Hi,
I'm running gam (mgcv version 1.3-29) and glm (logit) (stats R 2.61) on
the same models/data, and I got error messages for the gam() model and
warnings for the glm() model.
R-help suggested that the glm() warning messages are due to the model
perfectly predicting binary output. Perhaps th
Please do as the R posting guide asked you to do before posting, and
update your R. That will inter alia give you a factory-fresh version of
the file in question.
On Mon, 7 Jan 2008, Shen, Jill wrote:
> Dear friend,
>
> I am trying to use R in Excel, I installed R-(D)COM (RSrv250_pl1)., and
>
Suggest you read the posting guide and in particular
read the last line on every message to r-help. Also
the line wrapping on your email makes it unreadable.
On Jan 7, 2008 7:09 PM, Kondamani, Arjun (GMI - NY Corporate Bonds)
<[EMAIL PROTECTED]> wrote:
> Hi all:
>
> I have a directory of files as
On Jan 7, 2008 8:09 PM, Paul Smith <[EMAIL PROTECTED]> wrote:
> Thanks again, Gabor. Apparently, something is missing in your
> approach, as I tried to apply it to the original problem (without sin)
> and the result seems not being correct::
>
> f <- function(z) {
> a <- sum(-(999:1)*z)
> return(
Dear friend,
I am trying to use R in Excel, I installed R-(D)COM (RSrv250_pl1)., and
the rcom package (rcom_1.5-2.2). But when I try to start R in Excel, I
got an strange error "error in C:\Program Files\R\R-2.5.1\etc\RConsole
line 69". The line 69 is "language=". I modified it to "language=en"
You can say
table1$var2 %in% c("a","b","c","d")
--- Sebastian Leuzinger
<[EMAIL PROTECTED]> wrote:
> hi, this may be trivial, but we can't seem to find
> anything adequate,
> (although there is a work around with match() ). We
> are looking for something
> along the lines of
>
>
plot(table1
Thanks again, Gabor. Apparently, something is missing in your
approach, as I tried to apply it to the original problem (without sin)
and the result seems not being correct::
f <- function(z) {
a <- sum(-(999:1)*z)
return(a)
}
result <- optim(rep(0,999), f,
NULL,lower=rep(-1/1000,999),upper=re
I want to get the minimum arc (in degrees) needed to include a set of
compass directions. I would like to use the range.circular() function
of the package circular, because that package understands a compass-type
of angle convention, but it gives results I don't understand. Howver, I
can get the
On Mon, 7 Jan 2008, Seung Jun wrote:
> I'm looking for a way to improve code that's proven to be inefficient.
>
Jim was probably right on both counts (use Rprof and expect wtd.quantile
to be the place where the time is being used).
If following his advice doesn't get you what you need, try vect
Does this do what you want?
> x <- read.table(textConnection("AB C D
+ A1 B1 C1 D1
+ A1 B2 C1 D1
+ A1 B1 C1 D1
+ A1 B2 C2 D2
+ A2 B1 C1 D1
+ A1 B2 C2 D2
+ A1 B1 C2 D2
+ A2 B2 C1 D1
+ A1 B1 C2 D2
+ A2 B2 C1 D1"), header=TRUE)
> counts <- ave(seq(nrow(x)), x$B, x$C, x$D, FUN=lengt
> errors in the data usually because I know the data. I find errors
> because I can say things like
>
> library(Hmisc)
> datadensity(mydata) # show all raw data in small rug plots
> hist.data.frame(mydata) # postage-stamp size histograms of all
> variables in dataset
> latex(describe(my
John Sorkin wrote:
> Frank,
> I believe you are proving my point. The difference is not so much the
> language as the end users. I use SAS, R, and SPlus on a regular basis. For
> some analyses, SAS is easiest to use, for some R (or SPlus). I can be just as
> dangerous using SAS and I can be with
On 1/7/08, Emmanuel Charpentier <[EMAIL PROTECTED]> wrote:
> hadley wickham a écrit :
> > I followed the instructions at
> > http://cran.r-project.org/bin/linux/ubuntu/README.html, but I'm
> > getting the following error:
> >
> > ~: sudo apt-get install r-base
> > Reading package lists... Done
> >
hadley wickham a écrit :
> I followed the instructions at
> http://cran.r-project.org/bin/linux/ubuntu/README.html, but I'm
> getting the following error:
>
> ~: sudo apt-get install r-base
> Reading package lists... Done
> Building dependency tree... Done
> Some packages could not be installed. T
One thing to do is to use Rprof() on your script so that you can
determine where time is being spent. My guess it that most of the
time is in the wtd.quantile function. If your Counts don't get too
big, another way is to use 'quantile' directly:
> Index <- c(0,1,7,30)
> Count <- c(234,120,11,1)
Hi all:
I have a directory of files as in -> bunchafiles <-
list.files(path="/data/2.3/2006", pattern="returns",
full.names=T,recursive=T) Each file is a bunch of returns for a
particular date (unique). There are like 252 files or so.
