Could somebody share some tips on implementing multivariate integration and
partial differentiation in R?
For example, for a trivariate joint distribution (cumulative density function)
of F(x,y,z), how to differentiate with respect to x and get the bivariate
distribution (probability density fu
Hello, I'm switching over from SAS to R and am having trouble merging data
frames. The data frames have several columns with the same name, and each has a
different number of rows. Some of the values are missing from cells with the
same column names in each data frame. I had hoped that when I m
Hi All,
I'll appreciate your help on this. Do you know of any package that can be
used to solve optimization problems subject to general *non-linear* equality
constraints.
Thanks!
Lars.
[[alternative HTML version deleted]]
__
R-help@r-project
To r-help Forum:
I have downloaded R 2.8.1 and stalled it on my WinXP platform in folder:
"C:\Program Files\R\R-2.8.1\bin\R.exe"
R has worked properly with my data files and packages, which one call, past.
However, this week I decided to move the data files and packages to
different location:
Waldir Leôncio wrote:
Is there an easy way to add a thousand separator mark on the axis of a
plot? The best solution I've found so far is the following:
y <- seq(0, 10, 1)
plot(y, yaxt = "n", ylab = "")
axis(2, at = y, labels = formatC(y, big.mark = " ", format = "d"), las=2)
But that
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
> when I try to run an R command eg, "x <- rnorm(100,10,5)" vim returns this
> error:
> "~/.pipe" E212: can't open file for wirting"
Hi Andreas,
I'd recommend to use this script instead. It uses screen to
communicate R and vim, it works well.
http:/
One thing I notice immediately is a number of NA values for your
coefficients. If I were you, I would try a model with less parameters, and
use the anova() function to compare models, to see if the extra terms really
improve the model.
e.g.
fit1 <- glm(Y~X1+X2+X3,...)
fit2 <- glm(Y~X1+X2+X3+X1:X2,.
Hi list,
I have a real problem with plotting US state map. When I try to plot the
northern state, there will be some blank space in the top of graph (see case
1 example), and when I plot southern states, there will be a blank space in
the bottom of plot (see case 2). I spent almost 2 days to figure
Hi everyone,
Although this question is more related to ChIP and ChIP-seq, it could be
probably anchored in a more general statistical context.
The question is : what method is better to assess the significance of the
change in a signal (the signal can be DNA binding, for instance) given the
bac
?lag
?xtabs
songseq <- read.table(textConnection("NrSongtype
1 S1
2 S2
3 S3
4 S1
5 S1
6 S2
7 S4"), header=TRUE, stringsAsFactors=FALSE)
songseq$precsong <-c(NA,lag(songseq$Songtype)[1:6])
# need to drop last entry
xtabs( ~ precsong + Songtype,
Hi all,
I´m fitting GLM´s and I can´t interprete the coefficients when I run a
model with interaction terms.
When I run the simpliest model there is no problem:
Model1<-glm (Fishes ~ Year + I(Year^2) + Kind.Geographic +
Kind.Fishers + Zone.2 + Hours + Fishers + Month, family =
poisson(lo
Hello,
The example shows a sequence of songtypes a bird has sang.
The entire list contains several thousand songs which the bird has produced
consecutively.
Is there any convenient way in R to produce a contingency table which shows how
often a special song type was sang after a special other son
chaogai [Thu, Mar 05, 2009 at 07:04:19PM CET]:
> I'm having similar experiences on my Acer Aspire One. Everything will
> work good. Only thing that takes a lot of time is compiling R if you are
> in the habit of doing so.
>
On the Fedora version that came with my Acer Aspire One, I am even think
Dear all,
I have a dataset where the interaction is more than obvious, but I was asked
to give a p-value, so I ran a logistic regression using glm. Very funny, in
the outcome the interaction term is NOT significant, although that's
completely counterintuitive. There are 3 variables : spot (binary
On Fri, Mar 06, 2009 at 09:46:17AM -, Ted Harding wrote:
> On 06-Mar-09 09:25:26, Prof Brian Ripley wrote:
> > You might want to look into correspondence analysis, which has several
> > variants of PCA designed for categorical data.
>
> In particular, have a look at the results of
>
> RSit
Is there an easy way to add a thousand separator mark on the axis of a
plot? The best solution I've found so far is the following:
y <- seq(0, 10, 1)
plot(y, yaxt = "n", ylab = "")
axis(2, at = y, labels = formatC(y, big.mark = " ", format = "d"), las=2)
But that seems like quite a hassl
> "Liaw," == Liaw, Andy writes:
> Are you sure that's dual atoms? AFAIK it has a single Atom
> N270 (single core) at 1.6GHz. With hyper-threading, you
> may see "two cpus".
