I had hoped that
plot(c(0,24),c(0,-6),xlab="Time",ylab="Day",
type="n", main="This Week",axes=FALSE)
axis(2,at=0:(-6), labels =
c("Sun","Mon","Tues","Wed","Thurs","Fri","Sat"),hadj=TRUE)
axis(1,at=seq(0,24,4))
would give me horizontal tick labels.
It doesn't. What would?
Murray
--
Dr M
I was suggested to use
x = data.frame(string = c( "x1", "x2", "x3"))
x
apply(x,1,substr,1,4)
and it's worked good!
Thank you all
Wacek Kusnierczyk wrote:
>
> calpeda wrote:
>> thank you,
>> but I m importing data from a txt file and I have a matrix of n*1
>> The function str s
Greetings,
Could someone give me some help on how to solve problems with
packages? This worked just yesterday. Today I get an error. ?
> library(Rcmdr)
Error in structure(.External("dotTclObjv", objv, PACKAGE = "tcltk"),
class = "tclObj") :
[tcl] invalid command name "font".
Error : .onAt
Hi Stephen: If you want to take tem out of quotes, you can use deparse
substitute as in below. Maybe there are other ways
also but that's the one I often see used. I hope this email ends
up looking okay because my mailer has been acting strangely lately.
         Â
Hi again...
I will get an array with 5 matrix 2x2, with dimension 2x2x5
This code below works fine... And I am applying the same procedure...
The problem might be when I do the permutation of the array, but I checked
the dimension and its all right...
array1 <- array(1:30,dim=c(3,2,5))
array2 <-
Hi,
I'm trying to collapse a character vector to strings, but I am getting
unexpected behaviors in list context:
A <- "a"
B <- c("b","c")
xx <- list(A=A, B=B)
lapply(xx, paste, collaplse=".")
$A
[1] "a ."
$B
[1] "b ." "c ."
paste(B, collapse=".")
[1] "b.c" # this is what I
Hi listers,
I am having some trouble in a matrix multiplication...
I have already checked some posts, but I didn't find my problem...
I have the following code...
But I am not getting the right multiplication...
I checked the dimension and they are fine...
id_y <- array(1:10,dim=c(2,1,5))
id_yt<-
doBy and memisc packages have recoding functions as well, e.g.
> library(doBy)
> x<-c("A","B","C","D","E","A")
> out <- recodevar(x, list(c("A", "B"), c("C", "D", "E")), list("Treat 1",
> "Treat 2"))
> recodevar(x, list(c("A", "B"), c("C", "D", "E")), list("Treat 1", "Treat 2"))
[1] "Treat 1" "Tr
#excel like regression line but better
reg.line <- function(y,x, data)
{plot(data[,y]~data[,x], main=paste(x,"vs.",y, sep=" "), xlab=x, ylab=y, pch=20)
line <- lm(data[,y]~data[,x])
d <- summary(line)
mtext(3, line=1, adj=0.25 ,text=bquote(bold(R^2== ~ .(d$r.squared))), bty="n")
abline(line)}
x <
I think your problem is with R's behavior of dropping array indices
when dim==1.
Simulating that
> m1 <- matrix(1:2,nrow=2)
> m2 <- matrix(1:2,ncol=2)
> m1 %*% m2
[,1] [,2]
[1,]12
[2,]24
> m1[,] %*% m2[,]
[,1]
[1,]5
[1,]5
> m1[,,drop=FALSE] %*% m2[,,drop=
Tena koe Sueli
?abline
?text
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Sueli Rodrigues
> Sent: Thursday, 2 April 2009 2:35 p.m.
> To: r-help@r-project.org
> Subject: [R] Scatter plot
>
>
> Hi.
Hi. How do I plot the straight line and r² in a scatter plot using a
simple file x~y?