With a custom function myread (below), I def
Dear all,
I would like to do a goodness-of-fit test on my data to see if they follow a
mixture of 2 poisson distributions. I have small numbers for observed values.
Most of them <5. The chisq.test gives warning message: Chi-squared
approximation may be incorrect in: chisq.test(x , p = prob). Ho
Frank,
I believe you are proving my point. The difference is not so much the language
as the end users. I use SAS, R, and SPlus on a regular basis. For some
analyses, SAS is easiest to use, for some R (or SPlus). I can be just as
dangerous using SAS and I can be with R if I don't think about wha
Does this do what you want?
> x <- c(1,4,2,6,7,5)
> x.i <- c(1,4,5)
> partiMax <- function(vec, index){
+ # create a vector to identify the subvectors
+ x.b <- diff(c(index, length(vec) + 1))
+ # split up the vector
+ x.s <- split(vec, rep(seq_along(index), times=x.b))
+ unlist
names(alninc) <- c("F", "V", "G")
--- Wade Wall <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I am trying to name list objects, but am having
> trouble doing so. At the
> moment I have list alninc that has 3 objects and I
> can refer to them as
> alninc[[1]] . . .alninc[[3]]. I would like to
> re
I'm looking for a way to improve code that's proven to be inefficient.
Suppose that a data source generates the following table every minute:
Index Count
0 234
1 120
7 11
30 1
I save the tables in the following CSV format:
time,index,count
0,0:1:
?par and have a look at mfcol or mfrow.
Example
exp<-cbind(abs(round(rnorm(10),2)*10), seq(100, 200,
by=10))
ref<-cbind(abs(round(rnorm(10),2)*10), seq(100, 200,
by=10))
op <- par(mfrow=c(1,2))
plot(ref, col="red")
plot(exp, col="blue")
par(op)
--- [EMAIL PROTECTED] wrote:
> Hello everyone,
>
>
John Sorkin wrote:
> I fear I risk being viewed as something of a curmudgeon, but the truth must
> be stated. S-Plus, R, SAS, etc. are all similar in that they are all tools to
> an end and not an end in themselves. Any one of the three can do most
> statistical analyses one might want to do. I
?par and have a look at mfrow
par() controls a vast amount of plotting options for
regular graphics
Example
aa <- rnorm(100, 25, 4)
bb <- aa/2
par(mfrow=c(2,1))
hist(aa)
plot(bb)
--- Georg Ehret <[EMAIL PROTECTED]> wrote:
> Dear R community,
> I am plotting a histogram and would wish to
I fear I risk being viewed as something of a curmudgeon, but the truth must be
stated. S-Plus, R, SAS, etc. are all similar in that they are all tools to an
end and not an end in themselves. Any one of the three can do most statistical
analyses one might want to do. I could point out the strengt
Hi Mika03,
It would be useful to know what function you used to create your plot.