Yep - that is exactly what is going on.
Mike
__
R-help@r-pr
You might also look at http://www.portableapps.com for a portable version of
several editors, including gVim, Notepad++ and SciTE which all have some level
of support for R
--
"I like nonsense, it wakes up the brain cells. Fantasy is a necessary
ingredient in living, it's a way of looking at l
Hello,
please post to the list instead of posting me directly.
I am sorry to say, but I don't really know what you want to do. First
you asked about highly connected nodes, and now about clustering.
Perhaps you could give an example with the input and the expected
output.
Btw, if you know how to
Thanks! That solved it. I obviously had done something stupid. I just
redefined my options(digits=7) and they are all there. Sorry for the
silly question, and thanks for the suggestion towards options().
Thanks!
John
David Winsemius wrote:
> STX<-c(16.0962, 16.1227, 16.0921, 16.1498)
> ST
Hello,
I know I am forgetting to do something silly. I typed coordinates in
vectors (as below) but when I call them in R they come out as integers,
and I want them to be real numbers. I have tried using as.numeric,
as.real, etc... but they are still read by R as integers.
STX<-c(16.0962, 1
Dear all -
I do fret this to be a revealing beginner question - fortunately, this
mailing list have been good to me in the paste .-)
I am looking for a good R editor/environment in ubuntu. To that end, I have
decided to dive into gvim as the modality offered here seems to make sense
for editing.
To do that be sure to give each dimension, itself, a name when
the matrix is defined:
mat <-matrix(c(266, 359, 533, 313, 555, 504, 502, 242), nrow = 4,
dimnames = list(Region = c("Northeast", "Midwest", "South", "West"),
Type = c("Public", "Private")))
as.da
> STX<-c(16.0962, 16.1227, 16.0921, 16.1498)
> STY<-c(2.0387, 2.0214, 1.9877, 1.9846)
> STX
[1] 16.0962 16.1227 16.0921 16.1498
> STY
[1] 2.0387 2.0214 1.9877 1.9846
Did you perhaps redefine c()?
Or:
options()$digits
If not, then what do these say:
str(STX)
str(STY)
On Mar 6, 2009, at 1:47
There are real numbers:
> STX<-c(16.0962, 16.1227, 16.0921, 16.1498)
> STY<-c(2.0387, 2.0214, 1.9877, 1.9846)
> str(STX)
num [1:4] 16.1 16.1 16.1 16.1
> str(STY)
num [1:4] 2.04 2.02 1.99 1.98
>
So what is your question?
On Fri, Mar 6, 2009 at 1:47 PM, John Poulsen wrote:
> Hello,
>
> I know I
Hello,
I know I am forgetting to do something silly. I typed coordinates in
vectors (as below) but when I call them in R they come out as integers,
and I want them to be real numbers. I have tried using as.numeric,
as.real, etc... but they are still read by R as integers.
STX<-c(16.0962, 16.12
Answer to your first question:
> x<-matrix(c(266, 359, 533, 313, 555, 504, 502, 242),nrow=4)
> rownames(x)=c( "Northeast", "Midwest", "South", "West" )
> colnames(x)=c("Public", "Private")
> y <- as.data.frame(as.table(x))
> names(y)[1:2] <- c("Region", "Type")
> y
RegionType Freq
1 North
Nathan,
if you have a weighted adjacency matrix, then you don't need graph
packages for this, just do
rowSums(data)
or
rowSums(data != 0)
depending you want the sum of the weights of the adjacent edges, or
just the number of adjacent edges. Or optionally colSums instead of
rowSums if your grap
Hello R Help Team,
I have created graph from weighted adjecency matrix .Is there a way I can
find highly connected nodes in Igraph like the Package RBGL does.
nathan
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View this message in context:
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Sent from the
Dear Doug,
Here's my guess: the path to your file has an apostrophe in it -- I'm able
to duplicate the error in this case. If that's the case, then a work-around
would be to move the file. Meanwhile, I'll look into fixing the problem.
Regards,
John
> -Original Message-
> From: r-help-b
Dear Doug,
Can you send me the file, along with information about the versions of the
Rcmdr and R that you're using, and your OS? I suspect that the problem
doesn't have to do with the file, but with the command that the Rcmdr is
generating to read it, so some additional information would also be
Thank you for your reply, Dieter. I will try it.
