Naoki Irie-3 wrote:
>
> I am using rgl.sphere to visualize scatter plot data in three dimensional
> space. However, as I can not see the labels of each data point directly in
> RGL window, I usually look for the values of x, y, z axis to find out the
> label (or line number of the data poin
Steve Jaffe wrote:
> hmm, that is what I was afraid of. I considered that but thought to myself,
> surely there must be an easier way. I wonder why this feature isn't
> available. It's there in scripting languages, like perl, but also in
> "hardcore" languages like C++ where std::sort and sorted c
Stavros Macrakis-2 wrote:
>
> What is the recommended way of checking whether an RODBC connection is
> open?
>
> Since odbcValidChannel is not exported from namespace RODBC, I suppose I
> shouldn't be using it.
>
> This is the best I could come up with, but it seems a bit 'dirty' to be
> usin
William Dunlap wrote:
> Bill Dunlap
> TIBCO Software Inc - Spotfire Division
> wdunlap tibco.com
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk
>> Sent: Thursday, May 28, 2009 12:31 PM
>> To: Caro
Are there any particular licences under which R packages must be released or is
it the discretion of the author? The same question if the package is to be
destined for CRAN?
Kind regards,
Nathan
--
Dr. Nathan S. Watson-Haigh
OCE Post Docto
Hi,
Is there any way to sort a tabel with a colum with IP-address?
table:
id rank color status ip
138 29746 yellow no 162.131.58.26
138 29746 red yes 162.131.58.16
138 29746 blue yes 162.131.58.10
138 29746 red no 162.131.58.17
138 29746 yellow no 162.131.58.14
138 29746 red no 162.131.58.
Thank you, Duncan. While your answers didn't quite fix my problems, I
realized that I should probably "go with the flow". I installed gfortran
and everything worked fine. Thanks again.
Steve
sjnovick wrote:
>
> To all. I need your help. The big question is: How do I make an R
> l
Dear R list
I am using rgl.sphere to visualize scatter plot data in three
dimensional space.
Thanks to the rgl library, I can easily find out the data point that I
am interested in.
However, as I can not see the labels of each data point directly in
RGL window,
I usually look for the valu
Paul Murrel has an article about using grid graphics
to do this. He also mentions specialized R packages
that are specifically aimed at this task.
http://journal.r-project.org/current.html
On Thu, May 28, 2009 at 11:00 PM, George Chen wrote:
> Hi,
>
> I would like to represent the treatment cour
Hi,
I would like to represent the treatment course of a subject with an arrow
representing passage through time and various ticks, arrows, or bars
representing treatments done to the patient along the arrow. I would like to
generate this sort of schematic for each patient in a database taking
Of course, in MLE, if we collect more and more observations data, MLE
will perform better and better.
But is there a way to find a bound on parameter estimate errors in
order to decide when to stop collect data/observations, say 1000
observations are great, but 500 observations is good enough...
Dear all,
I want to set tool-tips for a gtkButton. I use the following code
which works under Windows. However, it doesn't work under Linux. Any
hints? Thanks.
library(RGtk2)
b<-gtkButtonNewWithLabel("OK")
gtkTooltips()$setTip(b,"Memo for a Button.")
gw <- gtkWindow(show=F)
gw$Add(b)
gw$Show()
>
Hello~
I have some ploblem.
How can I draw tree picture after plsda.
Please ,help me.
script
--
Result <- plsda(trainDescr, species, ncomp = 5, probMethod = "Bayes")
[[alternative HT
On 28/05/2009 8:17 PM, Stavros Macrakis wrote:
I couldn't get your suggested method to work:
`==.foo` <- function(a,b) unclass(a)==unclass(b)
`>.foo` <- function(a,b) unclass(a) < unclass(b) # invert comparison
is.na.foo <- function(a)is.na(unclass(a))
sort(structure(sample(5),class
Deepayan,
Many thanks for the quick response and suggested workaround.
