ma
Unfortunately, your code is not self-contained.
Please provide a completely self-contained set of commands which illustrate
the problem (as described in the posting guide).
If you mean a problem that the boundaries don't quite line up, as illustrated
by:
map("france")
map("world", col=2, add=T)
hie
1...i'm trying to carryout a relibility testusing cronbach's alpha what
fuctin do i use.
2.. this is more of a statistical question.if the alpha value for all the
variables is negative what does it mean. and if the alpha value is negative
for all tyha variables but is greate
You can get a slightly better X axis by converting your Date variable to chron:
Replace this:
plot(mydata[,1], ...)
with this:
library(chron)
plot(chron(unclass(mydata[,1])), ...)
and ignore the warning.
On Nov 13, 2007 9:08 AM, John Kane <[EMAIL PROTECTED]> wrote:
> I clearly spok
Gang Chen-3 wrote:
>
> Sorry to hijack this thread. I have a similar but slightly different
> situation. Using the original poster's example, how to elegantly get
> the mean of column V2 when column V1 is either A or C and F1 is 0?
>
>
I am not sure if this is elegant, but it works:
a=me
Lets take a look at your solution:
> mat1 <- matrix(0, nrow=10, ncol=3)
> dimnames(mat1) <- list(paste('row', 1:10, sep=''), LETTERS[1:3])
> mat2 <- matrix(1:3, ncol=1, dimnames=list(c('row3', 'row7', 'row5'), "B"))
> mat2
B
row3 1
row7 2
row5 3
> mat1[rownames(mat2)%in%rownames(mat1),"B"]=ma
Hi,
I am trying to run an analysis with the package maanova and I am not
getting success.
I suppose that I am wrong on set up the formula, so the issue may not
be related to R, properly.
I have two varieties of plants (V1 and V2). A group of each ones were
treated and another was not treated. Af
On Tuesday 13 November 2007, Prof Brian Ripley wrote:
> On Tue, 13 Nov 2007, Dylan Beaudette wrote:
> > Hi,
> >
> > I have setup a simple logistic regression model with the glm() function,
> > with the follow formula:
> >
> > y ~ a + b
> >
> > where:
> > 'a' is a continuous variable stratified by
>
Hi all,
For various reasons, I need to use ggplot instead of qplot for a complex
figure. Everything is working fine, except I cannot figure out how to rename
the axis labels in ggplot. I have pasted a simple example below. Any ideas
on what I am doing wrong?
Thanks for your help.
James
libra
Hi,
I want to illustrate when individuals had an event over time. I am using
dotplot from the lattice library. The plot itself works well, but as I have
many individuals, the labels on the y-axis for these individuals overlap. So
the y-axis is unreadable. Therefore, I want to suppress the labels o
Sorry to hijack this thread. I have a similar but slightly different
situation. Using the original poster's example, how to elegantly get
the mean of column V2 when column V1 is either A or C and F1 is 0?
Thanks,
Gang
On Nov 13, 2007, at 5:30 AM, Petr PIKAL wrote:
> Hi
>
> [EMAIL PROTECTED]
On Tue, 13 Nov 2007, Dylan Beaudette wrote:
> Hi,
>
> I have setup a simple logistic regression model with the glm() function, with
> the follow formula:
>
> y ~ a + b
>
> where:
> 'a' is a continuous variable stratified by
> the levels of 'b'
>
>
> Looking over the manual for model specification,
Here is how to wittle it down for the first two parts of your
question. I am not exactly what you are after in the third part. Is
it that you want specific DATEs or do you want the ratio of the
DATE[max]/DATE[min]?
> x <- read.table(textConnection("CODENAME
Hello R-gurus,
my research on archives gave on only result for the following query:
is there a way to output images (graphs) in not-buggy pgf or tikz
format?