Assuming you used boxplot, do this:
?boxplot
?boxplot.stats
Julian
mika03 wrote:
>
> http://www.nabble.com/file/p14668788/paragraphs.png
>
> Hi,
>
> R is is world full of wonders... I created the attached p
Try use args. 'position' and 'more'
h1<-hist()
h2<-hist()
print(h1,position=c(0,0,1,1),more=T)
print(h2,position=c(,,,)
Weidong Gu,
University of Alabama, Birmingham
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Georg Ehret
Sent: Monday, January 07,
?layout
check:
example(layout) to see its features.
b
On Jan 7, 2008, at 5:06 PM, Georg Ehret wrote:
Dear R community,
I am plotting a histogram and would wish to display another
variable of
the same dataset in a very narrow heatmap just below the x-axis.
Never mind
the specifics of m
Dear R community,
I am plotting a histogram and would wish to display another variable of
the same dataset in a very narrow heatmap just below the x-axis. Never mind
the specifics of my task: How can I draw a second graph/image just below a
first graph/image?
Thank you!
Georg.
***
Ido M. Tamir wrote:
> matchpt
Thanks for this hint. It is exactly what I'm looking for.
Cheers, Joh
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guid
[EMAIL PROTECTED] wrote:
> Hello everyone,
>
> I have an overlay plot it's nice but you can't see all the data. I would
> like to know if there is a way to get a plot that gives a side by side
> plot so that each plot would be next to each other. The two plots have
> the same data are of different
On Monday 07 January 2008, [EMAIL PROTECTED] wrote:
> Hello everyone,
>
> I have an overlay plot it's nice but you can't see all the data. I would
> like to know if there is a way to get a plot that gives a side by side
> plot so that each plot would be next to each other. The two plots have
> the
[EMAIL PROTECTED] said the following on 1/7/2008 2:59 PM:
> Hello everyone,
>
> I have an overlay plot it's nice but you can't see all the data. I would
> like to know if there is a way to get a plot that gives a side by side
> plot so that each plot would be next to each other. The two plots h
SAS programming is easy if everything you want to do fits easily into the
row-at-a-time DATA step paradigm. If it doesn't, you have to write macros,
which are an abomination. DATA step statements and macros are entirely
different programming languages, with one doing evaluations at "compile" time
Hello everyone,
I have an overlay plot it's nice but you can't see all the data. I would
like to know if there is a way to get a plot that gives a side by side
plot so that each plot would be next to each other. The two plots have
the same data are of different species. At the moment this is the
I apologize if this is somewhere in the archives, but I can't seem to find
a solution to this question.
I've been trying to plot a bwplot:
print(
bwplot( n.pareto ~ as.factor(gen) | mut.rate * n.pop,
data=p6,
horizontal=FALSE,
box.ratio=0.75,
Sorry the new column in DF2 should be called FreqD instead of FreqA.
How can I get DF2 with aggregate?
Gang
On Jan 7, 2008, at 2:38 PM, Gang Chen wrote:
> Yes, I misstated it when I said that I would keep B and C. I want to
> collapse column A, but count the frequency of D as a new column in
>
?table
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Gang Chen
Sent: Monday, January 07, 2008 11:38 AM
To: Duncan Murdoch
Cc: R-Help
Subject: Re: [R] Create a new dataframe from an existing dataframe
Yes,
Yes, I misstated it when I said that I would keep B and C. I want to
collapse column A, but count the frequency of D as a new column in
the new dataframe DF2 while collapsing A. The rows of columns B, C,
and D of course would be reduced because of A collapsing.
For example, if dataframe DF i
Johannes Graumann wrote:
> Hi all,
>
> I'm to inexperienced to come up with the matrix solution elusively
> appearing on the horizon for the following problem and would appreciate if
> you could give me a nudge ...
> I have two vectors a, and b and need to find the closest match for each
> value
Thanks Gabor! This is the sort of thing I was trying to devise, but I ran up
against my extensively documented brain power limitations ;-) Per your
remarks, it remains to be seen how it performs compared to a good
implementation with one of the apply functions.
