thanks so much! grapphics.off() works!
2010/12/16 Michael Bibo-2 [via R] <
ml-node+3090212-673909616-205...@n4.nabble.com
>
> kiotoqq googlemail.com> writes:
>
> >
> >
> > I'm pretty sure that I did everything right, but my R is just not drawing
> the
> > mosaicplot that I want, and there is al
Hi:
On Wed, Dec 15, 2010 at 10:05 PM, Anirban Mukherjee wrote:
> Hi all,
>
> Suppose:
>
> y<-rnorm(100)
> x1<-rnorm(100)
> lm.yx<-lm(y~x1)
>
> To predict from a new data source, one can use:
>
> # works as expected
> dum<-data.frame(x1=rnorm(200))
> predict(lm.yx, newdata=dum)
>
Yup.
>
> Suppos
Hi all,
Suppose:
y<-rnorm(100)
x1<-rnorm(100)
lm.yx<-lm(y~x1)
To predict from a new data source, one can use:
# works as expected
dum<-data.frame(x1=rnorm(200))
predict(lm.yx, newdata=dum)
Suppose lm.yx has been run and we have the lm object. And we have a
dataframe that has columns that don't
Hi Anirban,
You can do it like this...
lm.foobar <- lm(foo ~ bar)
some.data <- rnorm(200)
predict(lm.foobar, newdata=list(bar=some.data))
Hope that helps.
Michael
On 16 December 2010 17:05, Anirban Mukherjee wrote:
> Hi all,
>
> Suppose:
>
> y<-rnorm(100)
> x1<-rnorm(100)
> lm.yx<-lm(y~x1)
>
Hi all,
Suppose:
y<-rnorm(100)
x1<-rnorm(100)
lm.yx<-lm(y~x1)
To predict from a new data source, one can use:
# works as expected
dum<-data.frame(x1=rnorm(200))
predict(lm.yx, newdata=dum)
Suppose lm.yx has been run and we have the lm object. And we have a
dataframe that has columns that don't
Hello,
So I'm working on extracting IDs from a topGO result object. I have a
list of terms ranked by p values. However, does the result object
contain indices to the original list of gene ids per go term? The
documentation was a bit unclear.
Thanks,
Paul
_
Have your tried something like the following?
> # put some data to cluster in a data.frame
> d <- data.frame(x1=log(1:50), x2=sqrt(1:50), x3=1/(1:50))
> # put NA's in rows 1 and 3
> d[1,1] <- d[3,3] <- NA
> # cluster the non-NA rows
> tmp <- kmeans(na.omit(d), 3) # 3 clusters
> # add cluster id ve
Hi Jannis,
Thank you for answering my question. I saw the option called na.omit when
i used nnet() and tried to classify Iris data with that. I wanted to know if
there is a similar option available in kmeans which can omit or in some way
consider the null/NA values and cluster the observations.C
Hello,
So I'm working on extracting IDs from a topGO result object. I have a
list of terms ranked by p values. However, does the result object
contain indices to the original list of gene ids per go term? The
documentation was a bit unclear.
Thanks,
Paul
_
kiotoqq googlemail.com> writes:
>
>
> I'm pretty sure that I did everything right, but my R is just not drawing the
> mosaicplot that I want, and there is also no error messege, looks like this
> right now:
> > mosaicplot(arthritis)
> > mosaicplot(~ sex + treatment + improved, data = arthritis
Thanks Phil,
That is exactly what I was looking for.
Regards,
Mark
Phil Spector wrote:
Mark -
If I understand what you want, it can be done with a custom
panel function:
mypanel = function(x,subscripts,groups,...){
panel.densityplot(x,plot.points=FALSE,groups=groups,subscripts=subscript
previously I wrote...
THere's the peaks package which will do some peak detection, or
'peaks' in the simecol package. Or msc.peaks.find in caMassClass,
I'm madly trying to remember what spectral package I found which has a
tool similar to msextrema but returns a couple more useful lists
d
Hi Patrik,
If speed is an issue, an easy way to go a bit faster using Dr.