I only know the rgl package to do that, but pgf output is buggy (say,
I did't find the way to compile into a latex article the pgfpicture
r
Hi,
I have setup a simple logistic regression model with the glm() function, with
the follow formula:
y ~ a + b
where:
'a' is a continuous variable stratified by
the levels of 'b'
Looking over the manual for model specification, it seems that coefficients
for unordered factors are given 'aga
jim holtman wrote:
> Here is one way of doing it that uses the row and column names:
>
>> # create test data
>> mat1 <- matrix(0, nrow=10, ncol=3)
>> dimnames(mat1) <- list(paste('row', 1:10, sep=''), LETTERS[1:3])
>> mat2 <- matrix(1:3, ncol=1, dimnames=list(c('row3', 'row7', 'row5'), "B"))
>> ma
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear R-Users!
I am trying to fit a spline to a time-series with about 90 values.
I would like to specify about four to five knots and I would like the
values returned to be similar in fashion to the output of interpSpline -
I would only like the sp
I have had some problems with dates on axis in certain ranges. Here
is a version of axis.POSIXct that I have been using that seems to do a
better job:
my.axis.POSIXct <-
function (side, x, format, inc)
{
.diff <- diff(range(unclass(x), na.rm = TRUE))
if (.diff < 30) {
.by <- "sec
Hi Lauri,
A quick hack:
a <- lapply(g, 1, rle)
b <- sapply(a, function(x), x[[1]][x[[2]] == 1])
d <- cbind(rep(1:length(b), sapply(b, length)),
rep(sapply(b, length), sapply(b, length)),
unlist(b))
Best,
Jim
Lauri Nikkinen wrote:
> Hi R-users,
>
> I have a matrix si
I am trying to build a data.frame from some vectors and some matrices
and seem to be unable to find out how to do this without converting
everything to factors.
I can prevent conversion of the vectors by using I(), as explained in
help(data.frame). However, this doesn't help with the matrices: if I
Yip a écrit :
> Hello,
>
> I was trying a glm fitting (as shown below) and I got a warning and a fitted
> residual deviance larger than the null deviance. Is this the expected
> behavor of glm? I would expect that even though the warning might be
> warranted I should not get worse fitting with an
On Tue, 13 Nov 2007, Yip wrote:
> I was trying a glm fitting (as shown below) and I got a warning and a
> fitted residual deviance larger than the null deviance. Is this the
> expected behavor of glm?
It can be. One of two things is going on here:
A) There is complete separation between the c
Well, the obvious (but perhaps not the most elegant) solution is put
everything in a loop and run it 600 times.
coefficients=matrix(NA,ncol=3,nrow=600)
for (loop in 1:600){
[all your code here]
coefficients[loop,]=coef(log_v)
}
That will give you a matrix with the coefficients of each model
"Hans Werner Borchers" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> I've got some time series representing measurements from a physical
> process, like atomic decay data. These time series look almost
> random, but should hopefully be distinguishable as they were taken
> under dif
Thank you, I changed that and it's much more efficient.
Sigalit.
On 11/13/07, Phil Spector <[EMAIL PROTECTED]> wrote:
>
> You can use the replicate function to do your simulation. First,
> put the code to do one repetition in a function:
>
> dosim0 = function(n=500){
> x <- 0
> y <- 0
> z
On 13/11/2007, at 8:54 PM, Frede Aakmann Tøgersen wrote:
> R is case sensitive ;-) so in your list of start values you must
> have B=3 and not b=3.
Duhh!!! Typo.
> Perhaps you have an object named b of a different length than your
> data somewhere giving rise to your error messa
I want to repeat the simulation 600 times and to get a vector of 600
coefficients for every covariate: aps and tiss.
Sigalit.
On 11/13/07, Julian Burgos <[EMAIL PROTECTED]> wrote:
>
> And what is your question?
>
> Julian
>
> sigalit mangut-leiba wrote:
> > Hello,
> > I have a simple (?) simulati
Dear R users,
I have a huge database and I need to adjust it somewhat.