Looks like a split-block experiment. You should check # 1.6 at the bottom
of library\nlme\scripts\Ch01.R, the package nlme, and the book by Pinheiro/
Bates.
Dieter
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R-help@r-project.org mailing li
Dimitris, Domenico,
It seems I was using apply in a wrong way.
Thank you!
TK
2009/3/6 Domenico Vistocco :
> If I well understand, maybe you have still apply at your disposal:
>
> arrayA <- 1:60
> dim(arrayA) <- c(3,4,5)
> apply(arrayA, 2, sum)
>
> You have the same result of:
> res<-numeric(4);
Hello,
I have been noticing that some of my column headings are missing.
Can you give me a clue as to how to
1] replace Var1 and Var 2 by Region and type ie do it right the first time
x<-matrix(c(266, 359, 533, 313, 555, 504, 502, 242),nrow=4)
rownames(x)=c( "Northeast", "Midwest", "South", "
I'd like to report another instance...one of my students has had the exact
same error. I've checked the data, at it is a clean .csv file, with no
quotation marks at all. I am currently at a loss at how to figure out what
might be going wrong.
--
View this message in context:
http://www.nabble
You have pretty much exhausted my expertise on time series, you may want to
reask your question on the sig-finance list (with a more specific subject
line), there seem to be more time series experts hanging out on that list
(though someone please correct me if my impression is wrong).
--
Gregor
On 06.03.2009, at 16:48, soeren.vo...@eawag.ch wrote:
### example:start
v <- sample(rnorm(200), 100, replace=T)
k <- rep.int(c("locA", "locB", "locC", "locD"), 25)
tapply(v, k, summary)
### example:end
... (hopefully) produces 4 summaries of v according to k group
membership. How can I transf
soeren.vo...@eawag.ch wrote:
### example:start
v <- sample(rnorm(200), 100, replace=T)
k <- rep.int(c("locA", "locB", "locC", "locD"), 25)
tapply(v, k, summary)
### example:end
This one is better:
do.call(cbind, tapply(v,k,summary))
Ciao,
domenico
... (hopefully) produces 4 summaries of v acc
Dear Sören,
How about this?
do.call(cbind,tapply(v, k, summary))
HTH,
Jorge
On Fri, Mar 6, 2009 at 10:48 AM, wrote:
> ### example:start
> v <- sample(rnorm(200), 100, replace=T)
> k <- rep.int(c("locA", "locB", "locC", "locD"), 25)
> tapply(v, k, summary)
> ### example:end
>
> ... (hopeful
soeren.vo...@eawag.ch wrote:
### example:start
v <- sample(rnorm(200), 100, replace=T)
k <- rep.int(c("locA", "locB", "locC", "locD"), 25)
tapply(v, k, summary)
### example:end
Maybe this could be a solution:
t1 <- tapply(v, k, summary)
t2 <- sapply(t1, cbind)
rownames(t2) <- names(t1[[1]])
t2
### example:start
v <- sample(rnorm(200), 100, replace=T)
k <- rep.int(c("locA", "locB", "locC", "locD"), 25)
tapply(v, k, summary)
### example:end
... (hopefully) produces 4 summaries of v according to k group
membership. How can I transform the output into a nice table with the
croups as co
Hi,
I also use R under Eclipse/StatET. I found the following doc really
useful:
http://www.splusbook.com/Rintro/R_Eclipse_StatET.pdf
Regards
Alain
212-449-4894
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of John Fox
Sent: Frid
Dear Michael,
For what it's worth, I develop the Rcmdr under Eclipse, and it works fine
with Eclipse -- both under Windows and under Mac OS X. Of the IDEs that I've
used with R, I'm most impressed with Eclipse/StatET, but configuration is
non-trivial and documentation is sparse.
I have a little e
You can easily write a simple function to do that:
letters2num <- function(x) {
nletters <- 1:length(LETTERS)
names(nletters) <- LETTERS
nletters[x]
}
> x <- c("A", "X", "F", "W", "G", "V", "L")
> letters2num(x)
A X F W G V L
1 24 6 23 7 22 12
-Christos
> --
try this:
x <- c("A", "X", "F", "W", "G", "V", "L")
match(x, LETTERS)
I hope it helps.
Best,
Dimitris
Leandro Marino wrote:
Hi,
I know the function LETTERS, but, now I have some letters to convert it in
numbers, like A=1,B=2, etc... Is any function to do that?
