On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis wrote:
> I have a very large dataset with three variables that I need to graph using
> a scatterplot. However I find that the first variable gets masked by the
> other two, so the graph looks entirely different depending on the order of
> variables.
Hello R experts
I have two linear regressions for sexes (Male, Female, Unknown). All have a
good correlation between body length (response variable) and head length
(explanatory variable). I know it is not recommended, but for a good practical
reason (the purpose of study is to find a single c
Hi:
I can think of a couple: (1) size reduction of the points; (2) alpha
transparency; (3) (1) + (2)
>From your original plot in base graphics, I reduced cex to 0.2 and it didn't
look too bad:
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24), cex = 0.2)
points(rnorm(x,mean=20),rnorm(x),col=1,
Hello Greg,
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The matrix
'dumvar' is just added to the VECM as deterministic regressors and while you
are referring to case 5, this is basically what you are after, if I am not
mistaken. But we aware that this implies a quadrati
Hi.
You could also turn it into a 3D plot with some variation on the function
below:
plot4d<-function(x,y,z, u, main="", xlab="", ylab="", zlab="", ulab="")
{
require(rgl)#may need to install this package first
#standard trick to get some intensity colors
uLim<-range(u)
Hello,
I use nls.profile to compute confidence intervals of parameter estimates
of a non-linear model. When computing the profiles, the model function
produces an error for certain parameter combinations. Therefore nls
fails and so does nls.profile.
Is there a way to tell nls.profile to ignore er
Hi,
Yep, that was what it was doing. For a sum across week, try something like
get.week.flag <- function(dd) {
## get weekday from the date in dd and code it as Monday = 1, Tuesday = 2 etc
idd =
factor(weekdays(dd),levels=c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunda
Dear List!
I want to compare medians of non normal distributed data. Is it
possible and usefull to calculate 95% confidence intervals for
medians? And if so - how can this be achieved in R?
Thanks a lot!
Anna
--
Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zu
Hi there,
I'm trying to solve 2 nonlinear equations in 2 unknowns using the BB
package.
The first part of my program solves 3 ODEs using the deSolve package. This
part works. The output is used as parameter values in the functions I need
to solve.
The second part is to solve 2 equations in 2 unk
The default rank test in the quantreg package would look like this
summary(rq(y ~ d, tau = .5))
where d is a factor variable indicating which sample the elements of
y belonged to. Summary returns a confidence interval for the coef
of the factor variable -- if this interval excludes zero at the c
DeaR ComRades,
require(lattice)
data <- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,c
In addition to %...% operators one can define := (I haven't seen this
possibility documented anywhere but it's used in a package) which
seems to have different precedence.
> `:=`<-`%<-%` # the %<-% by John Fox
> a
[1] -1
> a := 1000 + 1
[1] 1001
> a
[1] 1001
> a %<-% 1000 + 1
[1] 1001
> a
[1] 10
Hi,
I had a piece of code which looped over a decimal vector like this:
for( i in where ){
thisdata <- subset(herde, herde$mlr >= i)
# do stuff with thisdata..
}
'where' is a vector like seq(-1, 1, by=0.1)
My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different if
'where' wa
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
d
On 31-Mar-11 11:24:01, Alexander Engelhardt wrote:
> Hi,
> I had a piece of code which looped over a decimal vector like this:
>
> for( i in where ){
>thisdata <- subset(herde, herde$mlr >= i)
># do stuff with thisdata..
> }
>
> 'where' is a vector like seq(-1, 1, by=0.1)
>
> My problem
Dear Ken and Bill,
Thanks for the comments. I obviously didn't think about the precedence issue
and was unaware that one could define :=.
Best,
John
> -Original Message-
> From: Kenn Konstabel [mailto:lebats...@gmail.com]
> Sent: March-31-11 7:12 AM
> To: William Dunlap
> Cc: John Fox;
Try this:
plot(1, main = bquote("Yield for " ~ italic(.(spp)) ~ "in management
region" ~ .(region)))
On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman wrote:
> Hi all!
