Hi all:
I have a question about one-way anova.
The data is ¡°sleep¡±which belongs to R.
code1:
summary(lm(extra~group))
Estimate of group2(1.58) is the difference between mean of group1 and
group2,and t value(1.861) and p value(0.0792) is the same as 2 sample t
test,and the code is ¡°
On Fri, 10 May 2013, ivo welch wrote:
I ended up wrapping my own new "ols()" function in the end. it is my
replacement for lm() and summary.lm. this way, I don't need to alter
internals. in case someone else needs it, it is included. of course,
feel free to ignore.
If you use NeweyWest() f
Hi,
dat1<- read.table(text="
x1 x2 x3
a1 b2 2
a1 b4 4
a2 NA 3
NA b2 6
a3 b1 NA
a1 b2 9
a1 b2 NA
a1 b4 2
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- dat1
dat2$x4<-with(dat2,ave(x3,x1,x2,FUN=f
What does ww[["waterused"]] have to do with Maji?
On Sat, May 11, 2013 at 10:07 AM, Smith Freeman wrote:
> Hello,
>
> I am attempting to use the "classInt" package in conjunction with
> "rworldmap" package in R to construct a chloropleth. I want to use the
> fixedBreaks argument to specify given
Hello,
I am attempting to use the "classInt" package in conjunction with
"rworldmap" package in R to construct a chloropleth. I want to use the
fixedBreaks argument to specify given breaks.
My data look like this:
> head(Maji)
Country waterused CC
1 Afghanistan 36
You may want to use C++ instead using Rcpp which string handling would
be easier:
http://gallery.rcpp.org/articles/strings_with_rcpp/
On 10 May 2013 15:51, cgenolin wrote:
> Hi the list,
> I include some C code in a R function using .C. The argument is a character.
> I find how to acces to the c
I ended up wrapping my own new "ols()" function in the end. it is my
replacement for lm() and summary.lm. this way, I don't need to alter
internals. in case someone else needs it, it is included. of course,
feel free to ignore.
docs[["ols"]] <- c(Rd= '
@TITLE ols.R
@AUTHOR ivo.we...@gmail.com
On 13-05-10 2:06 PM, William Dunlap wrote:
Thanks. I looks like my example worked because the value of the new
attribute had a length greater than 1. That difference in behavior
smells like a bug.
Yes, definitely looks like one. I'll submit the bug report; I hope
someone else will fix it.
On 13-05-10 1:21 PM, Paul A. Steckler wrote:
Hi,
I've used R just a little bit, and I couldn't find an answer to my
questions via Web search.
I'm aware of Luke Tierney's byte code compiler for R, and the
compile package, and I have some questions about its use. I want to
take some R code,
dear R experts:
I am contemplating the logic in R's lm() and summary(lm()) statements.
the reason is that I want to extend the functionality of lm to give
me both standardized coefficients and newey-west standard errors and
Ts. I have the code and can stick it at the end of the lm() function
(an
Hi,
vec1<-factor(rep(c("A","B","C"),c(19,8,11)),levels=c("A","B","C"))
sapply(split(vec1,vec1),length)
# A B C
#19 8 11
#or you could change gl() function
gl.new<-function (n, k, labels = 1:n, ordered = FALSE) {
out<- numeric()
for(i in 1:n){
out<- append(out,rep(i,length.out=k[i]))
That's great thanks
--- Original Message ---
From: "arun kirshna [via R]"
Sent: May 10, 2013 3:34 PM
To: "Tomos_D"
Subject: Re: Return a vector in Fibonacci sequence function
Hi,
May be this helps:
fibv =function(n)
{
f1 = f2 = 1
f3<- c(f1,f2)
for(i in seq(2, n-1)) {
if(n == 0 || n == 1) ret
On May 10, 2013, at 11:20 AM, Elizabeth Beck wrote:
This.
>
> SODIUM K CL HCO3 ANION CA P GLUCOSE CHOLESTGGTGLDH CK AST
> PROTEIN ALBUMIN GLOBULIN A_G UA BA CORTICO T3 T4 THYROID 145 3.3 102 24 22
> 2.9 2.45 9.8 5.7 3 3 678 5 34 15 19 0.79 180 6 70.97 1.31 12.77 0.102376
> 146
Hi,
i am tryng to run Bagging SVM but I have a problem. I do Bagging SVM based
on algoritm bagging Trees by Breiman. I get the following error and I think
I am wrong in calculating err.mean and err.vote. I also have difficulty in
making the plot.