Best,
Ben
Deepayan Sarkar wrote:
On Wed, May 27, 2009 at 4:38 PM, Benjamin Tyner wrote:
Hello,
Say I have a predictor taking a very wide value:
Data <- data.frame(pred="a",r
That seems to work for the toy data. How do I implement this change with my
real data, which are read from very large Stata and SPSS files and keep the
factor definitions? Won't I be losing information (and creating a larger
dataset) by not using the factor levels?
How do I recover the facto
I would assume that valsum is of zero length:
> valsum <- numeric(0)
> str(valsum)
num(0)
> if (valsum == 1) 1
Error in if (valsum == 1) 1 : argument is of length zero
check your objects with 'str'
On Thu, May 28, 2009 at 6:39 PM, Derek Lacoursiere wrote:
> Hi,
>
> I have the following:
>
>
I couldn't get your suggested method to work:
`==.foo` <- function(a,b) unclass(a)==unclass(b)
`>.foo` <- function(a,b) unclass(a) < unclass(b) # invert comparison
is.na.foo <- function(a)is.na(unclass(a))
sort(structure(sample(5),class="foo")) #-> 1:5 -- not reversed
What am I mis
Gabor Grothendieck wrote:
Try this:
sapply(1:n, function(i) sum(abs(outer(a, b, "-")-i)==0))
[1] 10 10 10 10 9
abs(outer(a, b, "-") - i) == 0 ==> outer(a, b, "-") == i.
sum(outer(a, b, "-") == i) asks how many times i is an element of
outer(a, b, "-"). This is
tabulate(outer(a, b, "-"),
On 28/05/2009 6:06 PM, Steve Jaffe wrote:
hmm, that is what I was afraid of. I considered that but thought to myself,
surely there must be an easier way. I wonder why this feature isn't
available. It's there in scripting languages, like perl, but also in
"hardcore" languages like C++ where std::
What is the recommended way of checking whether an RODBC connection is open?
Since odbcValidChannel is not exported from namespace RODBC, I suppose I
shouldn't be using it.
This is the best I could come up with, but it seems a bit 'dirty' to be
using a tryCatch for something like this:
odbcOpe
In this toy data, each of the tables should sum to
None of the tables shows NA columns or rows.
>
> ToyData <- read.table("C:/Data/R/Toy.csv", header=TRUE, sep=",",
> na.strings="NA", dec=".", row.names="ID_Num")
> ToyData
Data1 Data2 Data3 Weight
101
Try this:
> sapply(1:n, function(i) sum(abs(outer(a, b, "-")-i)==0))
[1] 10 10 10 10 9
On Thu, May 28, 2009 at 5:45 PM, KARAVASILIS GEORGE wrote:
> Hello, R users.
> I have the following code:
>
> a=1:10
> b=-3:15
> n=5
> x <- rep(0,n)
> for (i in 1:n) x[i] <- sum( outer(a,b, function(s,t) ab
Hi,
I have the following:
for(j in (y.raw+1):(rownum-1)){
valsum<-tstfframed[min.x,j]+tstfframed[min.x,j+1]
if(valsum == 1){
cat("valsum loop")
int.num<-int.num+1
}
}
I agree that it is surprising that R doesn't provide a sort function with a
comparison function as argument. Perhaps that is partly because calling out
to a function for each comparison is relatively expensive; R prefers vector
operations.
That said, many useful custom sorts are easy to define by
On May 28, 2009, at 5:45 PM, KARAVASILIS GEORGE wrote:
Hello, R users.
I have the following code:
a=1:10
b=-3:15
n=5
x <- rep(0,n)
for (i in 1:n) x[i] <- sum( outer(a,b, function(s,t) abs(a-b-i)==0) )
You don't seem to be doing anything with s and t? Did you mean:
for (i in 1:n) x[i] <- su
On Wed, May 27, 2009 at 4:38 PM, Benjamin Tyner wrote:
> Hello,
>
> Say I have a predictor taking a very wide value:
>
> Data <- data.frame(pred="a",resp=1)
>
> print(xyplot(pred~resp, data=Data)) # enough y-axis padding to accommodate
> the wide label
>
>
hmm, that is what I was afraid of. I considered that but thought to myself,
surely there must be an easier way. I wonder why this feature isn't
available. It's there in scripting languages, like perl, but also in
"hardcore" languages like C++ where std::sort and sorted containers allow
the user t
Daryl Morris wrote:
>
> Hi,
> Sorry, I was too hasty. I see that the tolerance is in the
> documentation (and that "random" is the default tie-breaker).