Spector's code is:
wh = which(is.na(mat), arr.ind=TRUE)
mat[wh] = colMeans(mat, na.rm=TRUE)[wh[,2]]
using that method, I can replace the NA values with column means on a
3 x 1 matrix with ~ 1% of values missing in
I'm pretty sure that I did everything right, but my R is just not drawing the
mosaicplot that I want, and there is also no error messege, looks like this
right now:
> mosaicplot(arthritis)
> mosaicplot(~ sex + treatment + improved, data = arthritis, color = TRUE)
>
--
View this message in conte
Mark -
If I understand what you want, it can be done with a custom
panel function:
mypanel = function(x,subscripts,groups,...){
panel.densityplot(x,plot.points=FALSE,groups=groups,subscripts=subscripts,...)
panel.rug(x,col=trellis.par.get('superpose.line')$col[groups[subscripts]])
}
Then
Hi, I know that the following question is a basic one,sorry in
advance. My problem is that I have a data sheet with 80 variables
which are unemployment rates of 80 countries. I call it as u1 to u80.
Besides I have for the same countries their productivity rates. I want
to do a kind of loop to selec
Hi All,
I'm trying to add a 'rug' representation of my data to a plot created
with densityplot(). While I can do this in the simple case, I can't do
it properly when I include the "groups" argument. I have an example
below. I am running a reasonably new version of R.
print(sessionInfo())
R v
Patrik -
I don't know if it's the fastest, but, assuming your
Matrix is called mat, this seems to work fairly quickly:
wh = which(is.na(mat),arr.ind=TRUE)
mat[wh] = apply(mat,2,mean,na.rm=TRUE)[wh[,2]]
- Phil Spector
Dear r-help members,
Could any of you help me with this model, please?
This model gives error when some value touch whatever border and I do not know
how to correct it. The 80% of the seeds produced by a plant will fell into the
parent cell, the 15% in the first ring according to the king movem
Dear Kraman,
MagickCoreTerminus() has been present in ImageMagick for quite a long
time. It seems likely that you're using a very old version of
ImageMagick. Please install a newer one.
Which versions of ImageMagick, R and EBImage are using ?
Best regards,
Greg
---
Gregoire Pau
EMBL Researc
This integral is NOT easy. Your solution is wrong. You cannot integrate
term-by-term when the polynomial is in the *denominator* as it is here.
I am not sure if there is a closed-form solution to this. I remember you
had posed a problem before that is only slightly different from this.
Previ
Hi,
I am trying to produce a levelplot using the following:
library(lattice)
df <- read.table("data", sep="\t", header=TRUE)
gr = levelplot(df$z ~ df$x * df$y,
xlim=0:1000,ylim=0:1000,aspect="iso",col.regions=heat.colors)
print(gr)
The produced result is: http://dl.dropbox.com/u/25473/bigplot.
Dear All,
Â
does anyone know which is the fastest way to replace NA values in a
Matrix by their column mean?
library(Matrix)
mat
Links:
--
[1] http://www.ownit.se
[[alternative HTML version deleted]]
__
I got it so far, but dont know if its right:
> x=c(120,40,75,85,55,75,55,90,90,55,155)
> Fn <- ecdf(x)
> plot(Fn,verticals = TRUE, ?
>
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-plot-empirical-distribution-function-tp3089821p3089845.html
Sent from the R help mailing
I have the table
company: 1 2 3 4 5 6 7 8 9 10 11
transaction volume:120 40 75 85 55 75 55 90 90 55 155
how can I plot the empirical distribution function of it? I only know there
is the ecdf, but I dont know how to use it, please help!
--
View this message in context:
http://r.789695.n4.nabble
Hi Steven and Peter,
I apologise for not providing the code for the sample
I now realise what I need may be a bit tricky...
my dataframe has hundreds of IDs in which case Steven's solution will not be
optimum
Peter's solution seems best, but how do I reverse this and use it to select the
dataf
You seem to be stuck on pixels and so forth. This should not be your concern.
You should (as Joshua said) specify the aspect ratio by indicating the real
dimensions of your graphic in your call to postscript (), but that image won't
become pixels until later, and your publisher will be the one t
On 2010-12-15 11:12, brianiphk wrote:
I'm trying to run an ANOVA in R and I have a problem:
I used this command: anova(lm(time~age*region*facility))
Age, region, and facility are all factor levels, which has 2, 4, and 3
levels respectively.
however, in the output (anova table), it states that
Hi Sebastien!