> library(e1071)
> gendata <- function(N=30,p=5,r=.
Hi,
I've used R just a little bit, and I couldn't find an answer to my
questions via Web search.
I'm aware of Luke Tierney's byte code compiler for R, and the
compile package, and I have some questions about its use. I want to
take some R code, create byte code for it, and distribute it, rath
Hi, you are very welcome!
I'm glad that it works now.
Irucka Embry
<-Original Message->
>From: Hiep [via R] [ml-node+s789695n4666747...@n4.nabble.com]
>Sent: 5/10/2013 6:27:23 AM
>To: iruc...@mail2world.com
>Subject: Re: Re: Tranfer data from exel into R
>
>Hi Irucka
>
>it works now,
On Fri, May 10, 2013 at 01:33:34PM +0100, Meenu Chopra wrote:
> Thanks to all
>
> The main problem with ma linux system is that its not able to install any
> software using sudo command.
> like i used command yum search libX11 , it shown that yum is not installed
> and when i use sudo apt-get in
Hi David,
You are right in that Bray-Curtis is not suitable for my dataset, and that
my variables are very different. Given your suggestions, I am struggling
with how to transform or standardize my data given that they vary so much.
Additionally, looking at the dist() package I am not sure which d
Thanks. I looks like my example worked because the value of the new
attribute had a length greater than 1. That difference in behavior
smells like a bug.
> { x <- 1 ; attributes(x)[["attrName"]] <- 2012 ; str(x) }
Error in attributes(x)[["attrName"]] <- 2012 :
attributes must be a list or
a) Please read the Posting Guide and post in plain text as it says to do. This
is an adjustment you must make in your email client software, we cannot advise
you how to do it here.
b) Not clear why you want a matrix. A vector is more typical in R; most of the
time using n*1 matrices is an unnec
sure, here's a series of commands/outputs (using R 3.0.0):
> x <- letters[1:4]
> y <- "foo"
> attributes(x)
NULL
> attributes(x)[[y]] <- "bar"
Error in attributes(x)[[y]] <- "bar" : attributes must be a list or NULL
> attr(x, y) <- "bar"
> attributes(x)
$foo
[1] "bar"
> attributes(x)[
On 5/10/2013 5:22 AM, Ott Toomet wrote:
Hi,
maybe we changed the convergence criteria a little bit? I cannot remember,
but we have played around with those a few times. You may try those,
perhaps it helps a bit.
Another recommendation---as you do not supply analytic gradient, you may
want use
Dear All
I have the following code:
datetime <-c("1/1/13 00:00","1/1/13 12:00","1/2/13 00:00","1/2/13 12:00")
datetime <-as.POSIXct(datetime,format='%m/%d/%Y %H:%M')
times
<-matrix(c(difftime(datetime[2],datetime[1],units="hours"),difftime(datetime[3],datetime[2],units="hours"),difftime(dateti
Meenu:
You have an elementary Linux setup and configuration problem to understand
first, before you worry about configuring and compiling your own R. I
agree strongly that this is something that all linux users should learn to
do, but compiling R itself is like climbing Mt Everest as your first
m
Try using a 2-column matrix, with columns formed from C and D, to subscript A.
E.g.,
> A <- matrix(101:132, nrow=8, ncol=4)
> C <- seq_len(nrow(A))^2 %% nrow(A) + 1
> D <- seq_len(nrow(A))^3 %% ncol(A) + 1
> for(j in seq_len(nrow(A))) print(A[C[j], D[j]])
[1] 110
[1] 105
[1]
Thanks, this one works!
