W dniu 24.05.2012 00:08, Ux.Seo pisze:
> Hi, all
> I got following errors when I tried to install "RCurl" in Ubuntu.
> my R version is
>
> na@ubuntu:~$ R --version
> R version 2.15.0 (2012-03-30)
> Copyright (C) 2012 The R Foundation for Statistical Computing
> ISBN 3-900051-07-0
> Platform: i68
Perhaps this?
X[, apply(X, 2, sd) < 1e-10]
HTH,
Jorge.-
On Wed, May 23, 2012 at 8:44 PM, Chris Burns <> wrote:
> How do I trim a matrix to exclude columns that have no standard deviation?
>
> Thanks,
>
> Chris
>
>[[alternative HTML version deleted]]
>
>
On May 23, 2012, at 8:44 PM, Chris Burns wrote:
How do I trim a matrix to exclude columns that have no standard
deviation?
mtx[ , as.logical(apply(mtx, 2, sd))]
--
David Winsemius, MD
West Hartford, CT
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R-help@r-project.org mailing list
ht
Dear Trying:
after
g <- read.table("ip.txt")
enter
str(g)
You will see that g is a dataframe with one variable, called V1. V1
contains the data you want to plot.
So type
plot(g$V1, type="l")
or
with(g, plot(V1, type="l"))
or, if you prefer to attach g:
attach(g)
plot(V1, type="l")
--
On May 23, 2012, at 6:11 PM, Trying To learn again wrote:
Hi all
Sorry in advance
I have this txt with data and I want to plot the data with a line
between
dots.
The thing is that if I run this
g<-read.table("ip.txt")
plot(g,type="l")
Try:
plot(g$V1,type="l")
I have prove to attac
On May 23, 2012, at 5:47 PM, Chris Burns wrote:
I'm trying to run a regression on certain columns. The columns are
named
"Cat_1", "Cat_2" etc. Is there a way to call upon all of the
columns that
contain "Cat"?
lm( y ~ . , data=dfrm[, c("y", grep("Cat", names(dfrm), value=TRUE) ] )
OR:
How do I trim a matrix to exclude columns that have no standard deviation?
Thanks,
Chris
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PLEASE do read the posting guide ht
Hi all,
I am working with a spatial data set for which I am only interested in high
concentration values ("leaks"). The low values (< 90th percentile) have
already been turned into NA's, leaving me with a matrix like this:
< CH4_leak
lonlatCH4
1 -71.11954 42.35068 2
Hi all
Sorry in advance
I have this txt with data and I want to plot the data with a line between
dots.
The thing is that if I run this
g<-read.table("ip.txt")
plot(g,type="l")
I have prove to attach(g) but doesn´ t work.
I know is a begginer question and sure is of the type of data or the f
I'm trying to run a regression on certain columns. The columns are named
"Cat_1", "Cat_2" etc. Is there a way to call upon all of the columns that
contain "Cat"?
Thanks,
Chris
[[alternative HTML version deleted]]
__
R-help@r-project.org mail
Hi, all
I got following errors when I tried to install "RCurl" in Ubuntu.
my R version is
na@ubuntu:~$ R --version
R version 2.15.0 (2012-03-30)
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i686-pc-linux-gnu (32-bit)
thanks in advance
* installing
On 2012-05-23 16:53, Duncan Murdoch wrote:
On 12-05-23 7:40 PM, Judith Flores wrote:
Hi,
I was unable to find documentation about how to include the letter 'n'
with tilde above it, like this: ñ.
Is it possible to do that?
Yes, just type it. The main problem will come if you write
Re:
coef(summary(glm(extra ~ group, data=sleep[ rep(1:nrow(sleep), 10L), ] )))
Your (corrected) suggestion is the same as one of mine, and doesn't do what I'm
looking for.
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Tuesday, 22 May 2012 3:37p
To: Stev
On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote:
> On 12-05-23 1:31 PM, Giannis Mamalikidis wrote:
>> Hello all.