How about this:
d<-data.frame(nest,LD,FD)
l<-apply(d,1,function(i){seq(i["LD"],i["FD"],by=1)})
jul<-60:273
active<-sapply(jul,function(i){sum(sapply(l,function(k){i %in% k}))})
ans<-data.frame(jul,perc_act=100*active/length(nest))
ans
When you say cumulative percentage active,
On Wed, Dec 15, 2010 at 01:29:16PM -0500, David Winsemius wrote:
>
> On Dec 15, 2010, at 6:01 AM, Nick Sabbe wrote:
>
> >Hi Felipe,
> >
> >gsub("[^0123456789]", "", "AB15E9SDF654VKBN?dvb.65")
> >results in "15965465".
> >Would that be what you are looking for?
>
>
> I tried figuring out how to
Sebastien -
Since you didn't provide a reproducible example, I may
be off track, but here's how I'd handle the problem.
Suppose the data set with nest_id, LD, and FD is called mydata.
First create a function to find the percentage of active nests for
a given time:
pctact = function(time)10
Phil,
this is exactly what do I want.
Thanks a lot.
--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
--
- Original Message -
From: "Phil Spector"
To: "Silvan
I'm trying to run an ANOVA in R and I have a problem:
I used this command: anova(lm(time~age*region*facility))
Age, region, and facility are all factor levels, which has 2, 4, and 3
levels respectively.
however, in the output (anova table), it states that the df is 1 for all
three factors, which
Dear Duncan and David,
For what it's worth I have no problem running rgl on R, R.app, or R64.app
under Mac OS X 10.6.5. I'd suspect an X11 issue, though David says that's
not the case.
Best,
John
John Fox
Senator William McMaster
Professor of Social Statisti
Hi All,
I've been struggling on that one for a couple of days. I've tried several
things, but nothing is worth publishing.
I'm working with a bird nests data frame for which I have a laying date (LD)
and a fledgling date (FD) (in calendar Julian date) by nest id. For the
period spanning between L
Silvano -
If you always have exactly one med1, one med2, and one med3
for each combination of Grupos, Dias, and Rato, you can use
aggregate(Esp.Inter.Trac~Grupos+Dias+ Rato,mean,data=x)
Grupos Dias Rato Esp.Inter.Trac
1 C31 100.55667
2 GFC31 87.62333
On 12/15/2010 7:18 AM, Amelia Vettori wrote:
Dear R-help forum members,
Suppose I have a data-frame having two variables and single data for
each of them, as described below.
variable_1 variable_2 10 20
I have written a function, say, 'fun' which uses input
David,
Thank you very much! Indeed Capitalizing names is very tricky, particularly
for people not having English -mother language (as I am). Hopefully, Using
your script will much better than simply having names in uppercase.
Happy Holidays!
Rock
--
View this message in context:
http://r.7896
Hi Aldi,
Perhaps this will help slightly:
Dots per inch (DPI) or points (pixels) per inch (PPI) = number of
pixels/number of inches.
there are several ways of specifying the size of a graphic in R. You
use the "width" and "height" argument regardless, but you can set a
third argument the "units
I want to analyse data with an unordered, multi-level outcome variable, y. I am
asking for the appropriate method (or R procedure) to use for this analysis.
> N <- 500
> set.seed(1234)
> data0 <- data.frame(y = as.factor(sample(LETTERS[1:3], N, repl = T,
+ prob = c(10, 12, 14))), x1 = sample
Hi,
I have a file, like below, and I want create a new
data.frame with variables:
Grupos Dias Rato Esp.Inter.Trac, but Esp.Inter.Trac will
be the mean of
med1 med2 and med3 for each Rato.
How can I do this?
Grupos Dias Rato blocos Esp.Inter.Trac
GFC 31 med1 85.99
G
On 12/15/2010 06:39 AM, kraman wrote:
>
> Hello, I've been having some trouble working with the aroma package and have
> realized that it's my ebimage is at the root of the problem. The package
> builds but during testing it fails because of an undefined symbol as seen
> below:
>
>
>
> install
Thank you Joshua and Duncan for your very quick responses.