On Thu, May 9, 2013 at 5:09 PM, Gabor Grothendieck
wrote:
> On Thu, May 9, 2013 at 8:09 PM, Gabor Grothendieck
> wrote:
> > On Thu, May 9, 2013 at 7:24 PM, Ye Lin wrote:
> >> Hey All,
> >>
> >> I want to duplicate the records but add a new "timestamp" columns as new
> >
Dear R-users,
I am looking for a solution to "parallelize" my PLSR predictions in order to
save processing time. I was trying to use the "foreach" construct with "doPar"
(cf. 2nd part of code below), but I was unable to allocate the predicted values
and the model performance parameters (RMSEP) t
hi,
is there a way to estimate the parameters of an IRT model (in my case, the
graded response model), when some item parameter are already known? what i have
heard is that other irt programs such as parscale allow to prespecify some of
the item difficulty and location parameters and estimate o
Hi,
I am trying to use the package 'SeqMiner" zhanxw.com/seqminer/, and I
would like to use any of the functions
readVCFToMatrixByRange()
readVCFToMatrixByGene()
readVCFToListByGene()
or readVCFToListByRange()
I have run the example code on the test datasets, and have got the
expected output.
I am currently using the PGLM package and I would like to implement a
within-model. Unfortunately, I do not succeed as I am not a big expert in
panel regression.
I am using the example data set from the PGLM package:
library(pglm)
data('Unions', package = 'pglm')
anb <- pglm(union~wage+exper+rura
Thanks to all
The main problem with ma linux system is that its not able to install any
software using sudo command.
like i used command yum search libX11 , it shown that yum is not installed
and when i use sudo apt-get install yum
its giving error
E: Unable to locate package yum
same problem with
Hi,
maybe we changed the convergence criteria a little bit? I cannot remember,
but we have played around with those a few times. You may try those,
perhaps it helps a bit.
Another recommendation---as you do not supply analytic gradient, you may
want use BHHH method instead (if this applies to yo
Dear all,
I want to compare a huge vector with selected element from a matrix.
A is a matrix and B is a vector. I want to compare each element of B with
selected element from A. C and D are selection criteria. They are vectors
of same length as B. C specifies the row number of A, and D specifies
Dear all,
we are computing maximum likelihood estimations using maxLik package
and we realized that the results of the estimation depend on the
version of the package installed
for example if we try to estimate this function with an old version of
maxLik under R 2.13.1 (32 bit version ins
This is a feature request. Or else a howto request. Can there be some
simple, automatic way to make fonts embed into pdf output? Could this be in
options() or par()?
Why?
I recently wanted to create a paper for a conference in a format called
AAAI (which I had never heard of before because I liv
> > attributes(my_obj)[[my_attr_name]] <- my_attr_value
>
> ...fails when my_obj doesn't already have an attribute named my_attr_name.
Do you have an example of this? In R-2.15.3 I don't see that problem:
> my_obj <- 37:41
> my_attr_name <- "myAttr"
> my_attr_value <- as.roman(2012:2013
Katharine,
Thank you for the pointer, and thanks to Josh Ulrich and William Dunlap for
good suggestions.
(Also thanks to Dirk Eddelbuettel for shaming me into modifying the code.)
William's fnTracer wrapper technique doesn't require rebuilding the package, so
it's easier in some ways,
especially
Hello,
R is open source, you can download the sources and then grep rlnorm to see
file: src/library/stats/distn.R
rlnorm <- function(n, meanlog=0, sdlog=1)
.External(C_rlnorm, n, meanlog, sdlog)
And file rlnorm.c is in src/nmath. It returns exp(rnorm(meanlog, sdlog))
So you'll have to wri
Hi list,
Does anyone know the code behind rlnorm(n, meanlog = 0, sdlog = 1)? I am
going to write it in c#.
thanks
Alireza
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PLE
thanks, both.
the only real difference between the two approaches that i can see is when
assigning _new_ attributes to an object where the attribute name is itself
variable.
something like this:
> attributes(my_obj)[[my_attr_name]] <- my_attr_value
...fails when my_obj doesn't already have an at
hi all --
i've written a series of procedures to annotate nodes in a dendrogram
object.
i'd like to be able to easily pluck out some of these attributes and view
them in tree form.
in a regular list, i'd do something like:
> sapply(my_list, function(x) x$my_entry)
and in a dendrogram, i can use
On 10/05/2013 10:50 AM, Rui Barradas wrote:
Hello,
There's at least one example where only the form attr(x, "foo") <- "bar"
would work, not the other form. If you want to set attributes
programatically, use the first form, like in the function below. Note
that the example is artificial.
setAtt
Hello,
There's at least one example where only the form attr(x, "foo") <- "bar"
would work, not the other form. If you want to set attributes
programatically, use the first form, like in the function below. Note
that the example is artificial.
setAttr <- function(x, attrib, value){
Anyone have a clue why assigning to a list field in a RefClass is so much
slower than assigning to a plain old list?
