Let F be the distribution function of Y, PSI the standard normal distribution
anf IPSI it's inverse. Let f(x) = IPSI(F(x)). It is not difficult to see that
f(Y) has standard normal distribution.
So you can replace F with the empirical distribution and IPSI is the qnorm
function of R.
--- On
On Wed, 25 Jun 2008, Moshe Olshansky wrote:
Let F be the distribution function of Y, PSI the standard normal
distribution anf IPSI it's inverse. Let f(x) = IPSI(F(x)). It is not
difficult to see that f(Y) has standard normal distribution. So you can
replace F with the empirical distribution
Hi all,
I have a simple question and couldn't find any post on this. When
plotting simple scatterplots (other plots as well), e.g.,
x-rnorm(30, 10, 1)
y-rnorm(30, 10, 1)
plot(x, y, pch = 15, cex = 1),
the points, even those close to each other, may have visibly different
sizes. Do you know
On 25 Jun 2008, at 19:45, Gabor Grothendieck wrote:
Try this:
plot(1, xlab = ~ alpha / V * m^-3 * kg ^-2 * l^4)
Thanks, I would never have expected this code to work, this is a
mystery to me! Actually, I thought xlab wanted an expression, but it
seems to be happy with a formula. Also,
Hi,
I'm trying to do a double for loop like this:
for (k in 1:1000){
for (i in 1:200){
y[i]-rbinom(1,1,0.8)
x1[i]-ifelse(y[i]==1,rnorm(1,mean=20, sd=2),rnorm(1,mean=16, sd=2.2))
}
for (j in 1:300){
}
}
Does anyone know a good reference about double loops?
Thank you,
Sigalit
You can at least get rid of the
for (i in 1:200){
y[i]-rbinom(1,1,0.8)
x1[i]-ifelse(y[i]==1,rnorm(1,mean=20, sd=2),rnorm(1,mean=16, sd=2.2))
loop with the following
y - rbinom(200, 1, 0.8)
y.1 - y == 1 # get logical vector of y == 1
x1 - numeric(200) # allocate the vector
x1[y.1] -
Thank you for your help
On 6/26/08, jim holtman [EMAIL PROTECTED] wrote:
You can at least get rid of the
for (i in 1:200){
y[i]-rbinom(1,1,0.8)
x1[i]-ifelse(y[i]==1,rnorm(1,mean=20, sd=2),rnorm(1,mean=16, sd=2.2))
loop with the following
y - rbinom(200, 1, 0.8)
y.1 - y == 1 #
On Thu, 26 Jun 2008, Marcin Kozak wrote:
Hi all,
I have a simple question and couldn't find any post on this. When
plotting simple scatterplots (other plots as well), e.g.,
x-rnorm(30, 10, 1)
y-rnorm(30, 10, 1)
plot(x, y, pch = 15, cex = 1),
the points, even those close to each other, may
oh, I'm sorry...
here's my affiliation:
Mikhail Spivakov, PhD [EMAIL PROTECTED]
Interdisciplinary Postdoctoral Fellowhttp://www.ebi.ac.uk/~spivakov/
European Bioinformatics Institute Tel: +44 1223 492660 (office)
Wellcome Trust Genome Campus+44 7985 09 6675 (mob)
Cambridge CB10
Thanks for getting back to me.
i meant that i am seeking for a transformation of an s-shaped response
(continuous data) to assure normally distributed residuals of the response.