>
> Would it be possible to allow tolerance to be a parameter?
>
> thanks, Daryl
>
>
Could you just use apply(x,1,which.max) ?
> x <
Dear R users,
Is there any package for Latent Class Analysis (to be used in a clustering
application) which supports mixed indicator variables (categorical and
continuous)?
Alternatively, is there any other clustering algorithm available that
supports this type of data?
Thanks in advance for you
On 28/05/2009 5:34 PM, Steve Jaffe wrote:
Sounds simple but haven't been able to find it in docs: is it possible to
sort a vector using a user-defined comparison function? Seems it must be,
but "sort" doesn't seem to provide that option, nor does "order" sfaics
You put a class on the vector (e
Hello, R users.
I have the following code:
a=1:10
b=-3:15
n=5
x <- rep(0,n)
for (i in 1:n) x[i] <- sum( outer(a,b, function(s,t) abs(a-b-i)==0) )
Can someone tell me if I could avoid the for command?
Thank you in advance.
__
R-help@r-project.org mai
The warning I met with is attached here:
Warning message:
In heatmap.2(as.matrix(val.sf.ns[-1:-3]), Rowv = F, Colv = T, dendrogram =
"column", :
Using scale="row" or scale="column" when breaks arespecified can produce
unpredictable results.Please consider using only one or the other.
What d
Sounds simple but haven't been able to find it in docs: is it possible to
sort a vector using a user-defined comparison function? Seems it must be,
but "sort" doesn't seem to provide that option, nor does "order" sfaics
--
View this message in context:
http://www.nabble.com/custom-sort--tp23770
Try reading the man page, which says:
Details
When ties.method = "random", as per default, ties are broken at random. In
this case, the determination of a tie assumes that the entries are
probabilities: there is a relative tolerance of 1e-5, relative to the
largest (in magnitude, omitting infinit
Hi,
Sorry, I was too hasty. I see that the tolerance is in the
documentation (and that "random" is the default tie-breaker).
Would it be possible to allow tolerance to be a parameter?
thanks, Daryl
Daryl Morris wrote:
Hi,
I think there's some rounding issue with returning the max column.
Dear All,
Many thanks for all your useful suggestions. Much appreciated.
Jos
2009/5/28 David Winsemius :
> Your last step will either be a single number (not really a sampling
> operation) or a
> non-positive number. So at best you really only have an number that depends
> entirely on the prior
Hi,
I think there's some rounding issue with returning the max column.
(running 2.9.0 on an Apple, but my buddy found it on his PC)
> x <- matrix(c(1234.568,1234.569,1234.567),1)
> max.col(x)
[1] 2
> x <- matrix(c(12345.568,12345.569,12345.567),1)
> max.col(x)
[1] 3
> x <- matrix(c(112345.568,
Thanks for your help Mike, it works like a charm now!!
--- On Thu, 5/28/09, Mike Lawrence wrote:
> From: Mike Lawrence
> Subject: Re: [R] ggplot2 legend
> To: "Felipe Carrillo"
> Cc: r-h...@stat.math.ethz.ch
> Date: Thursday, May 28, 2009, 1:06 PM
> First, your example didn't work
> because f
bogaso.christofer wrote:
I have following addition :
1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not understand how R adding those two unequal vector? Any help?
[[alternative HTML version deleted]]
__
R-help@r-
First, your example didn't work because fish_ByMuestreo didn't build
properly (deleting the first "structure(list(data = " bit solves
this).
To solve your plotting problem, note that as constructed, the Muestreo
column is numeric, whereas you seem to want to treat it as a factor.