>>
>> I would like to know: provided that I absolutely state that R is not mine
>> and I also include the R’s License which will be shown so people know R and
>> R’s license,
>> (pr
On May 23, 2012, at 2:52 PM, arun wrote:
> Hi Marc,
>
> Just to point out some difference,
>
>
> x <- 1:20
> y <- x + (x/4 - 2)^3 + rnorm(20, sd=3)
> names(y) <- paste("O",x,sep=".")
>ww <- rep(1,20); ww[13] <- 0
> summary(lmxy <- lm(y ~ x + I(x^2)+I(x^3) + I((x-10)^2),
On 12-05-23 7:40 PM, Judith Flores wrote:
Hi,
I was unable to find documentation about how to include the letter 'n' with
tilde above it, like this: ñ.
Is it possible to do that?
Yes, just type it. The main problem will come if you write it to a file
and read it somewhere else; the
On 12-05-23 1:31 PM, Giannis Mamalikidis wrote:
Hello all.
I would like to know: provided that I absolutely state that R is not mine
and I also include the R’s License which will be shown so people know R and
R’s license,
(provided the above) am I allowed to include R’s folder (the folder that h
Hi,
I was unable to find documentation about how to include the letter 'n' with
tilde above it, like this: ñ.
Is it possible to do that?
Thank you,
Judith
[[alternative HTML version deleted]]
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https:
Ok, I feel kinda silly for not having thought of that before...
Anyway, thanks, Peter.
De: Peter Ehlers
Cc: "r-help@r-project.org"
Enviadas: Quarta-feira, 23 de Maio de 2012 19:22
Assunto: Re: [R] Error message from optim
On 2012-05-23 09:39, Rubem Kaip
On 2012-05-23 09:39, Rubem Kaipper Ceratti wrote:
Dear list,
When running the script below, the optim function returns the message 'Error in function
(par) : could not find function "fn"'. I've re-read the code a few times and
just can't figure out why it's happening. Any help/suggestion is ap
This depends on what you mean by "distance between the 2 countries".
Do you want the shortest distance from a point on the border of each country?
The distance from the one capitol to the other?
Distance between centroids?
Weighted distance based on population distributions?
Etc.
Do you want grea
Yes, thanks Henrik. I neglected to mention that rowMedians could just
be plugged in instead of apply (..,1,...)
However, my main point is that that's probably not what matters,as
Benno points out. Maybe it's the data frames instead of the matrices,
but The process should execute in a few seco
Hi Marc,
Just to point out some difference,
x <- 1:20
y <- x + (x/4 - 2)^3 + rnorm(20, sd=3)
names(y) <- paste("O",x,sep=".")
ww <- rep(1,20); ww[13] <- 0
summary(lmxy <- lm(y ~ x + I(x^2)+I(x^3) + I((x-10)^2),
weights = ww), cor = TRUE)
> all.v
#If I have two lists as follows
a1<- array(1:6, dim=c(2,3))
a2<- array(7:12, dim=c(2,3))
l1<- list(a1,a2)
a3<- array(1:4, dim=c(2,2))
a4<- array(5:8, dim=c(2,2))
l2<- list(a3,a4)
#how can I create a new list with the mean across all arrays within the
list, so all components are included
Hello,
I am aware of the fact that the combine() function in the Random Forest package
of R is meant to combine forests built from the same training set, but is there
any way to combine trees built on different training sets? Both the training
datasets used contain the same variables and classe
Dear list,
When running the script below, the optim function returns the message 'Error in
function (par) : could not find function "fn"'. I've re-read the code a few
times and just can't figure out why it's happening. Any help/suggestion is
appreciated.
Regards,
Rubem
## R script
library
Hi,
I assume this is what you are looking.
?variable.names()
x <- 1:20
y <- x + (x/4 - 2)^3 + rnorm(20, sd=3)
names(y) <- paste("O",x,sep=".")
ww <- rep(1,20); ww[13] <- 0
summary(lmxy <- lm(y ~ x + I(x^2)+I(x^3) + I((x-10)^2),
weights = ww), c
Hello all.