however, i am unsure if the Let F be the distribution... is now the correct
thing for me to do. if this is the
Deepayan Sarkar [EMAIL PROTECTED] writes:
On 6/25/08, Franz Mueter [EMAIL PROTECTED] wrote:
As for your first problem, try:
xyplot(numbers~breaks|moltype, groups = type, data = alldata, type = l)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf
Hi
I'd like to write.table a dataframe, but with an specific order of
columns. Is there a direct way to do it? or I have to generate a new
dataframe as follows:
t - data.frame(c=1:10, b=11:20, a=letters[1:10])
t2 - data.frame(a=t$a, b=t$b, c=t$c)
write.table(t2, row.names=F)
Thanks for any
bquote is used like this:
alpha - 5
xlab - bquote(.(alpha) / V * m^-3 * kg ^-2 * l^4)
plot(1, xlab = xlab)
On Thu, Jun 26, 2008 at 4:21 AM, baptiste Auguié [EMAIL PROTECTED] wrote:
On 25 Jun 2008, at 19:45, Gabor Grothendieck wrote:
Try this:
plot(1, xlab = ~ alpha / V * m^-3 * kg ^-2 *
Try:
write.table(t[c(a, b, c)], row.names=F)
On Thu, Jun 26, 2008 at 6:28 AM, juli pausas [EMAIL PROTECTED] wrote:
Hi
I'd like to write.table a dataframe, but with an specific order of
columns. Is there a direct way to do it? or I have to generate a new
dataframe as follows:
t -
Mark Kimpel wrote:
Thanks for the pointers. I think the package is great, just want to
use it to its full potential without driving the list crazy with
questions. Below is my output to help and sessionInfo(). I don't see
the scaling functions, although I now see that they can be retrieved
with
On Thu, 26 Jun 2008, baptiste Auguié wrote:
On 25 Jun 2008, at 19:45, Gabor Grothendieck wrote:
Try this:
plot(1, xlab = ~ alpha / V * m^-3 * kg ^-2 * l^4)
Thanks, I would never have expected this code to work, this is a mystery to
me! Actually, I thought xlab wanted an expression, but
Dear All,
I'm having trouble ploting multiple xYplots (Hmisc) within the same pg
although I've done it before w/ xyplots. I've produced 4 similar plots
e.g.
rna.h-xYplot(Cbind(RNA,Lo,Up)~HO|factor(MO,labels=c(April,
May,June)),data=diel.data,method=bars,type=b,
Hi everyone !
I'm am trying to fit a kriging model to a set of data. When I just run the
likfit command I can obtain the results. However when I try to pass
additional arguements to the optimization function optim I get errors. That
is I want to obtain the hessian matrix so matrix
Dear all
Nobody responded to my previous post so far so I try with more offending
subject.
I just encountered a strange problem with nls formula. I tried to use nls
in cycle but I was not successful. I traced the problem to some parse
command.
Here is an example
DF-data.frame(x=1:10,
Dear Petr,
I think it's a feature. the formula interface also won't let you specify
the slots of S4 objects in the model spec.
How about
coef(nls(y~a*x^b, data=list(x=DF[,1], y=DF[,2]), start=list(a=3, b=.7)))
?
On Thu, 26 Jun 2008, Petr PIKAL wrote:
Dear all
Nobody responded to my
On Thu, Jun 26, 2008 at 12:14 AM, Wacek Kusnierczyk
[EMAIL PROTECTED] wrote:
sapply(dats,function(x){sapply(x,min)})
you can achieve the same with
sapply(dats, sapply, min)
Did you actually try it?
dats - data.frame(1:10,2:11)
sapply(dats,sapply,min)
X1.10 X2.11
[1,]
Try this and note, in particular, that the model: line
in the output has the correct variables substituted:
nm - names(DF)
eqn - sprintf(%s ~ a * %s, nm[2], nm[1])
nls(eqn, DF, start = c(a = 1))
Nonlinear regression model
model: y ~ a * x
data: DF
a
1.133
residual sum-of-squares:
Mzabalazo Ngwenya wrote:
Hi everyone !
I'm am trying to fit a kriging model to a set of data. When I just run
the likfit command I can obtain the results. However when I try to
pass additional arguements to the optimization function optim I get
errors. That is I want to obtain the hessian
Sorry. I'm using Windows XP and R 2.7.0, and the same problem occurred
for various graphics devices (windows, pdf, jpeg).
Marcin
On Thu, Jun 26, 2008 at 11:54 AM, Prof Brian Ripley
[EMAIL PROTECTED] wrote:
On Thu, 26 Jun 2008, Marcin Kozak wrote:
Hi all,
I have a simple question and
OK, thanks to both of you for the clarifications. I guess part of my
confusion came from the numerous functions and concepts involved in
producing such labels:
- call vs string vs formula vs expression ...
- substitute, bquote, expression, ~, .(), ...
I take it as a good thing once you
Hi, I wonder if you could help me with my barcharts, I am using the
barchart function from the lattice library.
I have data like this:
Complex,Organisms,Percentage
130,0,50
130,1,10
130,2,20
130,3,15
130,4,5
133,0,10
133,1,15
133,2,20
133,3,50
133,4,5
I draw barcharts using this command...