-- TMK --
212-460-5430ho
On 1/7/2008 1:28 PM, Gang Chen wrote:
> I have a dataframe DF with 4 columns (variables) A, B, C, and D, and
> want to create a new dataframe DF2 by keeping B and C in DF but
> counting the frequency of D while collapsing A. I tried
>
> by(DF$D, list(DF$B, DF$C), FUN=summary)
>
> but this is
Henrique Dallazuanna wrote:
> Try: sapply(paste("Data$x", 1:3, sep=""),
> function(x)eval(parse(text=x)))
>
> On 07/01/2008, Gregory Gentlemen <[EMAIL PROTECTED]> wrote:
>> Dear R users,
>>
>> I'd like to evaluate a vector of characters. For example, say I
>> have a data frame called Data includi
Try testing the performance of transforming your series to
one in which the values of each partition are larger than all
prior partitions and the untransforming back:
# test data
myseq <- c(1, 4, 2, 6, 7, 5)
part <- c(1, 4, 5)
M <- max(myseq)
# transform
myseq2 <- myseq + M * cumsum(replace(0 *
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf
> Of Wade Wall
> Sent: Monday, January 07, 2008 10:36 AM
> To: [EMAIL PROTECTED]
> Subject: [R] Naming list objects
>
> Hi all,
>
> I am trying to name list objects, but am having trouble doing so. At the
names(alninc) <- c("F","V","G")
See ?names
Gabor
On Mon, Jan 07, 2008 at 01:35:59PM -0500, Wade Wall wrote:
> Hi all,
>
> I am trying to name list objects, but am having trouble doing so. At the
> moment I have list alninc that has 3 objects and I can refer to them as
> alninc[[1]] . . .alnin
On Mon, 7 Jan 2008, [EMAIL PROTECTED] wrote:
> Jonas,
>
> In statistical sense polynomial is a linear regression fit. The function
> that handles linear fitting is called lm. Here is how you can reproduce
> your results:
>
> lm(y ~ x + I(x^2) + I(x^3))
>
> Unless you are really after the polynom
Hi all,
I am trying to name list objects, but am having trouble doing so. At the
moment I have list alninc that has 3 objects and I can refer to them as
alninc[[1]] . . .alninc[[3]]. I would like to refer to them as alninc$F,
alninc$V and alninc$G. How do I do this? This appears to be an elem
Talbot,
Try this:
PartMax <- function(x, breaks)
{
breaks <- c(breaks, length(x) + 1)
sapply(seq(length(breaks) - 1),
function(i) max(x[breaks[i]:(breaks[i + 1] - 1)],
na.rm = TRUE))
}
> PartMax(c(1,4,2,6,7,5),c(1,4,5))
[1] 4 6 7
> PartMax(6:1,c(1,4,
Thanks a lot! This is exactly what I wanted.
Gang
On Jan 5, 2008, at 2:20 PM, David Winsemius wrote:
> Gang Chen <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
>> Suppose I have a two-way table of nominal category (party
>> affiliation) X ordinal category (political ideology):
>>
>> p
I have a dataframe DF with 4 columns (variables) A, B, C, and D, and
want to create a new dataframe DF2 by keeping B and C in DF but
counting the frequency of D while collapsing A. I tried
by(DF$D, list(DF$B, DF$C), FUN=summary)
but this is not exactly what I want. What is a good way to do it
This is a fortune(106) situation (with a little fortune(77) in there as well).
I am unsure what exactly the original poster is wanting to do, but accessing
the columns of a data frame (or list) by character strings can be done using [[
]] rather than $ and should be able to work for this situati
Yes, Gabor, that is the interpretation that I meant. Thanks for the
clarification.
Paul
On Jan 7, 2008 5:59 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> I think what he meant is that x is a function of t so if dx/dt is regarded
> to be a function of t, which we shall call u(t), then u(t)'
Hi Joh.
To echo Bert, more detail would be helpful. But here is a stab at something.