Joshua, I will indeed read more about the OpenGL and GLU systems.
Duncan, I am running R via a console. I also tried via the GUI R.app and
the same problems with loading the package and opening a graphics device
occurred.
I find it
On Dec 15, 2010, at 11:19 AM, Gerrit Eichner wrote:
Hello, Vassilis,
maybe
with( test.df, ave( vector, factor, FUN = function( x) x - mean( x)))
does what you want.
THere is also the scale function which can be called with the
parameters set to center but not scale the results:
wit
On Dec 15, 2010, at 6:01 AM, Nick Sabbe wrote:
Hi Felipe,
gsub("[^0123456789]", "", "AB15E9SDF654VKBN?dvb.65")
results in "15965465".
Would that be what you are looking for?
I tried figuring out how to do this from a more positive perspective,
meaning finding a regular expression function
Thank you Jeff for your advice,
Maybe it was better my email subject to had been "high quality R graphs
for publication" instead of "300 dpi and eps".
a. In my latest response I was asking for any published paper/book
(written by anybody in the R list) that describes how R handles vector
grap
Hi David,
I had similar errors are linux awhile back. I ended up needing to
install/update some of rgl's non-R dependencies. I want to say it was
OpenGL and GLU in particular, but I was doing a lot of
install/updating at the time and I may be mixing one or both of those
up. Also that was linux,
On 15/12/2010 1:11 PM, David A. Johnston wrote:
Hello,
I have been experiencing an odd error with the RGL package that I cannot
figure out. I have searched without luck for fixes to this problem on
Rseek, Google, and the R-help-list.
RGL used to work on my machine about a month ago. It was a
Hi:
On Wed, Dec 15, 2010 at 4:24 AM, beatlebg wrote:
>
> Am I trying to perform multiple linear regressions on each 'VARIABLE2'. I
> figured out that there are different ways, using the following code:
> (data
> is given at the end of this message)
> reg <- lapply(split(TRY, VARIABLE2), function
On Dec 14, 2010, at 9:00 PM, RockO wrote:
Dear R world,
Do you know about a function that would capitalize in the correct
manner
first and family names?
I found in the cwhmisc only the CapLeading function, but it just
does not do
the job, taking care only to capitalize the first letter
Hello,
I have been experiencing an odd error with the RGL package that I cannot
figure out. I have searched without luck for fixes to this problem on
Rseek, Google, and the R-help-list.
RGL used to work on my machine about a month ago. It was a very useful
tool. The graphics device will not i
Hi Gerrit,
This does exactly what I want, thank you very much!
Even more, I notice that ave() uses the split/unsplit functions under the
hood, which are very useful tools as they allow to apply even more
complicated functions on a factor-by-factor basis.
best,
Vassilis
--
View this message
Now I have this:
trellis.par.set( peelTheme)
plot( spplot( thumbDf, "pri",
main= "Primary cover",
mapasp = "iso",
colorkey= FALSE),
panel.width= list( x=0.6, units= "npc"),
split= c( 1, 1, 3, 1),
position= c( 0, 0.2, 1, 0.8),
more= TRUE
On 2010-12-15 09:44, Hrithik R wrote:
Hi Steven and Peter,
I apologise for not providing the code for the sample
I now realise what I need may be a bit tricky...
my dataframe has hundreds of IDs in which case Steven's solution will
not be optimum
Peter's solution seems best, but how do I reverse
Dear Sir,
Thanks for your reply. Sir, the function I have defined (i.e. addition) was
just an indicative one. The function I have actually constructed is bit
complex. Basically, I wish to learn how the same function can be applied to a
dataframe where each row of the dataframe becomes new input
> I am trying to find the analytical solution to this differential equation
>
> dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
> If there is an analytial solution to this differential equation then it
It is a polynomial function of R, so just develop the expression and
when you get the two terms in R
> Date: Wed, 15 Dec 2010 11:46:40 -0500
> From: msamt...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Solution to differential equation
>
> Hello,
> I am trying to find the analytical solution to this differential equation
>
> dR/dt = k1*(R^k2)*(
On Dec 15, 2010, at 15:20 , gillian raab wrote:
> We work on a secure stand-alone server with no internet access, for security
> reasons. We are running Windows XP. SInce version 2.11 the R help files
> point straight to the internet. Is there any way we could have a local help
> that we coul
It is possible to embed a raster image inside eps, but AFAIK R does not do
this. Other than that, your questions do not apply to eps. Rendering resolution
only comes into play when you put it into a raster software (like photoshop) or
print it.