On Thu, 26 Jun 2008, Marcin Kozak wrote:
Sorry. I'm using Windows XP and R 2.7.0, and the same problem occurred
for various graphics devices (windows, pdf, jpeg).
I don't see it for pdf -- that's probably a viewer artefact and you need
to look at higher resolution.
For windows() and jpeg()
try:
barchart(Percentage~Organisms|factor(Complex),data=data,layout=c(1,2),col=rainbow(5),box.ratio=3,horizontal=FALSE,xlim=c(
0 , 1 , 2 , 3 , 4 ),ylim=c(1,100),xlab=)
On Thu, Jun 26, 2008 at 8:55 AM, Elizabeth Webb [EMAIL PROTECTED] wrote:
Hi, I wonder if you could help me with my barcharts,
Kenn Konstabel wrote:
On Thu, Jun 26, 2008 at 12:14 AM, Wacek Kusnierczyk
[EMAIL PROTECTED] wrote:
sapply(dats,function(x){sapply(x,min)})
you can achieve the same with
sapply(dats, sapply, min)
Did you actually try it?
dats - data.frame(1:10,2:11)
Hi all!
I would like to estimate confidence intervals for a non lm model.
For example, I use a mixed model of the form:
md=lme(y~x1+I(x1^2)+x2 ...)
Parameters x1+I(x1^2) are fixed effects and I would like to plot the predicted
(partial) curve corresponding to these ones, along with 90% CI
Hi all!
I would like to estimate confidence intervals for a non lm model.
For example, I use a mixed model of the form:
md=lme(y~x1+I(x1^2)+x2 ...)
Parameters x1+I(x1^2) are fixed effects and I would like to plot the predicted
(partial) curve corresponding to these ones, along with 90%
All,
I have data across 5 time points that I am graphing via xyplot, along with
error bars. For one of the variables I have missing data for two of the
time points. The code below is okay but I can't seem to get the lines to
connect across the missing time points. Does anyone now how to
Minor correction -- omitted comma. Should be:
write.table(t[ , c(a, b, c)], row.names=FALSE)
Also, using the name t should be avoided,
because t is a built-in function: t()
-Don
At 8:02 AM -0300 6/26/08, Henrique Dallazuanna wrote:
Content-Type: text/plain
Content-Disposition: inline
Hi,
I have two figures where
I stacked together as one PNG.
Top - scatter plot
Bottom - density plot.
However these two figures are squeezed
together as rectangle figures each.
(i.e. the y axes are compressed)
Is there a way I can resize the figure?
So that it can show proportional and
hi,
I'm student doing Msc in operational research and applied statistics in Salford
university ,now I'm doing my project which is conducting meta-analysis in R , i
was wondering how i can evaluate continuous outcome using rmeta package, i
really appreciate if you can help me in that matter.
Dear R community, I am using densityplot (lattice package) for a large
dataset and wish to print it to a jpeg (the pdf is huge). R crashes
consistently. Am I doing it wrong or is densityplot incompatible with jpeg?
I work on a Mac, R 2.7.0.
require(lattice)
jpeg(test.jpeg)
d[1:10]
[1]
One way to get at all the functions (although not
as easy as help() is
library(rgl); ls(pos=2)
which literally lists the available functions. As
Duncan said, there are multiple functions documented in each page and
only the page definitions show up (e.g. matrices is the page for
work[ing]
Sorry, my mistake, - it works both ways with a correct example and neither
way with the wrong example.
k
On Thu, Jun 26, 2008 at 4:25 PM, Wacek Kusnierczyk
[EMAIL PROTECTED] wrote:
sapply(dats,function(x){sapply(x,min)})
you can achieve the same with
sapply(dats, sapply, min)
looks like you need to have a look at the package description. If you want
help from this list you probably need to look at the posting guidelines, and
then do a little poking around to figure out what your specific problems
are.
Stephen
On Thu, Jun 26, 2008 at 10:19 AM, mohammed alawa [EMAIL
Maybe approx() will work?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
Sent: Thursday, June 26, 2008 10:26 AM
To: r-help@r-project.org
Subject: [R] Connecting lines across missing data points, xyplot
All,
I have data
Duncan Ben,
Thanks for giving me some tips as to how I can best investigate
packages. Ive been using ESS-Emacs and the help pages are HTML. Looks
like I need to figure out how to enable HTML help.