I would like to know: provided that I absolutely state that R is not mine
and I also include the R’s License which will be shown so people know R and
R’s license,
(provided the above) am I allowed to include R’s folder (the folder that has
its binaries) on my freeware program or not
andrija djurovic gmail.com> writes:
>
> Hi to all.
>
>..
>
> Thanks in advance
>
> Andrija
I am having the exact same problem. Any luck with this yet?
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R-help@r-project.org mailing list
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PLEASE do r
Seems like you have a solution by now, but I didn't see:
require(xlsx)
df1 <- data.frame(c1=1:2, c2=3:4, c3=5:6)
df2 <- data.frame(c21=c(10.10101010101,20, 3), c22=c(50E50,60, 3) )
outFile <- 'df12.xls'
wb <- createWorkbook()
sh1 <- createSheet(wb,'sheet1')
addDataFrame(df1,sh1)
sh2 <- createSh
You might want to look at
https://stat.ethz.ch/pipermail/r-help/2012-April/310582.html
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Marc Schwartz
> Sent: Wednesda
On 2012-05-23 08:19, Juan Antonio Balbuena wrote:
Hello
This is a simple question but I couldn't google an answer.
In the procrustes function of the vegan package, one uses
plot(procrustes_object, kind=2) to obtain a plot of the residual
differences. For in
On May 23, 2012, at 1:42 PM, Peter Ehlers wrote:
> On 2012-05-23 09:55, R. Michael Weylandt wrote:
>> I think the easiest that comes to mind is simply
>>
>> names(coef(myMod))
>>
>> But did you look at myMod$terms[[3]] ? That seems to be the RHS of the
>> formula input (in the few cases I tried
On 2012-05-23 09:55, R. Michael Weylandt wrote:
I think the easiest that comes to mind is simply
names(coef(myMod))
But did you look at myMod$terms[[3]] ? That seems to be the RHS of the
formula input (in the few cases I tried)
Best,
Michael
It depends a bit on just what the OP wants. In cas
On 2012-05-23 10:19, csrabak wrote:
Em 23/5/2012 11:59, evaD escreveu:
Hey, could I just ask you a quick question?
if i have two sets of numbers, eg:
list1 = 1,3,4,4,4 list2 = 1,2,4,5,6
and in R, i run the command:
wilcox.test(list1, list2, paired=TRUE, alternative="greater")
and say i g
On Wed, May 23, 2012 at 11:54 AM, peter dalgaard wrote:
>
> On May 23, 2012, at 19:30 , Preeti wrote:
>
> > Hmm.. that is interesting... I did this on our server machine which has
> > about 200 cores. So memory is not an issue. Also, building the dataframe
> > takes about a few minutes maximum fo
On May 23, 2012, at 19:30 , Preeti wrote:
> Hmm.. that is interesting... I did this on our server machine which has
> about 200 cores. So memory is not an issue. Also, building the dataframe
> takes about a few minutes maximum for me. My code is similar to yours but
> for the fact that I create m
Just adding a few cents to this:
rowMedians(x) is roughly 4-10 times faster than apply(x, MARGIN=1,
FUN=median) - at least on my local Windows 7 64bit tests. You can do
these simple benchmark runs yourself via the
matrixStats/tests/rowMedians.R system test, cf. http://goo.gl/YCJed
[R-forge].
/He
Hmm.. that is interesting... I did this on our server machine which has
about 200 cores. So memory is not an issue. Also, building the dataframe
takes about a few minutes maximum for me. My code is similar to yours but
for the fact that I create my dataframe from read.delim("filename") and
then I d
I wonder how you do this (or maybe on what kind of machine you execute it).