Here is a very little cut out from database:
CODENAME DATE
DATA1
4813ADVANCED TELECOM19870.013
3845ADVANCED THERAPEUTIC SYS LT
Hello,
I was trying a glm fitting (as shown below) and I got a warning and a fitted
residual deviance larger than the null deviance. Is this the expected
behavor of glm? I would expect that even though the warning might be
warranted I should not get worse fitting with an additional covariate in t
Hello,
I am trying to run a code (see reference section) and when I run the line:
fit<-lmFit(xen1dataeset,design),
I get the error message: Error in lm.fit(design, t(M)) : incompatible
dimensions
I will really appreciate if someone can help me understand this error
message and possibl
On Tue, 13 Nov 2007, Baize, Harold wrote:
> Tudor Bodea asked:
>
>> In this context, I try to get all the records for which market is atl-bos,
>> competitor is delta and dd is 2007-11-20 (first record above). To do this I
>> used
>
>>> # channel <- odbcConnectAccess("test.mdb")
>>> res <- sqlQuer
Baize, Harold a écrit :
> Tudor Bodea asked:
>
>> In this context, I try to get all the records for which market is atl-bos,
>> competitor is delta and dd is 2007-11-20 (first record above). To do this I
>> used
>
>>> # channel <- odbcConnectAccess("test.mdb")
>>> res <- sqlQuery(channel, "selec
Dear all:
I am trying to build a data.frame from some vectors and some matrices
and seem to be unable to find out how to do this without converting
everything to factors.
I can prevent conversion of the vectors by using I(), as explained in
help(data.frame). However, this doesn't help with the mat
And what is your question?
Julian
sigalit mangut-leiba wrote:
> Hello,
> I have a simple (?) simulation problem.
> I'm doing a simulation with logistic model and I want to reapet it 600
> times.
> The simulation looks like this:
>
> z <- 0
> x <- 0
> y <- 0
> aps <- 0
> tiss <- 0
> for (i in 1:5
What you are looking for is the findInterval() function.
Julian
W Eryk Wolski wrote:
> Hi,
>
> It's just some example code.. The application is uninteresting. I am
> searching for some functionality.
>
> X <- rnorm(100) //my data
>
> Y <- seq(-3,3,by=0.1) // bin boundaries.
>
> Now I would li
And here is a second solution, that differs in what happens if the variables
have differing lengths:
> var1 <- 1:4
> var2 <- 1:3
> sapply(ls(patt="^var[0-9]"), get)
$var1
[1] 1 2 3 4
$var2
[1] 1 2 3
> do.call("cbind", lapply(ls(patt="^var[0-9]"), get))
[,1] [,2]
[1,]11
[2,]2
Try this:
> A <- data.frame(a1 = c(1, 2, 1), a2 = c(2, 3, 3), a3 = c(3, 1, 2))
> B <- data.frame(b1 = 1:2, b2 = 2:1)
>
> library(sqldf)
> sqldf("select * from A, B")
a1 a2 a3 b1 b2
1 1 2 3 1 2
2 1 2 3 2 1
3 2 3 1 1 2
4 2 3 1 2 1
5 1 3 2 1 2
6 1 3 2 2 1
On Nov 13,
Try this:
sapply(ls(patt="^var[0-9]"), get)
On 13/11/2007, livia <[EMAIL PROTECTED]> wrote:
>
> Hello everyone,
>
> I would like to use the "cbind" function to construct a dataset, combining
> some variable I defined before.
> The codes are something like
>
> var1 <- ...
> var2 <- ...
> var2 <- .
Divaker,
Thanks for the data.
For me,
> summary(aov(Purity~Supplier/Batch, process))
gives exactly the same results for mean squares as
> aov(Purity~Supplier+Error(Supplier/Batch), process)
except that the latter gives no p-values (because Supplier appears as
both error term and fixed effect, th
Try this:
merge(B, A)
On 13/11/2007, sun <[EMAIL PROTECTED]> wrote:
> I have two data frame A and B adn want to cross them.
> A has format as:
>
> a1 a2 a3
> 1 23
> 2 31
> 1 32
> ...
>
> B:
>
> b1 b2
> 1 2
> 2 1
> ...
>
> the combine result shall be something like
>
> a1 a
Dear time-series specialist:
I've got some time series representing measurements from a physical
process, like atomic decay data. These time series look almost
random, but should hopefully be distinguishable as they were taken
under different conditions.