>
> I've written a handy script that uses a for loop to allow me to generate a
> large number of figures and statistical outputs f
Better:
plot(1, main = bquote("Yield for " ~ italic(.(as.character(spp))) ~
"in management region" ~ .(region)))
You have a factor, so you need convert it to character
On Thu, Mar 31, 2011 at 8:59 AM, Henrique Dallazuanna wrote:
> Try this:
>
> plot(1, main = bquote("Yield for " ~ italic(.(s
A very straightforward way to avoid this problem is to construct the
sequence by multiplying a sequence of integers by an approriate
constant. E.g. for your first example:
for( i in where ){
thisdata<- subset(herde, herde$mlr>= i)
# do stuff with thisdata..
}
'where' is a vecto
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea,
but still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(u
'time' was a trend variable from my.data set. Equivalent to the output of
the command 'matrix' you just gave me.
So now I did:
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
mat1 <- matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm <- ca.
On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
> this helps, thank you.
> But if this code is in a function, and some user supplies a vector, I will
> still have to round it in the function, I guess.
>
> It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want
> to read t
If you are doing multiple plots like this, there is no way I know of
to specify the fonts for labeling *once* for all such plots. However,
you can do something like this to save typing and keep things consistent:
my.largs <- list(
gp_labels = gpar(fontsize = 12, fontfamily = "calibri"),
Well, without further information, I do not know, but try the following
library(urca)
example(ca.jo)
trend <- matrix(1:nrow(sjf), ncol = 1)
colnames(trend) <- "trd"
ca.jo(sjf, type = "trace", ecdet = "const", K = 2, spec = "longrun", dumvar =
trend)
Best,
Bernhard
___
Am 31.03.2011 14:41, schrieb Sarah Goslee:
On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
this helps, thank you.
But if this code is in a function, and some user supplies a vector, I will
still have to round it in the function, I guess.
It's weird how 0.1 is different from round(0.1, dig
Thanks Henrique - that worked like a charm - I had tried lots of other
combinations before seeing your reply - wished I had asked sooner!
Alan
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, March 30, 2011 10:09 PM
To: Herbert, Alan G
Cc: r-he
n Thu, Mar 31, 2011 at 3:56 PM, Alexander Engelhardt
wrote:
> Am 31.03.2011 14:41, schrieb Sarah Goslee:
>>
>> On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt
>>>
>>> this helps, thank you.
>>> But if this code is in a function, and some user supplies a vector, I
>>> will
>>> still have to r
The code you gave me works fine with Finland, but the same for my data -
does not!
I do:
library(urca)
data(my.data)
dat1 <- my.data[, c("dY", "X", "dM")]
trend <- matrix(1:nrow(dat1), ncol = 1)
colnames(trend) <- "trd"
yxm.vecm <- ca.jo(dat1, type = "trace", ecdet = "const", K = 2, spec =
"longru
Dear Alexander,
Instead of testing 'somevector > 0.4', test 'abs(somevector - 0.4) <
some.small.number'
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
I am trying to do t-test to test whether the mean of one one column of the
data frame is greater then another. please help me out.
--
Arkajyoti Jana
M. Phil/ 2nd semester
Centre for Economic Studies and planning
School of Social Sciences
Jawaharlal Nehru University
New Delhi-67
[[alterna
Thank you so much, everyone, for your help.
Extremely valuable suggestions and extremely valuable learnings!
Dimitri
On Thu, Mar 31, 2011 at 5:03 AM, Martyn Byng wrote:
> Hi,
>
> Yep, that was what it was doing. For a sum across week, try something like
>
> get.week.flag <- function(dd) {
> ## g
On 11-03-30 7:00 PM, Carl Witthoft wrote:
I was cursing Matlab again today (what else is new) because the default
action for every Matlab command is to spew the result to the console,
and one must remember to put that darn ";" at the end of every line.
So I just wondered: was there ever a disc
On 11-03-31 7:24 AM, Alexander Engelhardt wrote:
Hi,
I had a piece of code which looped over a decimal vector like this:
for( i in where ){
thisdata<- subset(herde, herde$mlr>= i)
# do stuff with thisdata..
}
'where' is a vector like seq(-1, 1, by=0.1)
The solution to this problem is
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=dat
Dear List!
I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subsets and see
if they differ significantly. So what is my assumption - normal
distributed dat
Thanks Peters!
Just a few minor glitches now:
require(lattice)
data <- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,
Am 31.03.2011 15:46, schrieb Anna Lee:
Dear List!
I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subsets and see
if they differ significantly. So what
Folks,
I have a 'matching' matrix between variables A, X, L, O:
> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c("A", "X", "L", "O"), c("A", "X", "L", "O")))
> a
A X L O
A 1.00 0.4
Dear colleagues, working with the data frame below, trying to reverse two
variables I the error message below.
i searched through the help list but could not find any postings which could
help me solve the situation. I tried attaching and detaching the data frame to
no avail.