I tried it out of curiosity and get
> df = as.data.frame(lapply(1:300,function(x)sample(200,25,T)))
> colnames(df) = sample(letters[1:20],300,T)
> system.time(dfmed<-lapply(unique(colnames(df)), function(x)
+ rowMedia
Em 23/5/2012 11:59, evaD escreveu:
Hey, could I just ask you a quick question?
if i have two sets of numbers, eg:
list1 = 1,3,4,4,4 list2 = 1,2,4,5,6
and in R, i run the command:
wilcox.test(list1, list2, paired=TRUE, alternative="greater")
and say i get a p-value of 0.0001
Does this tel
Assuming your original matrix IS a matrix, call it yourmat, and not a
data frame (whose columns **must* have unique names if you haven't
messed with the check.names default) then maybe:
UNTESTED!!! ###
thenames <- unique(dimnames(yourmat)[[2]])
ans <- lapply(thenames, function(nm, {
apply
Dear useRs,
I have a question with respect to fitting a non-linearity using gam
(mgcv package, version 1.7-16).
In a study I'm currently conducting, I'd like to find out if there is
a breakpoint after which the effect of Age of Acquisition (AOA) of the
second language changes. I.e. if the slope o
I think the easiest that comes to mind is simply
names(coef(myMod))
But did you look at myMod$terms[[3]] ? That seems to be the RHS of the
formula input (in the few cases I tried)
Best,
Michael
On Wed, May 23, 2012 at 10:58 AM, jdub wrote:
>
> What is the best way to get the variable names use
Hello Everybody,
The code:
dfmed<-lapply(unique(colnames(df)), function(x)
rowMedians(as.matrix(df[,colnames(df) == x]),na.rm=TRUE))
takes really long time to execute ( in hours). Is there a faster way to do
this?
Thanks!
On Tue, May 22, 2012 at 3:46 PM, Preeti wrote:
> Thanks Henrik! Here i
Hi everyone,
I am looking for a package that would take "CountryName1" and
"CountryName2" as input and spit out the distance (approximate)
between the two countires. Does something like that exist?
I don't have R experience, so making one myself is out of the question...
Thanks in advance!
___
Hey, could I just ask you a quick question?
if i have two sets of numbers, eg:
list1 = 1,3,4,4,4
list2 = 1,2,4,5,6
and in R, i run the command:
wilcox.test(list1, list2, paired=TRUE, alternative="greater")
and say i get a p-value of 0.0001
Does this tell me that my values in list 1 tend to
Hi Joshua,
Thank you again for your very-generous help. I think I could follow most of
your explanations.
To give you some background about myself, I'm doing a PhD at the Institute of
Sound Recording (IoSR) at the University of Surrey in the UK. As I have no
statistics background, I've been te
Hello,
Would the following do it?
# make the dates vector
x <- seq(as.Date("2011-03-15"), as.Date("2011-09-09"), by="day")
# this is the important part
labs <- seq(min(x), max(x), by="month") # "month", not 30
labs <- format(labs, "%b")
labs.pos <- round(seq(1, length(x), length.out=length(labs
What is the best way to get the variable names used in lm() from its
results?
Stumbling around I found I could get the response variable name from
myMod$terms[[2]]
but using
myMod$terms[[1 ]]
gives a tilda.
I found the names buried in other places in the model object and in the
summary of t
I have a data frame like this:
T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h
C0h C0.25h C0.5h C1h C2h C3h C6h C12h C24h C48h
NM_001001130 68 9556 43 66 62 68 90 63 89 65
8558 49 81 49 76 73 48 77
NM_001001152 791
Dear R-Users,
Dr. Wood replied to a similar topic before where confidence intervals were
for a ratio of two treatments (
https://stat.ethz.ch/pipermail/r-help/2011-June/282190.html). But my
question is more complicated than that one. In my case, log(E(y)) = s(x)
where y is a smooth function of x.
You need to read the documentation for read.xlsx and data.frame.