For example, given a list of length N, if I want to assign a value to each
entry, then the running time is quadratic in N.
That is, refclass$a.list[[i]] <- value is much much slowe
Hi,
May be this helps:
fibv =function(n)
{
f1 = f2 = 1
f3<- c(f1,f2)
for(i in seq(2, n-1)) {
if(n == 0 || n == 1) return(n)
if(n == 2) return(1)
f = f1 + f2
f2 = f1
f1 = f
f3<- c(f3,f)
}
f3
}
fibv(0)
#[1] 0
fibv(1)
#[1] 1
fibv(3)
#[1] 1 1 2
fibv(10)
# [1] 1 1 2 3 5 8 13 21 34 55
A.K.
>I
Hi the list,
I include some C code in a R function using .C. The argument is a character.
I find how to acces to the characters one by one:
--- 8< --- C
void dealWithCharacter(char **chaine, int *size){
int i=0;
for(i=0;i<*size;i++){
Rprintf("Le caractere %i est %c\n",i,chaine
Awesome, exactly what I was hoping for! Thanks so much! You are right - I
put in the wrong end dates for the second day.
Best,
Kai
On Thu, May 9, 2013 at 4:01 PM, Rui Barradas wrote:
> Hello,
>
> Maybe the following will do it. Note, however, that in your data, for
> start day 2012-02-11,
Hi all, i used the function unionSpatialPolygons to dissolve polygons
according to an ID value. However, the results were not that i was
expecting. It created a SpatialPolygon and it grouped the objects according
to the ID value. i.e. one polygon had about 3 subpolygons with the same
value. How is
Katarzyna Kulma gmail.com> writes:
>
> Hiya,
>
> I'm using simple glm binomial models to test the effect of treatment
> (factor, 3 levels) on infection prevalence (infected/uninfected):
>
> ad3<-glm(Infection~ecs, family=binomial, data=eilb)
>
> but summary() function returns for each of the
Dear Alfonso
On 10 May 2013 12:51, wrote:
> we are computing maximum likelihood estimations using maxLik package and we
> realized that the results of the estimation depend on the version of the
> package installed
>
> for example if we try to estimate this function with an old version of
> maxL
Hiya,
I'm using simple glm binomial models to test the effect of treatment
(factor, 3 levels) on infection prevalence (infected/uninfected):
ad3<-glm(Infection~ecs, family=binomial, data=eilb)
but summary() function returns for each of the factor-level coefficients
against the control treatment:
2013/5/9 PIKAL Petr
> Hi
>
Thank you, that was the option I look for
>
> Use drop=TRUE argument in interaction for removing unused levels.
>
> Petr
>
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of Adel ESSAFI
>
On 10-05-2013, at 10:59, KatrinH Heimann wrote:
> Dear all,
>
> I would like to replace some values in a dataframe (including the columns
> Action, Interval and Artificiality) depending on a combination of different
> conditions
>
> the command I tried is
>
> for (j in 1:nrow(data))
>
> {
>
On 10-05-2013, at 10:59, KatrinH Heimann wrote:
> Dear all,
>
> I would like to replace some values in a dataframe (including the columns
> Action, Interval and Artificiality) depending on a combination of different
> conditions
>
> the command I tried is
>
> for (j in 1:nrow(data))
>
> {
>
Dear all,
I would like to replace some values in a dataframe (including the columns
Action, Interval and Artificiality) depending on a combination of different
conditions
the command I tried is
for (j in 1:nrow(data))
{
if ((data[j,data$Action]=="s") & (data[j,data$Interval]=="l") & (data[j,da
Hi,
@David: thank you for your fast response. I changed the universe to cover
the at least the total range of the values in the dataframe. However, this
does not save the problem for several combination of depth and velocity. I
attached the same example as before but with a dataframe of such
combi
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