Solution: convert
R recycles the shorter one to match the longer one:
1 2 3 4 5 6 7 8 9 10
+
1 2 1 2 1 2 1 2 1 2
=
2 4 4 6 6 8 8 10 10 12
R does this recycling in many cases, and it can sometimes trap the
unwary.
Sarah
On Thu, May 28, 2009 at 4:00 PM, bogaso.christofer
wrote:
> I have f
recycling rule: repeat the shorter element as many times as necessary,
all.equal(1:2 + 1:10 , rep(1:2, length=10) + 1:10)
# TRUE
HTH,
baptiste
On 28 May 2009, at 22:00, bogaso.christofer wrote:
I have following addition :
1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not unde
On 29/05/2009, at 8:00 AM, bogaso.christofer wrote:
I have following addition :
1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not understand how R adding those two unequal vector? Any
help?
Look at the help for ``+'' (?"+") and look at ``Value''. There you will
see:
Thes
Ah, that makes sense. But now another two issues have arisen.
Firstly, the error bars look like confidence intervals, and I'm pretty
sure that they are but does some document verify this? I suppose I
could check the code too.
Secondly, I just read about how dynamite plots should be avoided. It's
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk
> Sent: Thursday, May 28, 2009 12:31 PM
> To: Caroline Bazzoli
> Cc: r-help@r-project.o
I am trying to use a linear filter to reduce loops and thereby increase
the speed of an existing program. However, while the "filter" function
(stats package) should have reduced the looping by about 30-fold, the
time to complete the program remained about the same. This surprised
me, because I
substring(header, 2,) works for the purpose perfectly.
Xin Zheng
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/p
I have following addition :
> 1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not understand how R adding those two unequal vector? Any help?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.
Thank you guys. But I also want 'check.names' to help keeping colnames unique.
It seems there's no proper option for that in 'read.table'. I have to alter
colnames with strsplit or other similar functions.
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Thursday, May 28, 2009 3:40 PM
To: Zhe
This really depends on where the 'X' is coming from, which you did not tell us
(see the posting guide).
For example, if the 'data.frame' function is the one adding the 'X', then you
can use the 'check.names' argument to prevent the addition (or you can use
data.frame with the argument to explic
?read.table check.names=FALSE
On Thu, May 28, 2009 at 3:25 PM, Zheng, Xin (NIH) [C] wrote:
> Hi all,
>
> Is there any way to keep numeric colnames as is? Any hint will be
> appreicated.
>
> Xin Zheng
>
>
>[[alternative HTML version deleted]]
>
> _
Caroline Bazzoli wrote:
> Dear R-experts,
>
> I need to replace in an expression the character "Cl" by "Cl+beta"
>
> But in the following case:
>
> form<-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t)))
>
> gsub("Cl","(Cl+beta)",as.character(form))
>
> We obtain:
>
> [1] "((Cl+beta) - (
Hi all,
Is there any way to keep numeric colnames as is? Any hint will be appreicated.
Xin Zheng
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
Dear all
I am trying to install the "gRain" package on ubuntu 8.10,
running R Vers. 2.9. Unfortunately, I get a bad
Exit-Status. What could cause that problem?
Any help is appreciated.
Many thanks in advance.
A. Biedermann
install.packages("gRain")
Warnung in install.packages("gRain") :
Ar
Just place a point with an NA value between the two segments.
Here is one way to do it in zoo:
set.seed(123)
library(zoo)
# create sample data
tt <- c(1:1000, 1200:2000)
z <- zoo(rnorm(length(tt)), tt)
# this will fill in omitted values with NAs
z <- as.zoo(as.ts(z))
plot(z)
On Thu, May 28,
I have a time series of about 1500 measurements. There are sporadic data
gaps. I would like to plot the time series with type "line". I only want
a line to connect the periods with data; I don't want a line to connect
the points across a data gap. Is there a function or recommended method
t
Apologies once more, there was a slight error in our example. Here is a
correct version of how to color the points in qqmath according to another
variable.
qqmath(~ yield | variety, data = barley, groups=year,
auto.key=TRUE,
prepanel = function(x, ...) {
list(xlim = range(q
Hi:
I need some help with the legend. I got 14 samples(Muestreo) and I
am trying to plot a smooth line for each sample. I am able to accomplish that
but the problem is that the legend only displays every other sample. How can I
force the legend to show all of my Muestreos? Thanks in advance.