In it you would discover what you have done wrong:
1. rowIndex must be an array listing the columns you wish to read not a
character value
2. colIndex must be an array listing the rows you wish to read
[Deleting these arguments will
I'm trying to understand how a latent state matrix is updated by the MCMC
iterations in a WinBUGS model, using the package R2WinBUGS and an example
from Kery and Schaub's (2012) book, "Bayesian Population Analysis Using
WinBUGS". The example I'm using is 7.3.1. from a chapter on the
Cormack-Jo
Hello
This is a simple question but I couldn't google an answer.
In the procrustes function of the vegan package, one uses
plot(procrustes_object, kind=2) to obtain a plot of the residual
differences. For instance:
data(varespec)
vare.dist <- vegdist(wisconsin(var
Thank you Peter, works perfectly.
Funny how simple things are once someone tells you the answer =)
robbie
On Tue, May 22, 2012 at 9:37 PM, Peter Ehlers wrote:
> Robbie,
>
> Here's what I *think* you are trying to do:
>
> 1.
> y is a cubic function of x:
>
> y = b1*x + b2*x^2 + b3*x^3
>
> 2.
I think there is much to recommend this approach, but if you still
want to have separate functions in the workspace with the names having
successive integer designations then I have an approach. (There could
still be storage in a list for these functions with matching named
indices.)
> f
Hello,
I would be very grateful if somebody (or everybody:)) answers my question. I
am making ARMA-GARCH model with garchFit function and want in arma part make
lag list (like with arma models:
model=arma(data, lag=list(ar=c(2,6),ma=NULL), include.intercept=FALSE)
but I don't know how to do it
Hi everyone,
I have an issue with a data conversion. First, I tried it with the
reshape-package, but since it's quite a while that I used it, I feel kind
of rusty...
I have a data.frame like this:
id Sample.Name Marker Allele.1
Allele.2sample_id
You have spotted the problem correctly; indeed the problem was a local issue
(I am based in Sweden) and as per your suggestion adjustment of the local
parameters has solved it.
Many thanks
--
View this message in context:
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Thank you for the answer. I tried the first suggestion, but that didn't work
(length didn't match again), unless I'm doing something wrong.
The stacklab function is nice, but still I'm having way to much labels.
Maybe the actual graph makes it more clear:
http://r.789695.n4.nabble.com/file/n46310
On Wed, May 23, 2012 at 3:26 AM, Manish Gupta wrote:
> Thanks! It works but how can i remove x axis?
I have no idea "what works," except that it may in some way relate to
bar plots, but in general xaxt="n" will remove an x axis. See ?par for
details.
> --
> View this message in context:
> http:
Hello,
To keep related objects of the same nature in the same structure makes
very good sense but why not name the functions in the list after they
were created?
Using 'NextFunc' of a previous post,
f.list <- list()
f.list[[1]] <- function(x) x^2
f.list[[2]] <- NextFunc(f, 1)
f.list[[3]] <-
Michael Meyer yahoo.com> writes:
>
Check your logic. The following lines show that integrate *does* return the
correct values:
a = 0.08 # alpha
M <- function(j,s){ return(exp(-j*a*s)) }
A <- matrix(NA, 5, 5)
for (i in 1:5) {
for (j in i:5) {
f <-
On May 23, 2012, at 11:23 , Kony-2012 wrote:
> I am a new R user, but have searched the manuals on the topic , but found no
> suitable solution.
> I have imported a 20 years’ time-series data from csv format into R. The
> date variable was a factor and I used the command “ x$date <-
> as.Date(d
On 05/23/2012 05:00 AM, Soheila Khodakarim wrote:
Dear All
I can not download R for Linux.
I do not know which file I should install?
Best Regards,
Soheila
Can you provide the list with additional information such as:
What Linux distribution?
What have you tried?
For what it's worth, I al
On Wed, May 23, 2012 at 12:35:31AM -0700, kylmala wrote:
> Hi,
>
> I have a problem with R optimization. I try to copy excel solver acts to R.