Sueli Rodrigues
Eng. Agrônoma - UNESP
Mestranda - USP/ESALQ
PPG-Solos e Nutrição de Plantas
Fones (19)93442981
(19)33719762
__
R-help@r-project.org mailing lis
Ron @ Lecturemaker wrote:
Hi Ted and others,
I appreciate your time in sharing R user group experience with my video
player scheme. Over time, I hope to iron out all issues your views
encounter.
For now I think I have a good solution for flash 10 upgrade issue. See
screenshot "flash10warning.
This may help in understanding how to access the list: notice that I
am using numeric indices and using the '[[' operator
> # generate a list with a varying number of values
> myList <- list() # initialize
> for (i in 1:10) myList[[i]] <- seq(i)
> myList
[[1]]
[1] 1
[[2]]
[1] 1 2
[[3]]
[1] 1 2
MarcioRibeiro wrote:
>
> Hi listers,
> I am having some trouble in a matrix multiplication...
> I have already checked some posts, but I didn't find my problem...
> I have the following code...
> But I am not getting the right multiplication...
> I checked the dimension and they are fine...
>
Hello,
I am trying to install R and have it running through ESS on a Vista machine
that is not mine (I am only familiar with Windows XP and Ubuntu Linux) so
please pardon my unfamiliarity...
The Rgui.exe installation works fine and package installation also appears to
work fine. The base pack
Dear Andy,
There are two problems here: One is that you need to enclose character
values in quotes; the other is a bug in recode(), which is confused by the
values "Treat 1" and "Treat 2" and shouldn't be. Here's a work-around [until
tomorrow, when I'll fix recode()]:
> as.character(recode(x, "c(
I am puzzled to hear that repeating only plotting give different
results, if there were no further simulations. Both rt() and sample()
should give different results from call to call, so if either of those
functions were repeated, then varying results *should* occur. If that
is not the source of th
Oh ALRIGHT!
But as he didn't express the question completely, you had to view my response
as a *strategy* that could be used rather than a full answer to an incomplete
question.
Still, I never did make any claim to infalibility, and nor will I ever do. I'm
always happy, indeed grateful, to
Uhhh, Bill, he wanted E to be recoded as ``Treat 3''.
And he didn't say ***what*** he wanted to happen to D.
Fancy. A chance to ``correct'' Bill Venables for a second time
in two days! :-)
cheers,
Rolf
On 2/04/2009, at 12:39 PM, bill.venab...@csiro.au wrote:
Here i
On 2/04/2009, at 12:22 PM, Andrew McFadden wrote:
Hi all
I am trying to do a simple recode which I am stumbling on. I figure
there must be any easy way but haven't come across it.
Given data of A","B","C","D","E","A" it would be nice to recode this
into say three categories ie A and B becomes
Here is one way
> x <- c("A", "B", "C", "D", "E", "A")
> x
[1] "A" "B" "C" "D" "E" "A"
> f <- factor(x)
> levels(f)
[1] "A" "B" "C" "D" "E"
> levels(f) <- c(rep("Treat1",2), rep("Treat2", 3))
> f
[1] Treat1 Treat1 Treat2 Treat2 Treat2 Treat1
Levels: Treat1 Treat2
>
If you really want the charact
> Earlier I posted a question about memory usage, and the community's input was
> very helpful. However, I'm now extending my dataset (which I use when
> running a regression using lm). As a result, I am continuing to run into
> problems with memory usage, and I believe I need to shift to impl
Hi all
I am trying to do a simple recode which I am stumbling on. I figure
there must be any easy way but haven't come across it.
Given data of A","B","C","D","E","A" it would be nice to recode this
into say three categories ie A and B becomes "Treat1", C becomes "Treat
2" and E becomes "Treat 3"
I stand corrected, with thanks!
Gee you learn some fascinating stuff on this mailing listjust fascinating...
:-)
W.
Bill Venables
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: Greg Snow [mailto:greg.s...@imail.org]
Sent: Thursday, 2 April 2009 5:58 AM
To: Ve
Wacek Kusnierczyk wrote:
Stavros Macrakis wrote:
`->`
Error: object "->" not found
that's weird!