I am looking for statistical approaches th
Hello everyone,
I would like to use the "cbind" function to construct a dataset, combining
some variable I defined before.
The codes are something like
var1 <- ...
var2 <- ...
var2 <- ...
...
data <- cbind(var1,var2,var3...)
The problem is I would like some flexibity in the data, i.e. the numbe
I have two data frame A and B adn want to cross them.
A has format as:
a1 a2 a3
1 23
2 31
1 32
...
B:
b1 b2
1 2
2 1
...
the combine result shall be something like
a1 a2 a3 b1 b2
1 2 3 1 2
1 2 3 2 1
2 3 1 1 2
2 3 1 2 1
1 3 2 1 2
1 3
Tudor Bodea asked:
>In this context, I try to get all the records for which market is atl-bos,
>competitor is delta and dd is 2007-11-20 (first record above). To do this I
>used
>># channel <- odbcConnectAccess("test.mdb")
>>res <- sqlQuery(channel, "select * from test_table where market = 'atl-
do you mean something like this:
tapply(x, g, function (y) {
sapply(1:length(y), function (i) mean(y[-i]))
})
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuv
Hi
I am running an lmer which works fine with family=poisson
mixed.model<-lmer(nobees~spray+dist+flwabund+flwdiv+round+(1|field),family="poisson",method="ML",na.action=na.omit)
But it is overdispersed. I tried using family=quasipoisson but get no P
values. This didnt worry me too much as i thin
Answering to myself...
Emmanuel Charpentier a écrit :
> Casey Klofstad a écrit :
>> I need advice on how to create a variable that is the group mean of
>> another variable.
>>
>> For example, I have a variable called x for which each row in the data
>> set has a value. I also have a nominal variabl
Dear all,
I sample without replacement elements of a vector and generate a new
vector:
kl<-c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,8,8,
8,8,8,8,8,8,8)
the_index<-c(sample(40,35))
for(fs in
1:length(the_index)){if(fs==1){s<-c(kl[the_index[fs]])}else{s<-
append
I clearly spoke too soon.
With the actual data I am not getting sensible x-axis
units. The program with the actual data below. Graph
output is here:
http://ca.geocities.com/jrkrideau/R/hd.png .
I seem to be getting only a single entry for the
x-axis of "2007". However dates range from
Firs
I used the following to assign each row the group average, where w is
the new group average variable, x is the variabale to be averaged, and
g is the nominal group indicator:
w <- ave(x,g)
Now I want to calculate the group average, but WITHOUT each row's
value of x. Is there an easy way to do thi
On Tue, 13 Nov 2007, Muiser, EC wrote:
> Thank you very much Alberto! Atleast I can see there is an error now.
> The error however is strange. If I load the library in an R console it works
> perfectly but the package seems not to exist if I run R from a script. The
> library should be there.
>
I think you are looking for eval(quote(x), ...) or evalq(x, ...) .
Those do what you expect. The code you show evaluates
x before eval even gets a chance to deal with it.
On Nov 13, 2007 9:48 AM, Joerg van den Hoff <[EMAIL PROTECTED]> wrote:
>
> On Tue, Nov 13, 2007 at 09:04:25AM -0500, Duncan Mu
On Nov 13, 2007 1:02 AM, Rick Bilonick <[EMAIL PROTECTED]> wrote:
> On Tue, 2007-11-13 at 01:03 -0500, Rick Bilonick wrote:
>
> >
> > Is there some way to get ranef with postVar=TRUE to show what the
> > variances are, or what the lower and upper bounds are? qqmath makes nice
> > plots but I need t
On Tue, Nov 13, 2007 at 09:04:25AM -0500, Duncan Murdoch wrote:
> On 11/13/2007 8:39 AM, Joerg van den Hoff wrote:
> >my understanding of `eval' behaviour is obviously
> >incomplete.