> Solver has some problems with nonlinear optimisation.
>
> Suppose we have resources: (X,Y,Z)=(20, 30, 25) and services:
> matrix(10,11,0, 13,12,10, 0,24
Hi Joshua,
All sorted, turns out I just needed to add this "header=TRUE" and drop
those redundant statements you pointed out.
for anyone else:
# Libs
library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
library('quantmod')
library('TTR')
# Connect and get data
drv <- dbDriver('PostgreS
On 05/23/2012 09:34 PM, bets wrote:
Hey,
I am trying to create barplot of abundances over time (in days). The period
is over 171 days, so I don't want to have all labels there but only the
first day of the month. I couldn't find anything like this on the forum yet.
Mostrly it's about year to yea
Hey,
I am trying to create barplot of abundances over time (in days). The period
is over 171 days, so I don't want to have all labels there but only the
first day of the month. I couldn't find anything like this on the forum yet.
Mostrly it's about year to year data.
This is the relevant part of
Hi,
I am following the protocol outlined here for analysis of single channel
Agilent microarray data:
http://matticklab.com/index.php?title=Single_channel_analysis_of_Agilent_microarray_data_with_Limma
I keep getting the following error message when using Limma's read.maimages
function to load m
Wonderful!
Thanks very much for this: the step I'd clearly missed was to
1. put the rownames/colnames in a list object
2. *name* these with the titles separately ... allowing the expression
to be evaluated
names(dn)<-c(names(result[[1]]),names(result[[1]][2]))
3. apply this to the array
Ta v m
I'have tried to do something like that staring values:
pars=c(x1=1,x2=1,x3=1,x4=1,x5=1,x6=1,x7=1)
and the function
f=function(pars){
min(pars[1]/10,pars[2]/11)+min(pars[3]/20,pars[4]/22,pars[5]/25)+min(pars[6]/36,pars[7]/28)
}
optim(pars,f)
but how can I get the constraints in there also?
--
Hi,
I have a problem with R optimization. I try to copy excel solver acts to R.
Solver has some problems with nonlinear optimisation.
Suppose we have resources: (X,Y,Z)=(20, 30, 25) and services:
matrix(10,11,0, 13,12,10, 0,24,26), nrow=3.
Now we should optimize the problem: max( min(a/10,b/11)
I am a new R user, but have searched the manuals on the topic , but found no
suitable solution.
I have imported a 20 years’ time-series data from csv format into R. The
date variable was a factor and I used the command “ x$date <-
as.Date(date,format= "%d%b%Y") “ to convert to a date format. The
I want to measure the error in the estimation of a 2 dimensional density
function that is calculated using an integral but run into problems trying
to integrate with adaptIntegrate because the integrand also calls the
function adaptIntegrate. In particular I want
\int \hat{f}(x,y) - f(x,y) dx dy
Thanks! It works but how can i remove x axis?
Regards
--
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Sent from the R help mailing list archive at Nabble.com.
__
R
Greetings,
Sorry, the last message was sent by mistake! Here it is again:
I encounter a strange problem computing some numerical integrals on [0,oo).
Define
$$
M_j(x)=exp(-jax)
$$
where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products
$$
A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x
Greetings,
I encounter a strange problem computing some numerical integrals on [0,oo).
Define
$$
M_j(x)=exp(-jax)
$$
where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products
$$
A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx
$$
Analytically we have
$$
A_{ij}=1/(a(i+j)).
$$
In the co
A new 'dataframe' package is on CRAN. This is a modified version
of the data frame code in R, modified to make fewer copies of the inputs
and run faster, e.g.
as.data.frame(a vector)
makes 1 copy of the vector rather than 3,
data.frame(a vector)
makes 3 copies rather than 6, and
x[, "a"] <-
Having looked at this further, I've made some changes in mgcv_1.7-17 to
the p-value computations for terms that can be penalized to zero during
fitting (e.g. s(x,bs="re"), s(x,m=1) etc).