Why???
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen
I cannot read google spreadsheets. I get the following error:
assignment of an object of class "NULL" is not valid for slot "access"
in an object of class "GoogleSpreadsheet"; is(value, "character") is
not TRUE
RGoogleDocs is on the cusp of brilliance. How can I troubleshoot this
apparently last
Hi Ted and others,
I appreciate your time in sharing R user group experience with my video
player scheme. Over time, I hope to iron out all issues your views
encounter.
For now I think I have a good solution for flash 10 upgrade issue. See
screenshot "flash10warning.png" for example of flash ve
Dear R users,
I would be interested in using the lme() function to fit a linear mixed
model to a longitudinal dataset. I know this function allows for the
specification of a within-group covariance structure. However, does it allow
for the explicit specification of a between-group covariance stru
Anybody knows what could be the problem?
I am trying to use the class4 (I also tried the class2) JDBC driver for
DB2 and I get the following error. This driver is for db2 9.5, so it
does not require a matching license .jar file, and I tried redownloading
the driver.
I am running R 2.6 on Windows X
Hi,
I'm trying to follow the ggplot introduction here:
http://had.co.nz/ggplot/ggplot-introduction.pdf
I've installed ggplot2 with install.packages("ggplot2", dep=T)
but when I try to run
print(ggpoint(p, list(colour = sex)))
I get an error:
Error in print(ggpoint(p, list(colour = sex))) :
c
Hi,
thanks for all your hints.
> One of the points of my.symbols is that you can define your own symbols to
> use with it (hence the my).
I tried this and it works fine. I need all the symbols and I will
probably not trace them, but copy the svg code and modify them from
http://commons.wikimedia
Martin Morgan wrote:
>
>> x <- y = 1
>>
> Error in (x <- y) = 1 : object 'x' not found
>
this error does not make sense to me... if the precedence is as the
parentheses suggest, the error should be like in
(x <- y)
# error: object 'y' not found
vQ
__
Hi all! Does anyone know if a vector autoregression package is avaialable
that allows binary variables as part of the endogenous system? I'm looking
for something along the lines of what is implemented in "Dynamic Forecasts
of Qualitative Variables: A Qual VAR Model of US Recessions" by Michael
D
don't use the type argument. You are telling plot to plot points.
also please use dput() to make example code that can be cut and pasted
out of the email right into R, so that your problem is reproducible.
Stephen
On Wed, Apr 1, 2009 at 2:52 PM, Thomas Adams wrote:
> I have data that I read in
perhaps,
unlist(d, use.names=F)
baptiste
On 1 Apr 2009, at 22:15, oscar linares wrote:
Dear Rxperts
I have a data.frame as follows
ABCD
14 710
25 811
36 912
I want to convert it to a data frame with a single row (i.e., stack
the
columns w
I'm trying to follow the ggplot introduction here:
http://had.co.nz/ggplot/ggplot-introduction.pdf
I've installed ggplot2 with install.packages("ggplot2", dep=T)
but when I try to run
print(ggpoint(p, list(colour = sex)))
I get an error:
Error in print(ggpoint(p, list(colour = sex))) :
could n
Dear Rxperts
I have a data.frame as follows
ABCD
14 710
25 811
36 912
I want to convert it to a data frame with a single row (i.e., stack the
columns without the heading)
1
2
3
4
5
6
7
8
9
10
11
12
Any suggestions please.
Thanks in advance!
--
Hi Aaron,
Earlier I posted a question about memory usage, and the community's input was very helpful. However, I'm now extending my dataset (which I use when running a regression using lm). As a result, I am continuing to run into problems with memory usage, and I believe I need to shift to imp
Could you please include some code that demonstrates your problem?
Best wishes,
Ioannis
On Wednesday 01 April 2009 15:26:33 woodbomb wrote:
> I'm running brglm to do binomial loguistic regression.