Atenciosamente,
Leandro Li
Hi,
I know the function LETTERS, but, now I have some letters to convert it in
numbers, like A=1,B=2, etc... Is any function to do that?
Atenciosamente,
Leandro Lins Marino
Centro de Avaliação
Fundação CESGRANRIO
Rua Santa Alexandrina, 1011 - 2º andar
Rio de Janeiro, RJ - CEP: 20261-903
R (21)
that statistic would be appropriate for non-linear regression?. know how I
can calculate the VIF for a linear model?.
per243 wrote:
>
> How can a non-linear regression to calculate the statistical R-square,
> R-square adjusted, RMSE, VIF??
> Thanks
> jose
>
--
View this message in context:
Dear Sueli,
Assuming that your data is in a data frame called "mydata", something like
the following should work:
# splitting the data by SAMPLE
msp<-with(mydata,split(mydata,SAMPLE))
# linear models by sample
models<-lapply(msp,function(x) lm(KA ~ PA, data = x))
# linear models by sample
model
Hi,
i am using spatstat package for spatial data analysis and now i have a problem
to create a point pattern. The points are in file "points.txt" (first column
for Latitude and second column for Longitude ) and I imported them and
separated each columns in two arrays x and y. If I plots x and
> I have many files in my directory. I want to transfer each data into
> one which is readable.
> They have so many possibilities, i have collected(manually and
> visually) all possibilities and represent them as different numbers.
>
> Rep[grep('context_log',log1$Remain[1:length(log1$Date)]),]<-"
The docx file (which includes graphs) can be opened with either Microsoft
Office 2007 or if you have an earlier version: Google for the compatibility
pack ...OR use openoffice 3.
thefurryblur wrote:
>
> Uploaded the data and my comparison. Hopefully this will help illustrate
> and solve the prob
*** Travel and Accommodation Support ***
Funds from the U.S. National Science Foundation may be available to
provide partial support for travel and accommodation for some graduate
students and junior faculty at U.S. post-secondary institutions to
attend DSC 2009 and useR 2009. If
On 06-Mar-09 11:33:21, Swantje Löbel wrote:
> Dear all!
> I tried to fit Gompertz growth models to describe cummulative
> germination rates using nls. I used the following code:
>
> germ.model<-nls(percent.germ~a*exp(-b*exp(-k*day)),data=tab,
> start=list(a> =100,b=10,k=0.5))
>
>
janey ding hotmail.com> writes:
> Sorry!
>
> In my experiment, there are two type of soil ( soil F and soil D), each half
> of them were subjected to steam sterilize (result in FS and DS soil). A
> equal volume of soil from two of the four soil types (F, D, FS, DS) were
> mixed as follows: F+F,
Your methods were sent in a docx file. I could be wrong, but it seems
unlikely that very many people will bother to open such a file even if
they do have an M$ product that will allow them to do so. My advice:
First read the posting guide; then learn to sent plain text attachments.
This is
On 3/6/09, kenji_aoyagi wrote:
>
> Hi,
>
> I am using glm().
> I'd like to know what the command means.
>
> For example,
> glm(family=binomial(link=logit))
> means logit model.
Means : binomial response variable transformed with the logit
Then,
> glm(family=gaussian(link=logit)),
> does this me
On Thu, Mar 5, 2009 at 7:49 PM, per243 wrote:
> How can a non-linear regression to calculate the statistical R-square,
> R-square adjusted, RMSE, VIF??
It is not clear that these statistics are meaningful for a nonlinear
regression model. For example, an R^2 value is meaningful when the
model b
Uploaded the data and my comparison. Hopefully this will help illustrate and
solve the problem.
http://www.nabble.com/file/p22371555/data.csv data.csv
http://www.nabble.com/file/p22371555/arima%2Bmethods.docx arima+methods.docx
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On Fri, Mar 6, 2009 at 6:07 AM, Sueli Rodrigues wrote:
>
> Hi, I have the following file, and I need to work out the linear
> regression for each sample. I tried the model(*) and receive the error
> message (**):
>> data=split(mydata,rep(1:(nrow(mydata)/6),each=6))
>> arrang.linear=lapply(data,lm,
Try this:
library(gsubfn)
# convert date strings to dd-mm-yy
Dates <- gsub("/", "-", dat$Dates)
# regular expressiont to match dates
re <- "\\d\\d-\\d\\d-\\d\\d"
# extract dates and convert them to Date class
# giving a list d each of whose components is a vector of dates
d <- strapply(Dates, r
Dear All,
I have many files in my directory. I want to transfer each data into one which
is readable.