One additional question:
It seems that in practice the design effect is often calculated on the
response variable or on individual predictor variables. I have some OLS
models using observational data that are clustered. Does it make sense to
calculate the design effect on the residuals of one of
amor Gandhi wrote:
>
> Hi gues,
>
This should read: Hi, guess what I want
amor Gandhi wrote:
>
> Is there any function in R for boxplot with different time points?
> t1 <- c(rep(1,20),rep(2,20))
> t2 <- c(rep(1,10),rep(2,10),rep(1,10),rep(2,10))
> x <- rnorm(40,5,1)
> dat <- data.frame(t1
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Caroline Bazzoli
> Sent: Thursday, May 28, 2009 8:41 AM
> To: r-help@r-project.org
> Subject: [R] String replacement in an expression
>
> Dear R-experts,
>
> I need to replace in an expression the character
The package AMORE appears to be more flexible, but I got very poor
results using it when I tried to improve the predictive accuracy of a
regression model. I don't understand all the options well enough to be
able to fine tune it to get better predictions. However, using the
nnet() function in packa
On Thu, 28 May 2009 15:47:18 +0300 NOUF AL NUMAIR
wrote:
NAN> barplot(zz,width = 4,names.arg= xx,axes = TRUE, axisnames = TRUE,
NAN> main=title,xlab=xlab,ylab=ylab,ylim=c(0,1175),xlim=c(0,226),col =
NAN> c(for (i in zz){if i>70 col= "lightblue" else col= "mistyrose"))
NAN>
NAN>
NAN> i tried to
Dear All,
I would like to visualize a tree extracted from a random forest using
getTree {randomForest}.
I'm wondering if there is any way to do it directly or to convert my tree
to any other class of tree repesentation, which then can be plotted?
Thank you in advance,
Karoly
Try matching on word boundaries as well:
> gsub("\\bCl\\b","(Cl+beta)",as.character(form))
[1] "((Cl+beta) - (V * ka)) + (V * Vm * exp(-(Clm/Vm) * t))"
See ?regexp
On Thu, May 28, 2009 at 11:41 AM, Caroline Bazzoli
wrote:
> Dear R-experts,
>
> I need to replace in an expression the character "C
Dear R-experts,
I need to replace in an expression the character "Cl" by "Cl+beta"
But in the following case:
form<-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t)))
gsub("Cl","(Cl+beta)",as.character(form))
We obtain:
[1] "((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) *
On May 28, 2009, at 11:25 AM, eric lee wrote:
Hi,
I'm running R 2.7.2 on windows XP. I'd like to find the maximum of a
3-d array over it's third index to create a 2-d array. For example:
x <- array(c(1,2,3,10,11,12,3:8),c(2,3,2))
x
, , 1
[,1] [,2] [,3]
[1,]13 11
[2,]2
Your last step will either be a single number (not really a sampling
operation) or a
non-positive number. So at best you really only have an number that
depends
entirely on the prior sequence of draws.
--
David.
On May 28, 2009, at 8:21 AM, jos matejus wrote:
Dear Ritchie and David,
Thank
Hi,
I'm running R 2.7.2 on windows XP. I'd like to find the maximum of a
3-d array over it's third index to create a 2-d array. For example:
> x <- array(c(1,2,3,10,11,12,3:8),c(2,3,2))
> x
, , 1
[,1] [,2] [,3]
[1,]13 11
[2,]2 10 12
, , 2
[,1] [,2] [,3]
[1,]3
Will R be able to send one batch of commands to the remote machine, then wait
for the output? Or will there be back and forth with new commands based on the
output from the first commands?
You may want to look at the nws package for one way to have jobs run on a
remote machine.