>
> The perhaps multicollinearity-related feature(s) are:
>
> (1) the k IVs are all binary categori
Thanks Romain,
That code seems to be working pretty well, though for some reason it
seems to be dropping a few observations. I am not sure why.
No this isn't a April-fools joke :P
Francis
On Wed, Apr 1, 2009 at 1:03 PM, Rolf Turner wrote:
>
> This whole thing is an April Fool's joke. Isn't
Thanks! That did it. I should have seen that I needed the comma!
hadley wrote:
>
>> I tried messing with the line df$FixTime[which.min(df$FixInx)] changing
>> it
>> to df[which.min(df$FixInx)] or adding new lines with the additional
>> columns
>> that I want to include, but nothing seemed to
rkevinburton wrote:
>
> Thank you I had not considered using "gradient" in this fashion. Now as an
> add on question. You (an others) have suggested using SANN. Does your
> answer change if instead of 100 "variables" or bins there are 20,000? From
> the documentation L-BFGS-B is designed for a
Lorenzo Isella wrote:
>
> I am sure this is a one-liner, but I cannot find the R command to
> generate the LaTex symbols \perp and \parallel.
>
I took Dieter Menne's and Brian Ripley's examples, extended them slightly
with Hershey
font analogues, and posted the results at
http://wiki.r-pro
Lorenzo Isella wrote:
>
> I am sure this is a one-liner, but I cannot find the R command to
> generate the LaTex symbols \perp and \parallel.
>
I took Dieter Menne's and Brian Ripley's examples, extended them slightly
with Hershey
font analogues, and posted the results at
http://wiki.r-pro
Hello.
We have a question concerning the nonparametric analysis of a dataset, which
resulted in rejection of the null hypothesis (Kruskal-Wallis-test = H-test).
In order to find out which sample means actually are statistically
different, we want to do multiple comparisons with the Wilcoxon rank s
On 2/04/2009, at 7:04 AM, Thomas Levine wrote:
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-kT*pi/180)*kX
)
where kX,kY and kT are vectors of equal length. But I can't do that
with abline unless I use a loop, and I haven't figured out the least
unelegant way of writing the l
Stavros Macrakis wrote:
>
>> `->`
>>
> Error: object "->" not found
>
that's weird!
vQ
__
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/posting-guide.htm
pmatch() facilitates a very simple solution:
#Data
IA <- factor(c(1,2,2,3,3,4,3,5,5))
FixTime <- c(200,350,500,600,700,850,1200,1350,1500)
#First occurrence of each level
first. <- pmatch(levels(IA),IA)
#Use first occurrence to subscript a vector or data frame
FixTime[first.]
A simple way to a
Just to correct/expand/clarify the parenthetical below (how often is there a
chance to correct or clarify something posted by Bill Venables?), the ides are
the 15th of March, May, July, and October, but the 13th of the other months.
So if you want to use the ides as the date to use, you will ne
On 2/04/2009, at 8:37 AM, Jason Rupert wrote:
Is there any syntax in R that allows a "switch"-type condition to
be used?
switch(variable){
case CONSTANT_VALUE;
break;
default:
break;
}
?switch
##
Attention:
How about reading ?switch ?
Best,
Gabor
On Wed, Apr 1, 2009 at 9:37 PM, Jason Rupert wrote:
>
> Is there any syntax in R that allows a "switch"-type condition to be used?
>
> switch(variable){
> case CONSTANT_VALUE;
> break;
>
> default:
> break;
> }
>
>
> Thanks,
> Jason
>
> ___
Is there any syntax in R that allows a "switch"-type condition to be used?
switch(variable){
case CONSTANT_VALUE;
break;
default:
break;
}
Thanks,
Jason
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
One of the points of my.symbols is that you can define your own symbols to use
with it (hence the my).
I downloaded a graphic of the aries symbol (your original attempt in unicode I
belive) and used the following code to trace the left half of the symbol
(starting bottom center), then used that
Dear Allen,
On Wed, 1 Apr 2009 10:44:33 -0700 (PDT)
AllenL wrote:
>
> Dear R list,
> I've been attempting to interpret the results from a three-way ANOVA.