They have so many possibilities, i have collected(manually and visually) all
possibilities and represent them as different numbers.
Rep[grep('context_log',log1$Remain[1:length(log1$Date)]),]<-
Hi, I have the following file, and I need to work out the linear
regression for each sample. I tried the model(*) and receive the error
message (**):
> data=split(mydata,rep(1:(nrow(mydata)/6),each=6))
> arrang.linear=lapply(data,lm,formula=KA~PA)
Erro em storage.mode(y) <- "double" :
invalid to
?is.na
Manli Yan wrote:
>
>I have a 50*50 matrix,some entry are NAs,I want to replace these NA by
> 0,so can I use some syntax to do so other than using ifelse?
>I tried to use replace(a,NA,0),it didnt work~~(a is matrix name)
>
> Thanks~
>
> [[alternative HTML version delete
Hi, every body!
I am a new comer for R, so my question would unavoidablely sounds stupid.
Sorry!
In my experiment, there are two type of soil ( soil F and soil D), each half
of them were subjected to steam sterilize (result in FS and DS soil). A
equal volume of soil from two of the four soil typ
Dear All,
I am going through a worked example provided by Harrell, Lee and Mark
(1996, Stats in Medicine, 15, 361-387). I know that the code provided
is for S-PLUS and R but the languages don't differ enough for this to
be a problem.
I am using the Hmisc and Design libraries and have used the fo
Dear all!
I tried to fit Gompertz growth models to describe cummulative germination rates
using nls. I used the following code:
germ.model<-nls(percent.germ~a*exp(-b*exp(-k*day)),data=tab,start=list(a=100,b=10,k=0.5))
My problem is that I want that the fitted model goes through the origin, since
Hi,
I have the following dataframe:
IDDates
1 16-07-01 06-10-95
224/01/02 06-10-95
3 16/01/02 16/08/94 12/01/91
And I would like to extract the dates, but couple the ID's to the right
dates, eg:
ID Dates
116-07-01
1
If I well understand, maybe you have still apply at your disposal:
arrayA <- 1:60
dim(arrayA) <- c(3,4,5)
apply(arrayA, 2, sum)
You have the same result of:
res<-numeric(4);for (i in 1:4) res[i]<-sum(arrayA[,i,])
Ciao,
domenico
PS:
have a look at plyr package for more "slicing" and "applying"
Hi all,
Thanks very much for the suggestions. My experience for the benefit of
others is:
- use addtable2plot if you want a fancier legend or small table *within* a
plot area
- use textplot if you want combinations of tables and text listings in the
graphics device. The help for textplot gives a v
Hi,
I have a problem with the function ctree in the package party.
When I launch ctree with weights=NULL it works.
ctree(function~var1+var2, data=datalist, weights=NULL,
controls=ctree_control(mincriterion=0.95, maxdepth=4,
teststat="quad",testtype="Bonferroni"))
But when I try
ctree(function~va
well, you can still use apply(), e.g.,
A <- array(rnorm(3*4*5), c(3, 4, 5))
f <- sum
out <- numeric(4)
for (i in 1:4)
out[i] <- f(A[, i, ])
out
apply(A, 2, f)
I hope it helps.
Best,
Dimitris
Todor Kondic wrote:
Hello,
If I want to apply some f(x) to such chunks of the the array
dim(A)
Thanks a lot to everybody that helped me out with this.
Conclusions:
(1)
In order to edit arima in R:
>fix(arima)
or alternatively:
>arima<-edit(arima)
(2)
This is not contained in the "Introduction to R" manual.
(3)
A "productive" fix of arima is attached (arma coefficients printed out and
er
Hello,
If I want to apply some f(x) to such chunks of the the array
dim(A)==c(d1,d2,d3,..,dk,...,dn) which are defined by A[...,ik,...]
(ik belongs to {1,..,dk}), for now I use iteration via 'for (i in
dim(A)[k]) f(A[...,k,...])' . Is there any more elegant approach, e.g
like in 'apply' function w
HI,Barry.
Thanks a lot for your details reply. I have already rewritten the programs. Now
it totally works.
Your comment: " Note the use of comments and breaking the code up into small
independent, testable functions" is really Valuable! Thanks again.
Hi,David,
You are right, My previous c
Hi Maxl18,
>> error in mo...@fit(data,...) : object "var3" not found
>> What should I do?