Yours, Simon Kiss
Dear Simon,
Your intention appears to be to call the recode() function in the car package.
That function is not generic, and consequently has no default method, leading
me to believe that you've called a recode() function in some other package.
I hope this helps,
John
I'm trying to do AEM eigenfunctions, but I have a different number of points
into each reservoir (e.g.: reservoir 1-4 points, reservoir 2-6 points,
reservoir 3-2 points). Can someone help me?Besides this in the Legendre's
book exemple in the part:>length.edge<-vector(length=nrow(edges.b))>fo
The easiest way of changing the font used for labels by the windows
graphics device (opened by a call to windows()) seems to be the following:
Let us assume we want to use the font Consolas for all labels:
windowsFonts(myfont="Consolas")
par(family="myfont")
If one later on wants to change the d
If you haven't already received an answer, a careful reading of
?formula
will provide it.
-- Bert
On Wed, Mar 30, 2011 at 11:42 PM, Yusuke Fukuda wrote:
>
> Hello R experts
>
> I have two linear regressions for sexes (Male, Female, Unknown). All have a
> good correlation between body length (r
Hi Arkajyoti,
Please learn to help yourself. Start with
help.search("t-test")
Also read one of the many introductory R tutorials on the web.
Best,
Ista
On Thu, Mar 31, 2011 at 9:07 AM, arkajyoti jana wrote:
> I am trying to do t-test to test whether the mean of one one column of the
> data fra
Hi Murali.
I haven't compared, but this is what I would do:
bestMatch<-function(searchVector, matchMat)
{
searchRow<-unique(sort(match(searchVector, colnames(matchMat #if
you're sure, you could drop unique
cat("Original row indices:")
print(searchRow)
matchMat<-
On 2011-03-31 07:44, Alexander Engelhardt wrote:
Am 31.03.2011 15:46, schrieb Anna Lee:
Dear List!
I have a unverse (basic population) which is not normally distributed.
Now from this universe I take some subsets. Each subset is normally
distributed within itself. I now want to compare the subs
Here is a way to do it using just base graphics:
layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE))
plot(runif(10), type='b', ylim=c(0,1))
x.tmp <- grconvertX(4, to='ndc')
y.tmp <- grconvertY(0.9, to='ndc')
plot(runif(20), type='l', ylim=c(0,1))
par(xpd=NA)
segments( 10, 1,
grconvertX(x.tmp, from='n
Hi,
Thank you *very* much. I'll remember that in the future!
Cheers,
-Jeremy N
Undergraduate Researcher in Macroecology
University of Ottawa
Department of Biology
Ad astra per alia porci!
On Thu, Mar 31, 2011 at 8:01 AM, Henrique Dallazuanna wrote:
> Better:
>
> plot(1, main = bquote("Yiel
Just a note, Base graphics does support transparency as long as the device
plotting to supports it.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help
On 2011-03-31 06:58, Rubén Roa wrote:
Thanks Peters!
Just a few minor glitches now:
require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(
Folks:
I think the following may be somewhat faster, as it avoids sorting:
bmat <- function(mx,vec)
{
nm <- colnames(mx)
ivec <- match(vec,nm)
sapply(ivec,function(k){
if(k==1)NA else {
lookat <- setdiff(seq_len(k-1),ivec) ## only those to left and not
in search vector ##
nm[loo
Hello,
In the past I have tended to reside more in the ANOVA camp but am trying to
become more familiar with regression techniques in R. I would like to get the
F change from a model as I take away factors:
SO...
mod1<-lm(y~x1+x2+x3)...mod2<-lm(y~x1,x2)...mod3<-lm(y~x1)
I can
?drop1
-- Bert
On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker wrote:
>
> Hello,
>
> In the past I have tended to reside more in the ANOVA camp but am trying to
> become more familiar with regression techniques in R. I would like to get
> the F change from a model as I take away factors:
>
> SO
You could do bootstrapping for confidence intervals or permutation tests for
just comparing the medians. There are other tools as well, just be sure that
you understand what you are testing and what assumptions are being made in
whatever method you choose.
--
Gregory (Greg) L. Snow Ph.D.