> I
> think I understand contrasts and the R defaults for these (treatment
> contrasts). My question is: what is the intercept in this test?
On Wed, 2009-04-01 at 16:49 +0100, Jose Iparraguirre D'Elia wrote:
> Dear all,
>
> Say I have the following dataset:
>
> > DF
> x y z
> [1] 1 1 1
> [2] 2 2 2
> [3] 3 3NA
> [4] 4 NA 4
> [5] NA 5 5
>
> And I want to omit all the rows whi
calpeda wrote:
> thank you,
> but I m importing data from a txt file and I have a matrix of n*1
> The function str seems to work only from 1*n
>
>
you see, it would help if you provided more details from the start. you
may still need to do it; it seems that both solutions you were given
(mine
Hello,
Earlier I posted a question about memory usage, and the community's input was
very helpful. However, I'm now extending my dataset (which I use when running
a regression using lm). As a result, I am continuing to run into problems with
memory usage, and I believe I need to shift to impl
This whole thing is an April Fool's joke. Isn't it?
***Please***!!! (Let it be an April Fool's joke.)
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
I have data that I read in using:
data<-read.table("RAVK2.obs.data",sep="\t")
'data' looks like this:
V1V2
1 2009-03-25 06:00:00 12.86
2 2009-03-25 12:00:00 12.80
3 2009-03-25 18:00:00 12.76
4 2009-03-26 00:00:00 12.68
5 2009-03-26 06:00:00 12.66
6 2009-03-26 12:00:
Hello,
I have a problem in feature selection I would be thankful if you can help
me.
I have a dataset with limited samples (for example 100) and a lot of
features (for example 3000) and i have to do feature selection.
if i use cross validation (for example *10 fold*) i rank the features based
on 9
How can I performing Bootstrap Confidence Intervals for the estimates of
nonparametric regression y=f(x) such as loess and spline smoothing
Thanks in advance
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-kT*pi/180)*kX
)
where kX,kY and kT are vectors of equal length. But I can't do that
with abline unless I use a loop, and I haven't figured out the least
unelegant way of writing the loop yet. So is there a way to do this
without a loo
I'm trying to use snow in my dual-core (hopefully later this is going to
run in a cluster). So, at this moment I create a cluster using SOCK
connection (MPI in the future). However when I try to use
clusterApplyLB I got "Error in socketSelect(socklist) : not a socket
connection". Any ideas ? Do you
Dear r-help,
How can I add a strip to show group weights using lattice package? For
example, in the following code, I'd like to using "wt" variable in a trip
to demonstrate the relative size of groups.
(Following is just the simplest form to demonstrate the question. A
conditional variable will
Dear R list,
I've been attempting to interpret the results from a three-way ANOVA. I
think I understand contrasts and the R defaults for these (treatment
contrasts). My question is: what is the intercept in this test? As far as I
can tell, its NOT the expected value of a point that belongs to the
The company I work for require users to request what packages they want from
the IT department (user cannot download themselves). I intend to request
installation of the latest version of R plus the 23 Cran task views. As a
statistician what are the recommended packages or packages that
statisticia
First input the data frame:
> Lines <- "x y z
+1 1 1
+2 2 2
+3 3NA
+4 NA 4
+ NA 5 5"
>
> DF <- read.table(textConnection(Lines), header = TRUE)
> # Now uses complete.cases to get required rows:
>
> DF[complete.cases(DF[1:2]),]
x y z
1
Lorenzo Isella wrote:
>
> I am sure this is a one-liner, but I cannot find the R command to
> generate the LaTex symbols \perp and \parallel.