Given two numeric vectors, a and b, for each entry of a this finds the index in
b of the closest match. It's not efficient at all.
> bestmati <- function( a, b ) { return( sapply( a, function( x ) { which.m
I think what he meant is that x is a function of t so if dx/dt is regarded
to be a function of t, which we shall call u(t), then u(t)'s absolute value
at each value of t is less than or equal to 1 (as opposed to u(t) being
known).
On Jan 7, 2008 11:32 AM, Ravi Varadhan <[EMAIL PROTECTED]> wrote:
>
Ops:
b[which.min(abs(a-b))]
On 07/01/2008, Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> Try this:
>
> b[which.min(a-b)]
>
> On 07/01/2008, Johannes Graumann <[EMAIL PROTECTED]> wrote:
> > Vectors are numerics. I'd like to know for each a[i], which b[j] delivers
> > the smallest abs(a[i]-b[
Try:
sapply(paste("Data$x", 1:3, sep=""), function(x)eval(parse(text=x)))
On 07/01/2008, Gregory Gentlemen <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I'd like to evaluate a vector of characters. For example, say I have a data
> frame called Data including the field names x1, x2, x3, and I'd l
Try this:
b[which.min(a-b)]
On 07/01/2008, Johannes Graumann <[EMAIL PROTECTED]> wrote:
> Vectors are numerics. I'd like to know for each a[i], which b[j] delivers
> the smallest abs(a[i]-b[j]).
>
> Thanks for your time.
>
> Joh
>
> Bert Gunter wrote:
>
> > Vectors of what? How is "closest" defi
Vectors are numerics. I'd like to know for each a[i], which b[j] delivers
the smallest abs(a[i]-b[j]).
Thanks for your time.
Joh
Bert Gunter wrote:
> Vectors of what? How is "closest" defined? See e.g. ?dist
>
>
> Bert Gunter
> Genentch Nonclinical Statistics
>
>
> -Original Message
Dear R users,
I'd like to evaluate a vector of characters. For example, say I have a data
frame called Data including the field names x1, x2, x3, and I'd like to a list
out of paste("Data$x", 1:3, sep=""). How can I get list to evaluate
paste("Data$x", 1:3, sep="") as an R object rather than a
On Jan 7, 2008 4:32 PM, Ravi Varadhan <[EMAIL PROTECTED]> wrote:
> Your problem statement does not make much sense to me. You say that an
> analytical solution can be found easily. I don't see how.
>
> This is a variational calculus type problem, where you maximize a
> functional. Your constrain
Vectors of what? How is "closest" defined? See e.g. ?dist
Bert Gunter
Genentch Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Johannes Graumann
Sent: Monday, January 07, 2008 11:19 AM
To: [EMAIL PROTECTED]
Subject: [R] Efficient
Jonas,
In statistical sense polynomial is a linear regression fit. The function
that handles linear fitting is called lm. Here is how you can reproduce
your results:
lm(y ~ x + I(x^2) + I(x^3))
Unless you are really after the polynomial coefficients it is probably
better to use orthogonal poly
Hi all,
I'm to inexperienced to come up with the matrix solution elusively appearing
on the horizon for the following problem and would appreciate if you could
give me a nudge ...
I have two vectors a, and b and need to find the closest match for each
value of a in b.
How to do that efficiently?
Hi Marc.
Thank you for the swift response! I should have explained more about the
partitions, I hoped it would be clear from the code. I am supplying an index
vector to create the partitions. In my original example of v = c(1,4,2,6,7,5)
with v1 = v[1:3], v2 = v[4], v3 = v[5:6], I would spec
If your main goal is to find 10 colors that are easy to tell apart and
look good in a graph, then look at the RColorBrewer package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PR
Just looking at this again what I meant was that unless you can linearize it
you can't use linear programming. You still could use general nonlinear
optimization subject to constraints.