Stat
Try this:
bestMatch <- function(search, match) {
colnames(match)[pmax(apply(match[,search], 2, which.max) - 1, 1)]
}
On Thu, Mar 31, 2011 at 11:46 AM, wrote:
> Folks,
>
> I have a 'matching' matrix between variables A, X, L, O:
>
>> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0
Bert and anyone else with info,
First, Bert thank you for your quick reply. drop1 gives the results as a type
II anova. Is there a way to make drop1 give you type I anova (the args don't
appear to have a way to do so)? Another package/function perhaps?
Tyler
> Date: Thu, 31 Mar 2011 09
On Thu, Mar 31, 2011 at 4:02 AM, Daniel Kaschek
wrote:
> Hello,
> I use nls.profile to compute confidence intervals of parameter estimates
> of a non-linear model. When computing the profiles, the model function
> produces an error for certain parameter combinations. Therefore nls
> fails and so
dear lady and gentalmen:
i am gaoshan from kansas university.
i used such coding to deal with gel data
data <- ReadAffy()
Warning messages:
1: In file(out, "wt") :
cannot open file
'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd3f0b2e85': No such
file or directory
2: In file(ou
Hi,
My R version is old, and I would like to update it. How can I update R and keep
all the libraries? Thanks
Best,
Yunfei Li
--
Research Assistant
Department of Statistics &
School of Molecular Biosciences
Biot
On Thu, Mar 31, 2011 at 09:32:06AM -0700, wang peter wrote:
> dear lady and gentalmen:
> i am gaoshan from kansas university.
> i used such coding to deal with gel data
>
> data <- ReadAffy()
> Warning messages:
> 1: In file(out, "wt") :
> cannot open file
> 'C:\Users\gaoshan\AppData\
Dear R helpers,
I have a large data set with 36 variables and about 50.000 cases. The
variabels represent labour market status during 36 months, there are 8
different variable values (e.g. Full-time Employment, Student,...)
Only cases with at least one change in labour market status is
included i
Hi list,
I have a bunch of .csv files that are password-protected. I wonder if there is
a way to read them in in R without manually removing the password protection
for
each file?
Thank you very much!
...Tao
[[alternative HTML version deleted]]
_
This question has been asked by many people already. The easiest way is:
1) install the new version
2) copy all or the libraries that you installed later from the "library" folder
of older version to the new version
3) uninstall the old version
4) do a library update in the new version
Done!
.
Hi Yunfei,
It really depends to some extent on the version of R (and your OS) you
were using. Here is a link to a blog post that discusses some
potential strategies:
http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-the-r-code-to-do-it-on-windows/
There are also a number
Dear Hans,
clara doesn't require a distance matrix as input (and therefore doesn't
require you to run daisy), it will work with the raw data matrix using
Euclidean distances implicitly.
I can't tell you whether Euclidean distances are appropriate in this
situation (this depends on the interpre
Forgot to mention that my way of updating is for Windows only.
...Tao
From: Joshua Wiley
To: "Li, Yunfei"
Cc: r-help@r-project.org
Sent: Thu, March 31, 2011 11:05:29 AM
Subject: Re: [R] How to update R?
Hi Yunfei,
It really depends to some extent on the ve
Hi Tao,
You will need to give us more details. CSV files are pure text. How
are they password protected? Are they Excel files? In a password
protected compressed format? Encrypted?
Best Regards,
Josh
On Thu, Mar 31, 2011 at 11:00 AM, Shi, Tao wrote:
> Hi list,
>
> I have a bunch of .csv f
On Thu, 31 Mar 2011, Shi, Tao wrote:
Hi list,
I have a bunch of .csv files that are password-protected. I wonder if there is
There is nothing in the CSV standard about password protection. I
very much doubt that these actually CSV files. So please follow the
posting guide [*]
a way to
On Thu, Mar 31, 2011 at 11:00:48AM -0700, Shi, Tao wrote:
> Hi list,
>
> I have a bunch of .csv files that are password-protected. I wonder if there
> is
> a way to read them in in R without manually removing the password protection
> for
> each file?
I doubt that there is such a thing as a
Hi Josh and Prof. Ripley,
Thanks for the quick replies!
Sorry about the HTML email. It's the default for my yahoo account and forgot
to
switch.
The question was asked by a colleague of mine. After double-checking, yes,
they
are actually Excel files. Any idea on how to approach this?