>
As often, the most helpful "how-to" resource is by Prof. Brian Ripley
http://markmail.org/thread/kauzftprydrhqq5m
if you manage to get around the ma
On 01-Apr-09 15:49:40, Jose Iparraguirre D'Elia wrote:
> Dear all,
> Say I have the following dataset:
>
>> DF
> x y z
> [1] 1 1 1
> [2] 2 2 2
> [3] 3 3NA
> [4] 4 NA 4
> [5] NA 5 5
>
> And I want to omit all the rows which have NA, but o
I have a matrix of data. I need to scan the matrix and find every
sequence from maxima to maxima across a row. I can write a loop to do
this easily. Problem is, I can't figure out how to store the results.
Each result is a vector of widely varying lengths. Ideally I'd like a
vector of these
Bugzilla from rmh3...@gmail.com wrote:
>
> Responses:
> CompletionTIme
> VisitedTargets
>
> Fixed-factors:
> Targets (4-levels): 4, 9, 14, 19
> Entropy (3-levels): Low, Medium, High
>
> Random-factors:
> Participants: 31 total participants
> Replicates: 5 (this could also be viewed as a time
On 4/1/2009 11:39 AM, Stavros Macrakis wrote:
On Wed, Apr 1, 2009 at 10:55 AM, Duncan Murdoch wrote:
On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
As far as I can tell from the documentation, assignment with = is
precisely
equivalent to assignment with <-. Yet they call different primitives:
Feng, Jingyu wrote:
>
> I'am trying to develop some code if R, which would correspond to what I
> did in SAS.
> The data look like:
>
> TreatmentReplicategroup1 GSI
>
> ..
> The SAS code is:
> proc mixed data=data_name order=data method=ml; *scoring=10;
> classes group1;
>
David, thank you very much for the quick response:
The "sample" example helped and works fine for me. I'm sorry for not
providing an example. In order to explain my problem see following example:
test<-rt(1000,df=5)
Repeat only the following code to see how the second and third plot
cha
Dear All,
I am sure this is a one-liner, but I cannot find the R command to
generate the LaTex symbols \perp and \parallel. Consider for instance
the figure
(one can use any kind of data for the plot)
pdf("friction_linear_chain_perpendicular.pdf")
par( mar = c(4.5,5, 2, 1) + 0.1)
plot(data[ ,1],
On Wed, Apr 1, 2009 at 11:00 AM, hadley wickham wrote:
>> I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
>> to df[which.min(df$FixInx)] or adding new lines with the additional columns
>> that I want to include, but nothing seemed to work. I'll admit I only have a
>> mil
> I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
> to df[which.min(df$FixInx)] or adding new lines with the additional columns
> that I want to include, but nothing seemed to work. I'll admit I only have a
> mild understanding of what is going on with the function .fun.
I have another question regarding ddply. In my actual data.frame, I have many
other column variables corresponding to the type of the trial. I'm wondering
if I can have ddply include those in firstfixtime as well.
I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
to df[wh
Sorry if this is the wrong ml for this question, I am new to R. I am
trying to use R to analyze the data from my thesis experiment and I am
having troubles accounting for the pseudoreplication properly from
having each participant repeat each treatment combination (combination
of fixed factors) 5 t
Dear all,
Say I have the following dataset:
> DF
x y z
[1] 1 1 1
[2] 2 2 2
[3] 3 3NA
[4] 4 NA 4
[5] NA 5 5
And I want to omit all the rows which have NA, but only in columns X and Y, so
that I get:
x y z
1 1 1
2 2 2
3 3 NA
I'am trying to develop some code if R, which would correspond to what I did in
SAS.
The data look like:
TreatmentReplicategroup1 GSI
Control A 1 0.81301
Control B 1 1.06061
Control C 1 1.26350
Control
First of all I'd like to thank all those who answered me back teaching me
different ways to get the calledfunction modify global data rather than its
own. I fixed that.