Make sure that "var3" exists/is available
##
str(datalist$var3) ## is it here?
ls(pattern="var3") ## is it here?
Regards, Mark.
Maxl18 wrote:
>
> Hi,
> I have a problem with the function ctree i
See homals package in R. But also look documents for ade4 package.
Justin BEM
BP 1917 Yaoundé
Tél (237) 76043774
De : Galanidis Alexandros
À : "r-help@r-project.org"
Envoyé le : Vendredi, 6 Mars 2009, 10h09mn 18s
Objet : [R] PCA and categorical data
Hi al
John Sorkin grecc.umaryland.edu> writes:
>
> R 2.8.1
> Windows XP
> Fedora Linux.
>
> I would like a suggestion for an editor that will help format my R code that
can be used with Rcmdr. Is there
> anything I need to know about running or installing an editor when using
Rcmdr? I run R on both W
Hi Galandis,
dudi.mix() in package ade4 does PCA using categorical and/or quantitative
variables. Ordered cats are replaced by poly(x, deg=2). Squares of
categoricals can also be used. The method is a generalization by Chessel of
the method of Hill and Smith.
Regards, Mark.
Galanidis Alexandro
> > The best encoding depends upon which language you would like to
manipulate
> > the variable in. In R, genders are most naturally represented as
factors.
> > That means that in an external data source (like a spreadsheet of
data),
> > you should ideally have the gender recorded as human-
On 06-Mar-09 09:25:26, Prof Brian Ripley wrote:
> You might want to look into correspondence analysis, which has several
> variants of PCA designed for categorical data.
In particular, have a look at the results of
RSiteSearch("correspondence")
Ted.
> On Fri, 6 Mar 2009, Galanidis Alexandros
You might want to look into correspondence analysis, which has several
variants of PCA designed for categorical data.
On Fri, 6 Mar 2009, Galanidis Alexandros wrote:
Hi all,
I' m trying to figure out if it is appropriate to do a PCA having only
categorical data (not ordinal). I have only fin
On Fri, 6 Mar 2009, Rainer M Krug wrote:
Hi
I am using R scripts which are running remotely. To make debugging
easier, I would like to have the possibility to execute traceback()
automatically when an error occurs. Is this possible?
See options 'error' and 'showErrorCalls'. The latter is the
Hi,
I am using glm().
I'd like to know what the command means.
For example,
glm(family=binomial(link=logit))
means logit model.
Then,
glm(family=gaussian(link=logit)),
does this mean?
Thank you in advance.
Kenji. A
Analysis Manager
SPI - Strategy, Productivity, Insight., Japan
___
Hi all,
I' m trying to figure out if it is appropriate to do a PCA having only
categorical data (not ordinal). I have only find the following quote:
One method to find such relationships is to select appropriate variables and
to view the data using a method like Principle Components Analysis (PC
Hi
r-help-boun...@r-project.org napsal dne 05.03.2009 15:21:22:
>
> Hello Petr,
>
> In fact spec is data.frame with a column called code (containing
numerical
> values) and some other columns called data1, data2, ... containing data
for
> each equipment (that is for each code).
>
> But I don
Hi
I am using R scripts which are running remotely. To make debugging
easier, I would like to have the possibility to execute traceback()
automatically when an error occurs. Is this possible?
OS: Linux
Thanks
Rainer
--
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology
On Thu, 2009-03-05 at 02:20 +, Ben Bolker wrote:
> Sueli Rodrigues esalq.usp.br> writes:
> >
> > Olá. Tenho um arquivo que a cada 6 linhas corresponde uma amostra da qual
> > preciso dos coeficientes da regressão linear. Como faço para que o
> > programa distinga a cada 6 linhas como uma amos
Jani Lobo grupocomar.com> writes:
> I want to estimate the survival mean of a few specific teams. I'm trying to
> calculate it through a Kaplan Meier estimator. For doing so, I load the
> "survival" package and run the following instructions:
>
> "options(survfit.print.mean=TRUE)" allows s
Hi Viviana,
>> I am doing kernel density plots, and am trying to make the lines thicker.
You need to hack the code for sm.density.compare. See the code below. This
uses the same defaults as the original, but you can customize band colour,
line width, and so on using arguments to the function. Th
Viviana Ruiz gmail.com> writes:
> I am doing kernel density plots, and am trying to make the lines thicker. I
> comparing three groups, in sm.density.compare. I tried changing lwd to make
> the line sthicker right on the density compare call, but was not able to do
> it. There is not an option
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