> >
> >myquestion: usually `eval(expr)' and `eval(expr,
> >envir=parent.frame())' should be i
Stefano,
It did work - for some reason, the date components to be used in the sqlQuery
need to be placed in between hashes (i.e., #) for them to be interpreted
correctly.
Thank you so much.
Tudor
Quoting Guazzetti Stefano <[EMAIL PROTECTED]>:
> It seems that Access needs that you surround the
Dear all,
This is probably off topic, but I did not know where to turn and since
most of you have an excellent knowledge of statistics, I thought of
sending my question to this list.
I was wondering whether there is some literature on (or a way to
calculate) how many data points one would need to
Hi R-user,
I am new with R and I have a problem with the map and mapproj fonctions :
with the following code :
>test_proj=mapproject(LONG_d01_vec_t1, LAT_d01_vec_t1, projection="lambert",
parameters=c(30,60))
>longitude_vec=test_proj$x
>latitude_vec=test_proj$y
>longitude_mat=matrix(longitude_
http://www.openstreetmap.org/ might be one place to start.
Hadley
On Nov 13, 2007 6:30 AM, Albrecht Kauffmann <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> who knows how to get an ESRI Shapefile for the NUTS-2 Regions of the
> enlarged European Union? Particularly I want to draw maps of Germany,
> Pol
Casey Klofstad a écrit :
> I need advice on how to create a variable that is the group mean of
> another variable.
>
> For example, I have a variable called x for which each row in the data
> set has a value. I also have a nominal variable called g that
> indicates which of 100 different groups ea
Yes. The variances are an attribute, and you can grab them like this:
fm1 <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy)
attr(ranef(fm1, postVar = TRUE)[[1]], "postVar")
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Rick Bilonick
> Sent:
On 11/13/2007 8:39 AM, Joerg van den Hoff wrote:
> my understanding of `eval' behaviour is obviously
> incomplete.
>
> myquestion: usually `eval(expr)' and `eval(expr,
> envir=parent.frame())' should be identical. why does the
> last `eval' (yielding `r3') in this code
Thank you very much Alberto! Atleast I can see there is an error now.
The error however is strange. If I load the library in an R console it works
perfectly but the package seems not to exist if I run R from a script. The
library should be there.
Errorlog:
Error in library(tgp) : there is no
Thanks...
Thats exactly what I was looking for..
Eryk
On Nov 13, 2007 2:38 PM, Eik Vettorazzi <[EMAIL PROTECTED]> wrote:
> cut(X,Y,right=F)
>
> W Eryk Wolski schrieb:
>
> > Hi,
> >
> > It's just some example code.. The application is uninteresting. I am
> > searching for some functionality.
> >
Hi,
see ?cut
?pretty can also help to define "boundaries".
Regards,
Thibaut.
> Hi,
>
> It's just some example code.. The application is uninteresting. I am
> searching for some functionality.
>
> X <- rnorm(100) //my data
>
> Y <- seq(-3,3,by=0.1) // bin boundaries.
>
> Now I would like to gene
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Hash: SHA1
Hua Li wrote:
> Thank you very much for the help, Ben!
>
> As a follow up, is there a way to specify the labels,
> through the way the text is written, rather than
> reading the edge positions from the graph? For
> example,
>
> mytree =
> "((A:51.78
my understanding of `eval' behaviour is obviously
incomplete.
myquestion: usually `eval(expr)' and `eval(expr,
envir=parent.frame())' should be identical. why does the
last `eval' (yielding `r3') in this code _not_ evaluate in
the local environment of function `f' b
W Eryk Wolski wrote:
> Hi,
>
> It's just some example code.. The application is uninteresting. I am
> searching for some functionality.
>
> X <- rnorm(100) //my data
>
> Y <- seq(-3,3,by=0.1) // bin boundaries.
>
> Now I would like to generate a - list of factors, length as X...
> i.e.: all va
cut(X,Y,right=F)
W Eryk Wolski schrieb:
> Hi,
>
> It's just some example code.. The application is uninteresting. I am
> searching for some functionality.
>
> X <- rnorm(100) //my data
>
> Y <- seq(-3,3,by=0.1) // bin boundaries.