Define new variables z[i] = x[i] - x[i-1]
Then x[i] = z[0] + ... + z[i]
so maximize (we dropped the first and
putro nugroho yahoo.com> writes:
>
> Hii all, may I introduce myself. my name is Putro Nugroho i'am astudent of
Gadjah Mada University, i'am
> doing on my final exam about Parametric Regression Model with gamma and
log-gamma regression model.
.
I suggest that you generate a small, self-co
Talbot Katz wrote:
> Hi.
>
> Suppose I have a vector that I partition into disjoint, contiguous
> subvectors. For example, let v = c(1,4,2,6,7,5), partition it into
> three subvectors, v1 = v[1:3], v2 = v[4], v3 = v[5:6]. I want to
> find the maximum element of each subvector. In this example,
On Mon, Jan 07, 2008 at 05:25:44PM +0100, Lorenzo Isella wrote:
> Thanks for both the replies.
> I am now giving a try to the suggestion by Gabor since it looks easier
> (for me) to implement.
> I am testing it, but so far it does what I have in mind.
> I am going now through the documentation of t
"mika03" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> http://www.nabble.com/file/p14668788/paragraphs.png
>
> R is is world full of wonders... I created the attached plot, and I think
> it's exactly what I need! Well, actually I think it is more that wht I
> need...
>
> I wanted
Hi Paul,
Your problem statement does not make much sense to me. You say that an
analytical solution can be found easily. I don't see how.
This is a variational calculus type problem, where you maximize a
functional. Your constraint dx/dt=u(t) means that there exists a solution
(the anti-deri
You might want to google "box and whisker plot".
mika03 wrote:
>
>
> http://www.nabble.com/file/p14668788/paragraphs.png
>
> Hi,
>
> R is is world full of wonders... I created the attached plot, and I think
> it's exactly what I need! Well, actually I think it is more that wht I
> need...
Thanks for both the replies.
I am now giving a try to the suggestion by Gabor since it looks easier
(for me) to implement.
I am testing it, but so far it does what I have in mind.
I am going now through the documentation of the igraph package. I can
count the cluster number, but I also want to make
Hi.
Suppose I have a vector that I partition into disjoint, contiguous subvectors.
For example, let v = c(1,4,2,6,7,5), partition it into three subvectors, v1 =
v[1:3], v2 = v[4], v3 = v[5:6]. I want to find the maximum element of each
subvector. In this example, max(v1) is 4, max(v2) is 6
Dimitris Rizopoulos wrote:
> try this:
>
> y <- c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18,
> 11.32)
> x <- seq(3.75, 6, 0.25)
> coef(lm(y ~ x + I(x^2) + I(x^3)))
Or use the 'poly' function:
> coef(lm(x~poly(y,3)))
(Intercept) poly(y, 3)1 poly(y, 3)2 poly(y, 3)3
4.8750
Aha... I forgot to load the library.
Thanks everyone who responded my quaries.
With regards,
Abu
_
Fancy some celeb spotting?
__
R-help@r-project.org mailing list
https://sta
You may also wish to consider runmed(), which is a more robust running
median, and is automatically available as part of the stats package.
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Jared O'Connell
Sen
http://www.nabble.com/file/p14668788/paragraphs.png
Hi,
R is is world full of wonders... I created the attached plot, and I think
it's exactly what I need! Well, actually I think it is more that wht I
need...
I wanted R to show the mean values of the categories on the x-axis and maybe
the sta
Hii all, may I introduce myself. my name is Putro Nugroho i'am astudent of
Gadjah Mada University, i'am doing on my final exam about Parametric Regression
Model with gamma and log-gamma regression model. I have a problem about the
syntax program on R because it doesn't work with gamma distributi
Abu Naser wrote:
>
> Hi Jerad,
>
> Thanks for your quick response. I have installed the zoo using
>
> R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
> wondering why?
In R, don't forget to load "zoo" by
library("zoo")
Uwe Ligges
> With regards,
> Abu
>
>
>
>
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
Free games, great prizes - get gaming at
I'm assuming the package installed correctly (I've only ever installed
packages from within R)
Have you loaded the zoo package before calling rollmean?
ie.