Beyond that, we don't know what you are doing with
On 15/12/2010 9:20 AM, gillian raab wrote:
We work on a secure stand-alone server with no internet access, for
security reasons. We are running Windows XP. SInce version 2.11 the R
help files point straight to the internet. Is there any way we could
have a local help that we could access on our
Hello,
I am performing a meta-analysis using the metafor package. My data are
proportions and I used the Freeman Tukey double arcine (FT)
transformation to fit the random effects model. Now I want to create a
forest plot with my estimates backtransformed to the original scale of
proportions. Can t
We work on a secure stand-alone server with no internet access, for
security reasons. We are running Windows XP. SInce version 2.11 the R
help files point straight to the internet. Is there any way we could
have a local help that we could access on our server.
Thanks in advance for any soluti
Hello, I've been having some trouble working with the aroma package and have
realized that it's my ebimage is at the root of the problem. The package
builds but during testing it fails because of an undefined symbol as seen
below:
installing to /usr/local/lib64/R/library/EBImage/libs
** R
**
Ooops, you're right Jonathan and Ben, the problem is with
starting values the objective function evaluate to Inf, that's not normal.
Thank you for this help, now I will try to rosolve this and tell you.
Thanks a lot.
2010/12/15 Jonathan P Daily
> Your function is evaluating to Inf
Your function is evaluating to Inf. Is this expected behavior? I don't
know what starting values you are using, but it looks like the initial
values evaluate to Inf. If this is also true for values near your initial
conditions then I am not surprised the optimization returns initial
values.
Tr
Perhaps I have installed R first and after it gfortran. Should I
uninstall R and install once again?
--
pozdrawiam / regards
Tomasz Suchocki
W dniu 15 grudnia 2010 12:54 użytkownik Åukasz RÄcÅawowicz [via R]
napisaÅ:
> 2010/12/15 suchy <[hidden email]>
>
>> /usr/lib64/gcc/x86_64-suse-lin
Hi Steven and Peter,
I apologise for not providing the code for the sample
I now realise what I need may be a bit tricky...
my dataframe has hundreds of IDs in which case Steven's solution will not be
optimum
Peter's solution seems best, but how do I reverse this and use it to select the
dataf
ODEs don't get much easier than this... integration by parts is overkill.
See the posting guide re homework.
"mahesh samtani" wrote:
>Hello,
>I am trying to find the analytical solution to this differential
>equation
>
>dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
>
>k1 and k2 are parameters th
Amelia of Aukland writes:
> Suppose instead of having above dataframe having single data for
> variable 1 and variable 2, I have following data as
>
> variable_1 variable_2
>
>10 20
> 40 30
> 3
Hello,
I am trying to find the analytical solution to this differential equation
dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
k1 and k2 are parameters that need to fitted, while Ro and Rmax are the
baseline and max value (which can be fitted or fixed). The response (R)
increases
initially at an exp
On Wed, Dec 15, 2010 at 4:18 PM, Amelia Vettori
wrote:
> Dear R-help forum members,
>
> Suppose I have a data-frame having two variables and single data for each of
> them, as described below.
>
> variable_1 variable_2
> 10 20
>
> I have written a functi
Hello, Vassilis,
maybe
with( test.df, ave( vector, factor, FUN = function( x) x - mean( x)))
does what you want.