>
> Now I would like to generate a - list of factors, length as X..
one approach is the following:
res <- apply(g, 1, function (x) {
out <- rle(as.logical(x))
out <- out$lengths[out$values]
ln <- length(out)
c(rep(ln, ln), out)
})
cbind(rep(1:nrow(g), sapply(res, length) / 2),
do.call("rbind", sapply(res, matrix, ncol = 2)))
I hope it helps.
Hi,
It's just some example code.. The application is uninteresting. I am
searching for some functionality.
X <- rnorm(100) //my data
Y <- seq(-3,3,by=0.1) // bin boundaries.
Now I would like to generate a - list of factors, length as X...
i.e.: all values in the range [-3,-2.9) have the same f
Muiser, EC wrote:
>
> The php script executes the following command: "/usr/bin/R --quiet --
> slave < /perl/outfiles/Rscript19785065.R
> /perl/outfiles/error19785065.txt"
> Also there is no errorlog (in
> "/perl/outfiles/error19785065.txt")
This part is easy: the error output should go to 2>
Hi R-users,
I have a matrix similar to this. I would like to calculate the number
of 1 sequences in a row and also the length of the 1 sequence. For
instance, in the example below, row number one has four sequences of
number 1 (from left to right: 1; 1; 1,1; 1) and the corresponding
SeqCount is 4
sushi4u wrote:
> Dear R-help,
>
> I have to calculate the percent inclusion of each variable in a bootstrap
> validation of a cox proportional hazards model(described in Sauerbrei and
> Schumacher, Stat Med 11:1093, 1992).
>
This approach is not recommended. Collinearities can ruin the resul
Hello everyone,
I am busy making a website that uses R to plot certain biological data, but
there is a problem when I try load packages.
The php script executes the following command: "/usr/bin/R --quiet --slave <
/perl/outfiles/Rscript19785065.R > /perl/outfiles/error19785065.txt"
The R scri
Hi all,
who knows how to get an ESRI Shapefile for the NUTS-2 Regions of the
enlarged European Union? Particularly I want to draw maps of Germany,
Poland, Czech Republik, Hungary and Austria. I've found Shapefiles for the
US, Russia and other countries elsewhere in the web, but for Europe it
seems
We can do this using chron and zoo. First turn it into
a zoo series, z, using chron times, tt, which may not only
be useful for this but also for other purposes.
The ave(...) command gets a vector as long as the series with
the minimum dates in year each and then we aggregate z
by that using max.
i want to make survival plots for a coxph object using survfit
function. mod.phm is an object of coxph class which calculated results
using columns X and Y from the DataFrame. Both X and Y are
categorical. I want survival plots which shows a single line for each
of the categories of X i.e. '4' and
It seems that Access needs that you surround the dates with a # symbol.
You probably need something like.
res <- sqlQuery(channel, "select * from test_table where market = 'atl-bos'
and competitor = 'delta' and dd = #2007-11-20#")
Hope this helps,
Stefano
-Messaggio o
Thank you very much for your answers,you are all very kind!Excuse me, can I
make another question about time series?
When I use arimaId or best.arima function in package forecast, I can't
insert regressors, I can insert them only when I make predictions..
But isn't the model influenced by regresso
Thanks for the help received, I will test the proposal.
> R is case sensitive ;-) so in your list of start values you must have B=3 and
> not b=3. Perhaps you have an object named b of a different length than your
> data somewhere giving rise to your error message.
>
> Also brute force wont work
On Tue, 13 Nov 2007, Eric R. wrote:
>
> This may not be the smoothest way to produce the result, but here is a
> two-step solution:
>
> first find the formula within the object created by the call to glm():
>
> form <- qq["formula"] or equivalently form <- qq$formula
Oh, please , do use extractor
Hello to everyone, thank you for your answers, you are all very kind!
I'll have a look to the package you advised me.
Paul, I saw dse package, but I didn't understand how to use it..
For example, I have 2 sales series:
item_1 item_2 price_item1 price_item2
0 0 26
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