Than
Hi Tao,
There are a number of R packages to work with reading/writing Excel
files (e.g., xlsReadWrite). If Excel is on the system in question, as
long as you have opened the file in Excel, you should be able to use
RODBC with RDCOMClient or rcom, but I am not sure those will work if
it is passwo
Thanks, Josh. I use xlsReadWrite routinely. It would be nice if it has a
password option.
...Tao
- Original Message
> From: Joshua Wiley
> To: "Shi, Tao"
> Cc: r-help@r-project.org
> Sent: Thu, March 31, 2011 11:50:27 AM
> Subject: Re: [R] read password-protected files
>
> Hi Ta
Using RDCOMCliente:
library(RDCOMClient)
eApp <- COMCreate("Excel.Application")
wk <- eApp$Workbooks()$Open(Filename="your_file",Password="your_password")
tf <- tempfile()
wk$Sheets(1)$SaveAs(tf, 3)
DF <- read.table(sprintf("%s.txt", tf), header = TRUE, sep = "\t")
On Thu, Mar 31, 2011 at 3:33 P
On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote:
> Dear Hans,
>
> clara doesn't require a distance matrix as input (and therefore
> doesn't require you to run daisy), it will work with the raw data
> matrix using
> Euclidean distances implicitly.
> I can't tell you whether Euclide
Is there a grayscale equivalent to alpha levels in rgb?
Example: I have the following to make red transparent circles overlap
with previously plotted blue symbols.
symbols(x=sites$long,y=sites$lat,circles=log(sites$prop.nem
+1),add=T,inches=F,bg=rgb(red=1,green=0,blue=0,
alpha=0.5),fg=rgb(red=1,
On Thu, Mar 31, 2011 at 08:48:02PM +0200, Hans Ekbrand wrote:
> On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote:
> > Dear Hans,
> >
> > clara doesn't require a distance matrix as input (and therefore
> > doesn't require you to run daisy), it will work with the raw data
> > matrix
On Thu, Mar 31, 2011 at 11:48 AM, Hans Ekbrand wrote:
>
> The variables are unordered factors, stored as integers 1:9, where
>
> 1 means "Full-time employment"
> 2 means "Part-time employment"
> 3 means "Student"
> 4 means "Full-time self-employee"
> ...
>
> Does euclidean distances make sense on
Dear list,
Can anyone tell me how to obtain the rank of a sparse Matrix, for
example from package Matrix (class dgCMatrix)? Here is an example of
QR decomposition of a sparse matrix (from the sparseQR class help).
library(Matrix)
data(KNex)
mm <- KNex$mm
str(mmQR <- qr(mm))
Similarly, using the
Try this:
with(trees, symbols(Height, Volume, circles=Girth/24, inches=FALSE, fg
= "gray30",
bg = sapply(apply(rbind(col2rgb(gray(seq(0,1, l = 10)))/255, alpha =
0.5), 2, as.list), do.call, what = rgb)))
On Thu, Mar 31, 2011 at 2:52 PM, Frostygoat wrote:
> Is there a grayscale equivalent to al
Hey List,
does anyone know how I can generate a vector of random numbers from a
given distribution? Something like "rnorm" just for non normal
distributions???
Thanks a lot!
Anna
--
Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zugesandt worden sein, bitte ich
See
http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf
?update.lm
-- Bert
On Thu, Mar 31, 2011 at 10:27 AM, Tyler Rinker wrote:
> Bert and anyone else with info,
>
> First, Bert thank you for your quick reply. drop1 gives the results as a
> type II anova. Is there a way to make drop1 give you
There are several non-normal distributions built in that you could just use
their generationg functions. There are packages (most with dist somewhere in
the name) that allow for functions on standard distributions including
combining together distributions.
If your distribution is not one of t
On 31-Mar-11 19:23:33, Anna Lee wrote:
> Hey List,
> does anyone know how I can generate a vector of random numbers
> from a given distribution? Something like "rnorm" just for non
> normal distributions???
>
> Thanks a lot!
> Anna
SUppose we give your distribution the name "Dist".
The generic a
I tried multiple imputation with aregImpute() and
fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1.
The warning message below suggests that summary(f) of
fit.mult.impute() would only use the last imputed data set.
Thus, the whole imputation process is ignored.
"Not using a Design fitti
On Mar 31, 2011, at 8:41 AM, Peter Ehlers wrote:
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data<- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
Hello,
I try to plot the marginal effect by using package "effects" (example of the
graph i want to get is in the attached picture).
All variables are continuous.