Now I have a similar pproblem. Whenever the caller passes a matrix to the
called function I thought the called function would
Hi
r-help-boun...@r-project.org napsal dne 01.04.2009 11:16:26:
> Hello,
>
> A nice guy call Jun Shen was helping me out with this, but I require a
bit
> more help. Below is my data set or list called 'test'. I'm trying to
calculate
> the %RSD for each pair of index and keep it in cronologica
On Wed, Apr 1, 2009 at 10:55 AM, Duncan Murdoch wrote:
> On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
>
>> As far as I can tell from the documentation, assignment with = is
>> precisely
>> equivalent to assignment with <-. Yet they call different primitives:
>>
>
> The parser does treat them dif
Duncan Murdoch writes:
> On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
>> NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
>> which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
>> (e.g. f(a=1), which is equivalent to
>> eval(structure(quote(f(1)),names
Hi: I have had a similar issue so below are ways that I deal with
that.  i don't think there are manuals/ documentation for
becoming more of a developer ( someone can correct me if I'm wrong and I'd
be happy to be wrong ) but there are other ways:
1) Staying on this list
Hi,
When I type library("tcltk") under R 2.8.1 I get the error message:
Loading Tcl/Tk interface ...Error in inDL(x, as.logical(local),
as.logical(now), ...) :
unable to load shared library
'C:/PROGRA~1/R/R-28~1.1/library/tcltk/libs/tcltk.dll':
LoadLibrary failure: The specified module co
thank you,
but I m importing data from a txt file and I have a matrix of n*1
The function str seems to work only from 1*n
Wacek Kusnierczyk wrote:
>
> calpeda wrote:
>> hi
>> I ve a list of item x = ( "x1"
>>"x2"
>>"
R-help,
I am trying to perform a basic anlaysis of the BreastCancer data from
"mlbench" using the svm() function in "e1071". I use the following code
library("e1071")
library("mlbench")
data(BreastCancer)
BC <- subset(BreastCancer, select=-Id)
pairs(BC)
model <- svm(Class ~ ., data=BC, cross=1
I'm running brglm to do binomial loguistic regression.
The perhaps multicollinearity-related feature(s) are:
(1) the k IVs are all binary categorical, coded as 0 or 1;
(2) each row of the IVs contains exactly C (< k) 1's; (I think this is the
source of the problem)
(3) there are n * k unique r
Hi everyone,
I intend to do a discriminant analyse for 2 measures(eye diameter and body
length) and for different areas to show differences between those areas if
there are any. The raw data (eye diameter, body length) make one cloud of
points so it seems there aren't any differences between tho
On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell f
Hi Stephen:
It's the inputs given to me by a end-user. Ultimately trying to fit a student-t
copula to a bunch of simulated price returns
while maintaining the structure of the estimated correlation matrix.
The other challenge is I use R to test and work a solution but then have also
done in mat
Neil,
pls tell why do you need the correlation matrix? if you are trying to
simulate correlated variables then you can go around the cholesky by using
svd.
if you really need the correlation ( I think it is always possible to avoid
it ) then Rmetrics have a function to turn yo
Hi R users,
I apologize for a seemingly trivial question, but I felt this forum would be
the best place to seek advice.
I have been an R user for a year now, but I am limited to using R and its
various contributed packages. I strongly feel that users of a free and open
source software tool must e
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell from the documentation, assignment with = is prec
Look at the nearPD() function in the package "Matrix".
require(Matrix)
?nearPD
In particular, pay attention to the arguments "eig.tol" and "posd.tol",
which you can tweak to define how much "positiveness" you would like to
have.
Ravi.
--
Thank you I had not considered using "gradient" in this fashion. Now as an add
on question. You (an others) have suggested using SANN. Does your answer change
if instead of 100 "variables" or bins there are 20,000? From the documentation
L-BFGS-B is designed for a large number of variables. But
Dear celpeda,
Try this:
x = c( "x1", "x2", "x3")
substr(x,1,4)
[1] "" "" ""
See ?substr for more details.
HTH,
Jorge
On Wed, Apr 1, 2009 at 9:30 AM, calpeda wrote:
>
> hi
> I ve a list of item x = ( "x1"
> "x2"
>
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