--
Gregory (G
> I want to plot data such that the 3 time points(a,b,c) lie on the X-axis
and
> the values of these times points are on Y-axis for n samples (e.g.100).
>
> So, I have an object x, dim 100 4, it is a dataframe (when checked the
> class)
> x =
> name a b c
> 10.11 1.11 0.8
Dear R-list,
I know is going to be weird, but I need to do something about and I know
the R-list is big group of good people. Im moving to Philadelphia in
August 2009 (my fiancé is there) and I need to find a job. I know to do so
many things, but my skill is in science and research. I would like
Hi,
I am new to R and am using quilt.plot (from fields package) to plot gas
concentrations in Europe.
I am using the following code to plot my data
library(maps)
library(fields)
test = read.csv("%change 1996_2005.txt", sep="\t")
colnames(test) = c("Station", "Measurement_BaseO3", "Model_BaseO3"
Hello,
I am an R amateur.
I want to plot data such that the 3 time points(a,b,c) lie on the X-axis and
the values of these times points are on Y-axis for n samples (e.g.100).
So, I have an object x, dim 100 4, it is a dataframe (when checked the
class)
x =
name a b c
10.11
Dear all ...
I need your help for the following purpose :
I would like to create regularly spaced points on a multi-segment line (a
'psp class' object).
A function of the spatstat library( = pointsOnLines() ) is dedicated to that
objective but the problem is that the probability of falling on a
Using the package rcom (by Thomas Baier) you can do everything you want
since you have the full power of COM at your disposal.
Dieter Menne wrote:
>
>
> Hans-Peter Suter wrote:
>>> If you only have the sheet names, you should use package xlsReadWrite
>>> which
>>> is rather fast, but has some l
Hans-Peter Suter wrote:
>
>> If you only have the sheet names, you should use package xlsReadWrite
>> which
>> is rather fast, but has some limitations in the non-commercial version.
>
> what limitations, i.e. features do you miss?
>
>
Reading of named ranges.
Dieter
--
View this message
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
> -Original Message-
> From: r-devel-boun...@r-project.org
> [mailto:r-devel-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk
> Sent: Thursday, May 28, 2009 5:30 AM
> Cc: R help project; r-de...@r-project.org; A
2009/5/28 Dieter Menne :
> If you only have the sheet names, you should use package xlsReadWrite which
> is rather fast, but has some limitations in the non-commercial version.
what limitations, i.e. features do you miss?
Cheers,
Hans-Peter
__
R-help@r
On 5/28/2009 8:58 AM, Mario Valle wrote:
Good morning!
Which is the preferred method to leave a sourced script and returning back to the
'>' prompt?
For example I search for certain files to be processed, but nothing should be
done if they
are not present. Normally I do in my script:
f<-dir(pa
hi there :)
i want to use barplot with if else but i dont know how to do it ?
i tried this but it is not working with me
SNP <- read.table("my.txt")
>SNP[,2]
[1] 1175 483 240 170 99 79 76 45 38 35 21 16 14 19 16
[16]333 1021686
> >> From: Jeff Newmiller
> >> To: "R Heberto Ghezzo, Dr"
> >> Date: Wed, 27 May 2009 09:00:44 -0700
> >> Subject: Re: [R] R in Ubunto
> >> R Heberto Ghezzo, Dr wrote:
> >>
> >> Hello , I do not know anything abount Ubunto, but I found a Portable
> >> Ubunto for Windows and since so many peop
Good morning!
Which is the preferred method to leave a sourced script and returning back to
the '>' prompt?
For example I search for certain files to be processed, but nothing should be
done if they
are not present. Normally I do in my script:
f<-dir(pattern="qq")
if(length(f) > 0) {
...proc
My goal is for a user to sit down at a Linux laptop, get to an Rcmdr
type screen, submit jobs on a remote system and then get the results
back in R.
We will assume that the user is naive, and the only thing he/she can
do is get to the Rcmdr screen.
The Rcmdr plugin will have a "submit jobs" menu.