As for a draft vignette, sure, I would be happy (and honored) to
contribute as best I can. I would
Hello all,
Working with a typical xyplot figure (for example the first from
example(lattice) )
I would like to place the tick marks inside the panels instead of outside.
I believe working with the axis argument might be the way to do it, but do not
see how?
Could anyone help me out in doing
Dear list,
I am trying to use the 'mvrnorm' function from the MASS package for
simulating multivariate Gaussian data with given covariance matrix.
The diagonal elements of my covariance matrix should be the same,
i.e., all variables have the same marginal variance. Also all
correlations between
Hi Geord,
I run the code bellow without problem on R
R version 2.6.2 (2008-02-08)
i386-pc-mingw32
require(lattice)
setwd(c:\\temp)
jpeg(test.jpeg)
d-runif(10)
d[1:10]
densityplot(~d[1:10])
dev.off()
Good luck with Mac!
miltinho
On 6/26/08, Georg Ehret [EMAIL PROTECTED] wrote:
Dear R
Dear UseRs,
I would like to know the way to create a new column by naming it
simultaneously.
For example, in for() loop I want to create columns named as
paste(test, i, sep = ), as shown below.
dt - data.frame(a = c(1, 2, 3), b = c(3, 2, 2), c = c(1, 3, 5))
Try this:
xyplot(rivers ~ rivers, scale = list(x = list(tck = -1)))
On Thu, Jun 26, 2008 at 12:02 PM, GOUACHE David
[EMAIL PROTECTED] wrote:
Hello all,
Working with a typical xyplot figure (for example the first from
example(lattice) )
I would like to place the tick marks inside the
Is this what you want
dt - data.frame(a = c(1, 2, 3), b = c(3, 2, 2), c = c(1, 3, 5))
for (i in 1:2){
+ dt[[paste('test',i,sep=)]] - rep(i,3)
+ }
dt
a b c test1 test2
1 1 3 1 1 2
2 2 2 3 1 2
3 3 2 5 1 2
On Thu, Jun 26, 2008 at 12:23 PM, Dong-hyun Oh [EMAIL
Dear Dong-hyun,
What about
dt - data.frame(a = c(1, 2, 3), b = c(3, 2, 2), c = c(1, 3, 5))
test=matrix(rep(c(1,2),each=3),ncol=2)
colnames(test)=paste('test',1:2,sep=)
cbind(dt,test)
?
HTH,
Jorge
On Thu, Jun 26, 2008 at 12:23 PM, Dong-hyun Oh [EMAIL PROTECTED] wrote:
Dear UseRs,
I would
Hi,
Fortunately, I found a way.
--
dt - data.frame(a = c(1, 2, 3), b = c(3, 2, 2), c = c(1, 3, 5))
for(i in 1:2){
dt[, paste(test, i, sep = )] - rep(i, 3)
}
-
Thanks.
Best,
Dong-hyun Oh
On Jun 26, 2008, at 6:34 PM, Jorge Ivan Velez wrote:
Hi.
I want to do an lmer model but have doubts of what family I should use.
My response variable was originally a proportion, however I standarized it
for each year of data collection (20 in total). After standarizing it I
checked for normality with the Kolmogorov-Smirnov test, and it turns out
Well, if you think about the geometry, all correlations equal usually
won't work. Think of the SDs as the sides of a simplex and the
correlations as the cosines of the angles between the sides (pick one
variable as the 'origin'.) Only certain values will give a valid
covariance or correlation
I have inherited some S-Plus code and objects that I am trying to get
to work in R.
The object files (3 __i files and one ___nonfi file) are stored in a
folder (named _Model), but when I try to use attach(c:/_Model) I
get an error saying Error in readChar(con, 5L, useBytes = TRUE) :
Yes, *but* (because you keep on reporting misleadingly), jpeg() on Windows
is a completely different device to jpeg() on Mac OS X.
jpeg() on Mac OS X with R 2.7.0 is based on Quartz, so this is a Mac
OS-specific issue, for r-sig-mac and not here.