Here is regression function, results and error effect function gives:
> mreg01 = lm(a90$enep1 ~ a90$enpres + a90$proximity1 + (a90$en
Hi
I am conducting a meta-analysis on proportions and I'm interested in using the
exact binomial likeliood for the within-study effects rather than the
approximate approaches and I don't want to use the quasi-likelihood approach
in R's glmmPQL.
Does anyone know how I can do this using nlme? I c
Dear Pfaff,
Would that be possible to fit a Time varying VECM using urca?
Yours,
Renoir
On Thursday, March 31, 2011, Grzegorz Konat wrote:
> The code you gave me works fine with Finland, but the same for my data -
> does not!
> I do:
>
> library(urca)
> data(my.data)
> dat1 <- my.data[, c("dY",
Hi,
I want to create variable names from within my code, but can't find any
documentation for this.
An example is probably the best way to illustrate. I am reading data in from a
file, doing a bunch of stuff, and want to generate variables with my output.
(I could make a "list of lists" and n
Take a look in ?assign
On Thu, Mar 31, 2011 at 5:42 PM, Noah Silverman wrote:
> Hi,
>
> I want to create variable names from within my code, but can't find any
> documentation for this.
>
> An example is probably the best way to illustrate. I am reading data in from
> a file, doing a bunch of s
Thanks, Henrique! I'll try that.
...Tao
- Original Message
> From: Henrique Dallazuanna
> To: "Shi, Tao"
> Cc: Prof Brian Ripley ; r-help@r-project.org
> Sent: Thu, March 31, 2011 12:00:45 PM
> Subject: Re: [R] read password-protected files
>
> Using RDCOMCliente:
>
> library(RDCO
Perfect.
Thank You
On Mar 31, 2011, at 1:47 PM, Mark Leeds wrote:
> hi noah: assign(thing you paste, 123, envir=whatever) should work I think.
>
>
> On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman
> wrote:
> Hi,
>
> I want to create variable names from within my code, but can't find any
Dear all,
I'm deep into Chambers' SoDA and R-exts.html but I can't find all
answers. The thing is that I would like to run my unit tests right
after a package installation.
That is while the command "R CMD check " runs all files
named /tests/*.R (so unit tests can be placed there) I
wonder if it is
On 01/04/11 08:50, Ted Harding wrote:
On 31-Mar-11 19:23:33, Anna Lee wrote:
Hey List,
does anyone know how I can generate a vector of random numbers
from a given distribution? Something like "rnorm" just for non
normal distributions???
Thanks a lot!
Anna
SUppose we give your distribution the
I have a model Q=K*A*(R^r)*(S^s)
A, R, and S are data I have and K is a curve fitting parameter. I
have linearized as
log(Q)=log(K)+log(A)+r*log(R)+s*log(S)
I have taken the log of the data that I have and this is the model
formula without the K part
lm(Q~offset(A)+R+S, data=x)
What is the fo
The best thing to do is to understand how to use 'list' for this
purpose. Much easier to handle the information.
On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman wrote:
> Hi,
>
> I want to create variable names from within my code, but can't find any
> documentation for this.
>
> An example is p
Hi, I am stuck on this: how to specify a match pattern that means not to
include
"abc"?
I tried:
grep("^(abc)", "hello", value=T)
should return "hello".
while
grep("^(abc)", "hello abcd foo", value=T)
should return character(0).
But both returned character(0).
Thanks
John
[[alte
Hmm, looks like I'm 'answering' my own question here...
library(Matrix)
data(KNex)
mm <- KNex$mm
str(mmQR <- qr(mm))
# new stuff:
R <- mmQR@R
Rdiag <- diag(R)
rank <- sum(Rdiag > max(dim(mm))*.Machine$double.eps*abs(R[1,1])) #
this is the matlab default I think?
# 712
for comparison, rankMatrix f
Am 31.03.2011 19:31, schrieb array chip:
Hi, I am stuck on this: how to specify a match pattern that means not
to include "abc"?
I tried:
grep("^(abc)", "hello", value=T) should return "hello".
> grep("[^(abc)]", "hello", value=T)
[1] "hello"
HTH,
Bernd
___
Thanks Bernd! I tried your approach with my real example, sometimes it worked,
sometimes it didn't. For example
grep('[^(arg)]\\.symptom',"stomach.symptom",value=T)
[1] "stomach.symptom"
grep('[^(arg)]\\.symptom',"liver.symptom",value=T)
character(0)
I think both examples should return the text
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