> library(zoo)
On Jan 7, 2008 4:31 PM, Abu Naser <[EMAIL PROTECTED]> wrote:
>
>
> Hi Jerad,
>
> Thanks for your quick response. I h
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
Fancy some celeb spotting?
1. ?factanal says:
==
The factor analysis model is
x = Lambda f + e
...
but the major restriction is that the scores be uncorrelated and of unit
variance, and
that the errors be independent with variances Phi, the uniquenesses.
===
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
[[replacing trailing spam]]
Hello,
I am planning to implement Genetic Algorithms in solving classification
problems. As far as I can see there is only genalg package that I can use. I
can see that there are gafit and rgenoud packages that can be used for
regression and optimisation problems but not for classification. Is
try this:
y <- c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18,
11.32)
x <- seq(3.75, 6, 0.25)
coef(lm(y ~ x + I(x^2) + I(x^3)))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Ad
Thanks you for your answers.
I finally managed to have odfweave functionning well by downloading Rtools
available at http://www.murdoch-sutherland.com/Rtools/ and selecting the
option that allows you to update the variable PATH; I re-start my computer and
miracle !! it works
Sylvie
> Dat
look at ?"%in%"; I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
ht
I wonder how one in R can fit a 3rd degree polynomial to some data?
Say the data is:
y <- c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18, 11.32)
x <- seq(3.75, 6, 0.25)
And resulting degrees of polynomial are:
5.8007 -91.6339 472.1726 -774.2584
THanks in advance!
--
Jonas
Try this:
plot(table1[table1$var2 %in% c("a","b","c","d"),"var1"])
On 07/01/2008, Sebastian Leuzinger <[EMAIL PROTECTED]> wrote:
> hi, this may be trivial, but we can't seem to find anything adequate,
> (although there is a work around with match() ). We are looking for something
> along the lin
hi, this may be trivial, but we can't seem to find anything adequate,
(although there is a work around with match() ). We are looking for something
along the lines of
plot(table1[table1$var2==or("a","b","c","d"),"var1"])
would be handy, with the potential or() function leading to what
plot(
Dear Lorenzo,
if I understand your posting correctly, this is exactly what Single
Linkage clustering does if you cut the dendrogram tree at your threshold
distance.
Therefore you can use hclust with method="single" (which produces the full
dendrogram; you have to generate the Euclidean distanc
Achaz von Hardenberg pngp.it> writes:
>
> Dear all,
> I am performing a binomial glmm analysis using the lmer function in
> the lme4 package (last release, just downloaded). I am using the
> "Laplace method".
>
> However, I am not sure about what I should do to test for the
> significance
Have a look at ?staxlab library(plotrix)or
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-create-rotated-axis-labels_003f
--- mika03 <[EMAIL PROTECTED]> wrote:
>
> http://www.nabble.com/file/p14664173/at-modality.png
>
>
>
> I created the above image with R and I have one
> problem l
Lorenzo, why can't you actually generate the graph to find the
connection components? With the 'igraph' package this is something like:
g <- graph.adjacency( DIST < 0.5, mode="undirected" )
g <- simplify(g)
no.clusters(g)
assuming you have your distance matrix in 'DIST'. If N is too big
then you
See ?rollmean in the zoo package.
On Jan 7, 2008 3:31 PM, Abu Naser <[EMAIL PROTECTED]> wrote:
>
> Hi all R users,
>
> Can anyone please let me know how to do the moving average with R?
>
>
>
> With regards,
> Abu
>
>
>
>
>
>
>
> _
>
Hi all R users,
Can anyone please let me know how to do the moving average with R?
With regards,
Abu
_
Free games, great prizes - get gaming at Gamesbox.
__
R-help@r-project.org
Dear All,
I hope I am not asking a FAQ. I am dealing with a problem of graph
theory [connected components in a non-directed graph] and I do not
want to rediscover the wheel.
I saw a large number of R packages dealing for instance with the
k-means method or hierarchical clustering for spatially dist
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