Folks, please do remember that there are lots
To make David's approach a little more concrete:
You can always have correlations all equal to 1 --
the variables are all the same, except for the names
you've given them. You can have two variables
with correlation -1, but you can't get a third variable
that has -1 correlation to both of the
I'd like to thank those who contacted me with ideas on how to solve
this little problem. I learned something from looking through each
snippet of code, even if it wasn't doing quite what I'd hoped it would
do. Mark Leeds deserves special thanks, for helping me debug my
several attempts to improve
On Thu, 26 Jun 2008, Morten Hønsen wrote:
I have inherited some S-Plus code and objects that I am trying to get to work
in R.
The object files (3 __i files and one ___nonfi file) are stored in a folder
(named _Model), but when I try to use attach(c:/_Model) I get an error
saying Error in
Hello,
I am trying to wrap my head around the coordinates systems associated with
the layout() function ...with the end goal of simply drawing a decorative
line in the upper margin of my figure, which is composed of three plots.
My output is defined as this:
Hello list,
I'm new to R and I have a problem :-) Below is what my data file that looks
like. I tried to import and contour this data by doing this:
cv_data - read.table(cv_data.csv,sep=,,header=TRUE)
attach(cv_data)
contour(x,y,z)
I get the error Error in contour.default(x,y,z) :
If I understand you correctly, then to paraphrase what Brian Ripley has
stated in recent posts, it is not the (possibly transformed) response that
you want to be normal, but rather the error distributions. Your response
presumably contains systematic variation due to your covariates (your
model).
Hi
I am changing my databases from S-plus to R. I want to know how can I use
that?
May I only attach the folder _Data equal I do in S-plus?
Thanks!
Best Regards,
Leandro Marino
__
R-help@r-project.org mailing list
I am looking for a way to generate a data matrix that contains all possible
response patterns for 10 binary items. This should produce a matrix with 10
rows (representing 10 items) and 1024 columns (representing 2^10 possible
response patterns). Does anyone know of code that would produce such
Hi List,
In Windows, if I do R CMD build mypkg, then I'll get 'mypkg_1.0.tar.gz'.
Any option in R CMD build lets me to change the version, i.e. gives me
'mypkg_2.0.tar.gz? It seems -version option doesn't do anything for me.
Is it OK if I just change the version number in the file name
this is probably a cludge, and there may be a neater way to do this,
but... here's one:
a = 0:1
for (i in 1:9){ a= merge(unname(a), 0:1) }
a = t(a)
after the for loop, 'a' will contain a 1024 row by 10 col dataframe.
putting it through a transpose, gives you the 10 rows by 1024 cols
Change the version on DESCRIPTION file
On Thu, Jun 26, 2008 at 3:36 PM, Tao Shi [EMAIL PROTECTED] wrote:
Hi List,
In Windows, if I do R CMD build mypkg, then I'll get 'mypkg_1.0.tar.gz'.
Any option in R CMD build lets me to change the version, i.e. gives me
'mypkg_2.0.tar.gz? It seems
I can't seem to find just what I'm looking for in R help, Everitt and
Hothorn HSAUR, Murrell's book, or the R graphics gallery at
http://addictedtor.free.fr/graphiques/. Probably not looking
efficiently, but anyway,
If my data look like this:
head(data)
cat startyear studentid
1 other
Try this also:
t(expand.grid(rep(list(0:1), 10)))
On Thu, Jun 26, 2008 at 3:18 PM, SARAH A DEPAOLI [EMAIL PROTECTED] wrote:
I am looking for a way to generate a data matrix that contains all possible
response patterns for 10 binary items. This should produce a matrix with 10
rows
Please **READ** ?contour. x and y are the location of grid lines, z must be
a matrix. Your data are **not** of that form.
Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Bob and Deb
Sent: Thursday, June 26, 2008 11:01 AM
To:
Got it! Thank you very much!
...Tao
Date: Thu, 26 Jun 2008 20:41:29 +0200
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
CC: r-help@r-project.org
Subject: Re: [R] a question regarding package building
Tao Shi wrote:
Hi List,
In Windows,
I have read the contour help. I guess what I am asking is how to transform
my data into a form that contour (or maybe some other functions?) so I can
contour my x,y,z data. I know I can do it in SyStat. I'm hoping that R
would allow me to do something like this.
Bob
On Thu, Jun 26, 2008 at
Tao Shi wrote:
Hi List,
In Windows, if I do R CMD build mypkg, then I'll get 'mypkg_1.0.tar.gz'. Any option in R
CMD build lets me to change the version, i.e. gives me 'mypkg_2.0.tar.gz? It seems
-version option doesn't do anything for me.