Farley, Robert wrote:
>
> I can't get the syntax that will allow me to show NA values (rows) in the
> xtabs.
>
> lengthy non-reproducible example removed
>
If you want a reproducible answer, prepare a reproducible result. And check
that the
syntax is
na.action=na.pass
Dieter
--
View
simeon duckworth wrote:
>
> I'd like to be able to read multiple sheets from an excel workbook and use
> the sheet name to name the resulting dataframe using RODBC.
>
In Microsoft theory, something like the below should be ok (note the $), but
never managed to get this to work. The same met
(diverted to r-devel, a source code patch attached)
Wacek Kusnierczyk wrote:
> Allan Engelhardt wrote:
>
>> Immaterial, yes, but it is always good to test :) and your solution
>> *is* faster and it is even faster if you can assume byte strings:
>>
>
> :)
>
> indeed; though if the speed is
This is one way of sampling all the values in the matrix that meet your
criteria:
> set.seed(1)
> (x <- matrix(rnorm(100), 10))
[,1][,2][,3][,4] [,5]
[,6][,7] [,8] [,9] [,10]
[1,] -0.6264538 1.51178117 0.91897737 1.35867955 -
Dear Ritchie and David,
Thanks very much for your advice. I had thought of this potential
solution, however it doesn't really fullfill my second criteria which
is that once a particular cell has been sampled, the row and column of
that cell can't be sampled from subsequently. In other words, the n
Both of those worked, but hierobarp looked a bit easier, so I used that. The
one annoying thing is that it sorts alphabetically.
Tom
On Thu, May 28, 2009 at 6:46 AM, Jim Lemon wrote:
> Thomas Levine wrote:
>
>> I want to plot quantitative data as a function of three two-level factors.
>> How do
Wacek Kusnierczyk wrote:
> Gavin Simpson wrote:
>> The "View message source" would be a more direct way of viewing the full
>> email, not just the bits TBird shows you in the preview pane. I forget
>> how it is named exactly and under which menu it is found as it has been
>> quite a while since I u
> How can I find out what survreg generates: the full likelihood or a
> likelihood with "unnecessary" constants dropped?
Survreg returns a log-likelihood, not a deviance (LL with parameters dropped);
and the result is labeled as such.
It turns out that the deviance is not as easily defined f
I'd like to be able to read multiple sheets from an excel workbook and use
the sheet name to name the resulting dataframe using RODBC. at the moment
i've figured out how to do it the long way (see below) but feel sure that
there is a speedier & possibly automatic way to do it in R. i've tried to
Will respond privately, since the thread becomes uninteresting for
R-help, I guess.
Uwe
Romain Francois wrote:
Hello Uwe,
I have recently taken over maintenance of the package, which is now
version controlled at r-forge
(http://r-forge.r-project.org/projects/r2html/), and I am planning the
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell
http://cran.r-project.org/web/views/Survival.html
On May 11, 2009, at 1:46 PM, Nancy Tejerina wrote:
*Dear R Users;*
* *
*I´m writing to ask you how can I do Survivals Curves using Time-
dependent
covariates? Which packages I need to Install?*
*Thanks for your help, I look foward you reply
> I have a matrix of both negative and positive values that I would like
> to randomly sample with the following 2 conditions:
>
> 1. only sample positive values
> 2. once a cell in the matrix has been sampled the row and column of
> that cell cannot be sampled from again.
>
> #some dummy data
>
On 28/05/2009 6:03 AM, Madan Mohan wrote:
Hi Friends,
I am working on R-2.9.0 and i want to connect oracle through R, but i could
not find ROracle (zip file for windows) on CRAN Packages. Can anyone suggest
me how to connect Oracle from R in windows platform? I would be grateful if
u can provide
On 28/05/2009 3:20 AM, bogaso.christofer wrote:
Hi,
Recently I come through those R-expressions and understood that "1L" means
"1" and "0L" means "0". Why they are so? I mean, what the excess meanings
they carry, instead writing simple "1" or "0"? Is there any more this kind
of expressions in R
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