-- Gerrit
On Wed, 15 Dec 2010, Vassilis wrote:
I would like to de-mean the 'vector' column of the following dataframe by
factor:
set.seed(5444)
vector <- rnorm(1:10)
facto
Hi Peter,
Thx for your comment - I guess you're right... I made a drawing to
better understand which quadrants overlap
So I ende up with this Code:
library(mnormt) # can handle multivariate normal distributions
uX = 2
uY = 1
uZ = .5
mean = c(uX, uY, uZ)
LX = matrix(c(1,-1,0,1,0,-1), 2, 3, b
I have come around several times from R to A. Illustrator, or A.
photoshop, and between them with PowerPoint. It is possible that the
last one I reported was from PowerPoint.
So from your postings it was made clear that postscript plot from R
produces a vector graph.
Can someone recommend some
Dear dror,
If I understand correctly what you want to do, the setup is similar to the
O'Brien-Kaiser example in ?Anova, only simpler.
First, you have to reorganize your data in "wide" format so that there's one
record for each subject, with response variables (say) present.C and
present.D.
Then
I would like to de-mean the 'vector' column of the following dataframe by
factor:
set.seed(5444)
vector <- rnorm(1:10)
factor <- rep(1:2,5)
test.df <- data.frame(factor, vector)
which is:
factor vector
1 1 -0.4963935
2 2 -2.0768182
3 1 -1.584
4 2 0.8025474
Dear R-help forum members,
Suppose I have a data-frame having two variables and single data for each of
them, as described below.
variable_1 variable_2
10 20
I have written a function, say, 'fun' which uses input 10 and 20 and gives me
desired result.
RockO gmail.com> writes:
>
>
> Dear R world,
>
> Do you know about a function that would capitalize in the correct manner
> first and family names?
> I found in the cwhmisc only the CapLeading function, but it just does not do
> the job, taking care only to capitalize the first letter of a wor
kamel gaanoun gmail.com> writes:
>
> I appologize for the lack of information in my previous mail, but it is my
> first one.
>
> Yes I red -in part- the Port documentation to learn what does the message
> X-convergence (3) mean.
>
> so this is my nlminb and the results :
>
> res1<-nlminb(par
I appologize for the lack of information in my previous mail, but it is my
first one.
Yes I red -in part- the Port documentation to learn what does the message
X-convergence (3) mean.
so this is my nlminb and the results :
res1<-nlminb(parametres[,3], V, lower=c(rep(0.01, 18), rep(0.01, 3),
rep
How are we supposed to help you if we have no idea how you are using it?
Did you read the PORT documentation? Did you look at the control
parameters outlined in the function help page? Often times for me a return
of the initial conditions is an issue with either inappropriate tolerance
or havin
On Wed, Dec 15, 2010 at 2:22 PM, Jim Maas wrote:
> I get a list object from an iterative function. I'm trying to figure out
> the most efficient way to calculate the mean of one element, across all
> components of the overall list.
>
> I've tried
>
> output <- mean (listobject[[1:5]]$element)
>
>
Hi Everyone,
Can anyone help me resolve a problem that i'm having with nlminb. The
problem is that it stops after just one iteration and returns the same
values as "start" ones.
Thank you very much for your help.
Sincerely.
--
Kamel Gaanoun
(+33) (0)6.76.04.65.77
[[alternative HT
Try this:
mean(unlist(sapply(listobject[1:5], '[', 'element')))
On Wed, Dec 15, 2010 at 11:22 AM, Jim Maas wrote:
> I get a list object from an iterative function. I'm trying to figure out
> the most efficient way to calculate the mean of one element, across all
> components of the overall lis
Hi
r-help-boun...@r-project.org napsal dne 15.12.2010 14:22:15:
> I get a list object from an iterative function. I'm trying to figure
> out the most efficient way to calculate the mean of one element, across
> all components of the overall list.
>
> I've tried
>
> output <- mean (listobject
Dear all,
Roger, Georg, thanks a lot for your suggestions.
I am working in R (time series…), but I am truly beginner in spatial
statistics. How to create a shapefiles including my data about some
socio-demographic indicators (7), for about 30 countries around the world?
I know it is a stupid
Dear all,
I am using Cairo version 1.4-5 for plotting R graphs on a Gentoo system
that does not have X11 installed. I have full control on the fonts when
running R in interactive mode. But when running R in batch mode I cannot
change e.g. font size. Letters (including symbols) are, however, dr
I get a list object from an iterative function. I'm trying to figure
out the most efficient way to calculate the mean of one element, across
all components of the overall list.