Is it OK if I just change the version number
Bob and Deb [EMAIL PROTECTED] writes:
I'm new to R and I have a problem :-) Below is what my data file that looks
like. I tried to import and contour this data by doing this:
cv_data - read.table(cv_data.csv,sep=,,header=TRUE)
attach(cv_data)
contour(x,y,z)
I get the error
mat - outer( 0:9, 0:(1024-1), function(x,y) y %/% (2^x) %% 2 )
On Thu, 26 Jun 2008, Daniel Folkinshteyn wrote:
this is probably a cludge, and there may be a neater way to do this, but...
here's one:
a = 0:1
for (i in 1:9){ a= merge(unname(a), 0:1) }
a = t(a)
after the for loop, 'a'
Christopher W. Ryan cryan at binghamton.edu writes:
I can't seem to find just what I'm looking for in R help, Everitt and
Hothorn HSAUR, Murrell's book, or the R graphics gallery at
http://addictedtor.free.fr/graphiques/. Probably not looking
efficiently, but anyway,
If my data look
G'day Petr,
On Thu, 26 Jun 2008 13:57:39 +0200
Petr PIKAL [EMAIL PROTECTED] wrote:
I just encountered a strange problem with nls formula. I tried to use
nls in cycle but I was not successful. I traced the problem to some
parse command.
[...]
I am not sure if this behaviour is a bug or
Ben--
worked perfectly. Thank you. My first real experience with lattice
graphics.
--Chris
Christopher W. Ryan, MD
SUNY Upstate Medical University Clinical Campus at Binghamton
40 Arch Street, Johnson City, NY 13790
cryanatbinghamtondotedu
PGP public keys available at
Hi,
I was reading e-mail about one doubt that you were when you read files
.sas7bdat in R.
Now, I am with the same problem. But I don´t know how can i do that. I have
download de Sas Viewer and i am using this sintax in r:
read.ssd(X:\\users\\Anresc07,que0411.sas7bdat)
SAS failed. SAS program
Hi
Take a look at ...
http://www.stat.auckland.ac.nz/~paul/R/basegraphics.pdf
... it's getting old, but a lot of what it says should still hold,
especially the stuff about coordinate systems and what you can draw where.
Paul
[EMAIL PROTECTED] wrote:
Hello,
I am trying to wrap my head
This is about R 2.7.0 and related packages on windows NT.
I have a mixture of numeric and character data and empty cells in an
Excel spreadsheet with several tabs that I'm trying to read with
sqlFetch from RODBC.
The data that is returned by sqlFetch is unfortunately not identical
to the source
Hi everyone,
I'm looking for some function using R to do the relative mortality
function (Klein Moechberger EXAMPLE 6.3) Im working with this datas,
from 26 psychiatric patients.
Someone can help with this?
O__ José Bustos M.
c/ /'_ --- Master Apllied Stat Program
(*) \(*) --
You could try read.xls or xls2csv in the gdata package and see
how that works.
On Thu, Jun 26, 2008 at 4:44 PM, Al Pöhi [EMAIL PROTECTED] wrote:
This is about R 2.7.0 and related packages on windows NT.
I have a mixture of numeric and character data and empty cells in an
Excel spreadsheet
This is not exactly an R question but the R code below may make my
question more understandable.
If one plots sin(x) where x runs from -pi to pi , then the curve hovers
around zero obviously. so , in astationary in the mean sense,
the series is stationary. But, clearly if one plots the acf,
You need to specify the path to SAS, not Sas viewer. From the help page:
sascmd: character string giving full path to SAS executable.
I don't think you have done that.
On Thu, 26 Jun 2008, Leandro Marino wrote:
Hi,
I was reading e-mail about one doubt that you were when you read files
Prof Brian Ripley wrote:
You need to specify the path to SAS, not Sas viewer. From the help page:
sascmd: character string giving full path to SAS executable.
I don't think you have done that.
Or, put differently, it is not going to work unless you have SAS itself,
not just the viewer.