The Wald statistic based p-values from summary.gam and anova.gam (i.e.
what you get from e.g. anova(a) wher
Hi!
2012/5/23 Soheila Khodakarim :
> I can not download R for Linux.
Usually in Linux you need to give an installation command such as
"apt-get install " The OS will then grab it from a repository for
you and install it. It depends on the Linux distribution.
Best,
Henri-Paul
--
Henri-Pau
Hello,
http://cran.r-project.org/index.html
Regards.
- Mail original -
De : Soheila Khodakarim
À : "r-help@r-project.org"
Cc :
Envoyé le : Mercredi 23 mai 2012 18h00
Objet : [R] linux
Dear All
I can not download R for Linux.
I do not know which file I should install?
Best Regard
Dear All
I can not download R for Linux.
I do not know which file I should install?
Best Regards,
Soheila
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posti
Martijn,
I don't think there is one right answer to this. If you look at things
in the way that one would usually view a smooth model then m2 is both
simpler (lower EDF) and fits better, so is simply a better model (if the
simpler model fits better then why would you not use it?).
But of cou
Try
for( i in 1:10 ){
...
}
That should resove your problem 1.!
Rgds,
Rainer
On Wednesday 23 May 2012 09:23:04 RH Gibson wrote:
> "blap.txt" is a numeric vector of length 64.
>
> I am using the following code:
>
>
> bd<-scan("blap.txt")
> output<-matrix(0,64,10)
> s<-sum(bd)
> for (i in 10
Hi
What is your intention?
basically one output column can be made by
cumsum(bd)
Then you can shuffle bd by let say
bd <- sample(bd, 64)
and repeat cumsum for new bd.
bd<-scan("blap.txt")
output<-matrix(0,64,10)
for (i in 1:10) {
bd<-sample(bd, 64)
cs<-cumsum(bd)
output[,i]<-cs
}
If you in
"blap.txt" is a numeric vector of length 64.
I am using the following code:
bd<-scan("blap.txt")
output<-matrix(0,64,10)
s<-sum(bd)
for (i in 10){
while (s>0)
{x1<-sample(bd,(length(bd)-1), replace=F)
s<-sum(x1)
bd<-x1
output[i]<-s
}
}
write.table(output, file="res.txt")
This code is not do
On 05/22/2012 06:08 PM, HAOLONG HOU wrote:
Dear list,
The name of R-language is too short and is not friendly to search engines.
Do you think it can be renamed to something like "Rsio" or "Radio" ?
Thank you so much for this useful software!
Aw, come on guys, here is somebody praising R and try
Yes, in the .Rd file.
\eqn is my cure.
Thank you very much.
Best
ozgur
-
Ozgur ASAR
Research Assistant
Middle East Technical University
Department of Statistics
06531, Ankara Turkey
Ph: 90-312-2105309
http://www.stat.metu.edu.tr/people/assistants/ozgur/
>
> I'm not sure what you are trying to prove with that example - the
loopless
> versions are massively faster, no?
In some languages loops are integral part of programming habits. In R you
can many things do with whole objects without looping - vectorisation
approach. See R-Inferno from Patri
Do you mean in the .Rd files of your package?
On 23/05/2012 08:04, Özgür Asar wrote:
Dear all,
I have some trouble with special characters while building my R package. I
tried to follow the usual LATEX format, but could not fix the problem:
For instance, for "greater than or equal", I tried "\
Dear all,
I have some trouble with special characters while building my R package. I
tried to follow the usual LATEX format, but could not fix the problem:
For instance, for "greater than or equal", I tried "\geq", but R says that
this is an unknown macro.
Could anyone direct me how to solve thi
On 2012-05-23 01:11, zgu9 wrote:
> Hi everyone,
>
> I have an R problem about Input.
> I want to read an arbitrage function from keyboard. But I don't know how.
> I tried scan(), readLines() but fail to make the output as a function, or a
> list that I can use as.function to transform.
> Is it po
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