I've tried
output <- mean (listobject[[1:5]]$element)
to get the mean of "element" in the first five components?
I
> testpred
Call:
coxph(formula = Surv(time, status) ~ x, method = object$method)
coef exp(coef) se(coef) zp
karno -0.00756 0.992 0.0364 -0.208 0.84
Likelihood ratio test=0.89 on 1 df, p=0.344 n= 7
> newd1 <- list(testall$x[1,])
> survtest <- survfit(testpred,newdat
On Wed, Dec 15, 2010 at 11:08:06AM -0200, Henrique Dallazuanna wrote:
> Try this:
>
> gsub("[^0-9]", "", "AB15E9SDF654VKBN?dvb.65")
Consider also
strsplit("AB15E9SDF654VKBN?dvb.65", "[^.0-9][^.0-9]*")
[[1]]
[1] """15" "9" "654" ".65"
PS.
> On Wed, Dec 15, 2010 at 6:55 AM, Luis Fe
I do not really understand your question. You can use use kmeans but
without the observations that include the NA values (e.g. by deleting
whole rows in your observation matrix). If you want to keep the
information in the valid observations of those rows, I fear you need to
look for a clusterin
Try this:
gsub("[^0-9]", "", "AB15E9SDF654VKBN?dvb.65")
On Wed, Dec 15, 2010 at 6:55 AM, Luis Felipe Parra <
felipe.pa...@quantil.com.co> wrote:
> Hello, I have stings which have all sort of characters (numbers, letters,
> punctuation marks, etc) I would like to stay only with the numbers in th
On Dec 15, 2010, at 11:40 , Nicolas Berkowitsch wrote:
> Dear list member,
>
> I struggle with the problem, why the probabilities of choosing one of three
> mutually exclusive alternatives don?t sum up to 1!
>
> Let?s assume we have three alternatives X, Y, and Z. Let?s further assume we
> kn
On 14.12.2010 23:48, Erik Lukac wrote:
Hi Mike, Hi Uwe,
thanks for your reply.
The main thing is that we have mixed OSs. On the one hand we have Linux,
on the other WinXP. I need the same version of packages and R on both
systems. Otherwise I cant explain my colleagues why error messages are
r
Am I trying to perform multiple linear regressions on each 'VARIABLE2'. I
figured out that there are different ways, using the following code: (data
is given at the end of this message)
reg <- lapply(split(TRY, VARIABLE2), function(X){lm(X2 ~ X3, data=X)})
lapply(reg, summary)
Which produces
On Dec 15, 2010, at 4:18 AM, Ivan Calandra wrote:
Hi,
Just to note that which() is unnecessary here:
DF2 <- DF[DF$ID==2 | DF$ID==5, ]
And to further note that it is only unnecessary of you have no NA's in
that ID column.
> DF[4,1] <- NA
> DF[8,1] <- NA
> DF2 <- DF[DF$ID==2 | DF$ID==5, ]
2010/12/15 suchy
> /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../x86_64-suse-linux/bin/ld:
> cannot find -lgfortran
> collect2: ld returned 1 exit status
> Similar situation I have for other packages.
>
> Could You help me?
>
>
Did you install gfortran first?
--
Mi³ego dnia
[[alte
Hi Felipe,
gsub("[^0123456789]", "", "AB15E9SDF654VKBN?dvb.65")
results in "15965465".
Would that be what you are looking for?
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove
-Origin
Dear helpers,
I have a 2x2 mixed design with two "groups" (between-subjects) and two
presentation-types (within-subjects).
The difference between groups is in the order of manipulations:
group.CD having first a block of present.type.C and then a block of
present.type.D, each block containing 31 t
Hi Heinz
OK, Point taken. I must say I do not do concatenation of factors very
often so this feature does not bothers me much.
Best regards
Petr
Heinz Tuechler napsal dne 13.12.2010 13:52:17:
> Hello Petr,
>
> don't want to convince you. If you like the following:
>
> x <- factor(1:4, labe
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