On 6/26/08, David Afshartous [EMAIL PROTECTED] wrote:
All,
I have data across 5 time points that I am graphing via xyplot, along with
error bars. For one of the variables I have missing data for two of the
time points. The code below is okay but I can't seem to get the lines to
I would like a sequence of dates with a time step of 15 minutes
starting:
1/1/2006 00:00:00 - 12/31/2006 23:45:00
function(x) {
chron(sub( .*, , x), gsub(.* (.*), \\1:00, x))
}
this is the piece of code I use to read in zoo objects
for any help I would be grateful I have tried sequence and I
Dear All,
I'm having trouble ploting multiple xYplots (Hmisc) within the same pg
although I've done
it before w/ xyplots. I've produced 4 similar plots e.g.
rna.h-xYplot(Cbind(RNA,Lo,Up)~HO|factor(MO,labels=c(April,
May,June)),data=diel.data,method=bars,type=b,
Hi,
I am a Newbie for R. I just installed R-base on my notebook with
openSuSE 11. However, I always got compilation errors in installing
add-on packages. For example, when installing igraph I got the
following error:
___
* Installing
Try this:
library(chron)
t1 - chron(1/1/2006, 00:00:00)
t2 - chron(12/31/2006, 23:45:00)
deltat - times(00:15:00)
tt - seq(t1, t2, by = times(00:15:00))
Note that if you have some data and your intention is simply to
create the index for it so you can create a zoo or zooreg object
then the
Dear All,
I am having trouble in using R function constrOptim to do constraint
optimization. It seems that constrOptim calls function optim when it
does the optimization, and optim allows us to set method to be SANN
if we want to use simulated annealing. In optim, the function allows us
to
Gabor this is better than using
x = seq(ISOdate(2006,1,1, 0, 0, 0), by = 15 min, length.out=35040)
this works but your code below is exactly what I needed instead of almost.
thankyou very much
Stephen
On Thu, Jun 26, 2008 at 6:06 PM, Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Try this:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leandro Marino
Sent: Thursday, June 26, 2008 12:50 PM
To: r-help@r-project.org
Subject: [R] Read sas7bdat
Hi,
I was reading e-mail about one doubt that you were when you read files
.sas7bdat in
First the R question. I have the results of a rather large survey
(thousands of forms, each with dozens of questions) with some existing
weights and expansion factors. I wish to add additional weighting
factors, based on new information that elements of certain variables
should appear in certain
Not sure if this is sufficient but nearcor in the sfsmisc package
will find the nearest correlation matrix to a given matrix.
On Thu, Jun 26, 2008 at 12:11 PM, Mizanur Khondoker
[EMAIL PROTECTED] wrote:
Dear list,
I am trying to use the 'mvrnorm' function from the MASS package for
If the main diagonal element of matrix A is 1 and the off diagonal element is a
then for any vector x we get that t(x)*A*x = (1-a)*sum(x^2) +a*(sum(x))^2 . If
we want A to be positive (semi)definite we need this expression to be positive
(non-negative) for any x!= 0. Since sum(x)^2/sum(x*2) = n
I have trying to figure this out all day so hopefully the answer isn't too
obvious. I am able to view a graph in the viewer window. However, I need to
export graph outside of the viewer window. Here is the script I am using:
png(Compare.png)
plot(compare$DepthSLI, compare$DischargeSLI,
not sure why it doesn't work, but try the following:
first, plot to a regular window, then run:
dev.copy(device=png, file=yourfilename.png)
dev.off()
see if that produces a file you want.
another note: what do you mean you can't just copy and paste the graph
in ubuntu? doesn't pressing
Hi,
this is my code:
model-glm(Risk ~ Status + Duration + Credit_History + Credit_amount + Savings
+
Employment + Inst_rate + Residence + Property + Age + plans +
Respected All,
I am writing a program in R and facing some problem with applying if statment.
Program first draw random numbers from bivariate normal distribution
then compute variable say Pi and Pij from that sample and
then further computation
..
.
.
In some samples Pij is
Try:
RSiteSearch(rejection sampling)
On Thu, Jun 26, 2008 at 11:52 PM, Nadeem Shafique
[EMAIL PROTECTED] wrote:
Respected All,
I am writing a program in R and facing some problem with applying if
statment.
Program first draw random numbers from bivariate normal distribution
then
Hi everyone.
I installed affylmGUI and oneChannelGUI package on R 2.7.1 with the
latest version of BioC.
After I start R, I do get an error that says Error in
loadNamespace(name) : there is no package
called 'affylmGUI' and a pop-up window with a message fatal error :
unable to restore saved
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