Greg Orm wrote:
Dear r-help list members,
I am quite new to R, and hope to seek advice from you about a problem I have
been cracking my head over. Apologies if this seems like a simple problem.
I would not call it exactly simple... but mostly because your data
representation is so obscure.
Le 11/05/10 04:16, Elizabeth Lawson a écrit :
Hi,
I just bought new macbook pro 10.6.3. I am trying to use some old c code
that used to work. I tried to recompile the code. In the directory with
code I used R CMD SHLIB hello.c and get the error
make: Nothing to be done for `all'.
I have t
On 11 May 2010 00:52, Achim Zeileis wrote:
> On Mon, 10 May 2010, RATIARISON Eric wrote:
>> I'm using maxlik with functions specified (L, his gradient & hessian).
>>
>> Now I would like determine some robust standard errors of my estimators.
>>
>> So I 'm try to use vcovHC, or hccm or robcov for e
Apologies.
Let me clarify. I have included my code below :
data.table.b represents the medical nomenclature, whereas data.table.a is a
patient derived database.
data.table.b$CATEGORY categorizes features (e.g. 'cardiac', 'respiratory'),
whereas data.table.b$FEATURE is a corresponding disease (e.g
To my knowledge, fixed and random effect models may be estimated for the
logit model and only the random effect model for the probit model
(because of the incidental parameter problem).
I think clogit in the survival package fits the model that is called the
fixed effect logit model in the econome
Hi, even after rereading, I have little of a clue what it is exactly that you
are trying to do. It'd help if you provided a more concise, step-by-step
description and/or the smallest unambiguous example of the two tables AND of
what should come out at the end. Also, unless for relatively trivial
p
There is the plm package for linear panel models. Further, since estimation
of fixed effects models rests on the within-subject or -object variance, the
R-squared of interest is typically the within R-squared, not the overall or
between R-squared. Read up about it before you use it though.
Daniel
Hi:
Perhaps this will do; the data were read into a data.frame object named xx.
# Melt the data - constructs a factor 'variable' whose levels are the
variable names
# and a numeric variable named 'value' containing, oddly enough, the numeric
values.
# Requires package reshape.
library(reshape)
l
Bimal,
in the "memisc" packages:
?panel.errbars
This might be a good option for you.
HTH,
Sam
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Who is working on it, using it etc. I use both.
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Hi there,
Maybe people who know both R and econometrics will be able to answer
my questions.
I want to run panel regressions in R with fixed-effect. I know two
ways to do it.
First, I can include factor(grouping_variable) in my regression equation.
Second, I plan to subtract group mean from my va
Hi,
I just bought new macbook pro 10.6.3. I am trying to use some old c code
that used to work. I tried to recompile the code. In the directory with
code I used R CMD SHLIB hello.c and get the error
make: Nothing to be done for `all'.
I have tried reinstalling Xcode and R but I am still hav
Rhelpers:
I'd like to modify this RSQLite statement:
rs_stations<-dbSendQuery(con_stations, "select * from stations")
so that stations is actually an R variable, e.g.:
stations=c("stationA","stationB")
How would I modify the above statement to query from stations[[1]]
(aka "stationA")?
--j
_
I would like to obtain a matrix
2010/5/8 Tal Galili :
> Here is one way for doing that:
> nombreL <- c("Alvaro Perez", "Sebastian García","Sebastian 2", "Luis Gomez",
> "Jorge Rial", "Ronaldo Apud", "Ana María Bianco")
> nombreC <- c("Alvaro", "Ana","Jorge","Ronaldo", "Sebastian")
> func1 <- funct
Dear r-help list members,
I am quite new to R, and hope to seek advice from you about a problem I have
been cracking my head over. Apologies if this seems like a simple problem.
I have essentially two tables. The first (Table A) is a standard patient
clinicopathological data table, where rows cor
Hi r-sers,
I have a data of relative frequencies for the interval of 0-20,
20-40,...380-400. I would like the two data on the same graph using the same
x-axis label. My question is how to get a smooth curve using kernel density
code if it possible for this data.
> cbind(rel_obs,rel_gen)
There is definitely a difference. The pcr program is Principal Component
Regression, but document says
"The formula argument should be a symbolic formula of the form response ~
terms, where response is the name of the response vector or matrix (for
multi-response models) and terms is the name of
You need to provide an 'str(yarn)' so we can see what the structure is. I
don't know what the 'pcr' program is expecting, but it looks like from what
you have provided that 'yarn' might be a dataframe and "X" is a vector.
Look at the documentation for pcr and see what it expects.
On Mon, May 10,
Jim - thanks.
It worked, however pcr program is still not accepting it there is a
sample yarn data loaded and I am comparing. My data looks like this
X1 X2 X3 X4 X5 X6 X7
X8 X9 X10 X11 X12 X13 X14 X15 X16 X17
1K
Is this what you want:
> x <- paste("X", 1:10, sep='')
> x
[1] "X1" "X2" "X3" "X4" "X5" "X6" "X7" "X8" "X9" "X10"
> sub("X", "X.", x)
[1] "X.1" "X.2" "X.3" "X.4" "X.5" "X.6" "X.7" "X.8" "X.9" "X.10"
>
On Mon, May 10, 2010 at 8:53 PM, Ravi Ramaswamy wrote:
> Hi - a newbie qu
Hi - a newbie question, if someone can please help
I want to change X1, X2,,.to X.1 X.2 etc in the names below. I am using
the Principal Component Regression function (pcr) and it seems to want it
this way
> datap3.pcr <- pcr(water ~ X, 10, data = datap3, Validation ="cv")
Error in mode
May I ask for further illumination of my dumbness?
I had been merrily plotting ..density.. and using xlim() to constrain the
x-axis like this and everything was working fine:
df<-data.frame(names=c("Bob","Mary","Joe","Bob","Bob"))
ggplot(df,aes(names,..density..,group=1))+geom_histogram()+xlim("B
On May 10, 2010, at 7:36 PM, John Rauser wrote:
I'm learning ggplot and am a little confused. Sometimes discrete
scales work
like I'd expect, and sometimes they don't. For example...
This works exactly like one would expect:
df<-data.frame(names=c("Bob","Mary","Joe","Bob","Bob"))
ggplot(df
I'm learning ggplot and am a little confused. Sometimes discrete scales work
like I'd expect, and sometimes they don't. For example...
This works exactly like one would expect:
df<-data.frame(names=c("Bob","Mary","Joe","Bob","Bob"))
ggplot(df,aes(names))+geom_histogram()
But this yields an erro
On May 10, 2010, at 4:54 PM, Laura S wrote:
Dear all:
I would like to create a landscape of environmental values that
follow a
uniform frequency distribution and also have spatial autocorrelation
in the
landscape.
Perhaps:
X <- distribution on one dimension
Y <- distribution on another
Perhaps you should be using the chron package. It has no time zones
in the first place.
On Mon, May 10, 2010 at 2:26 PM, Garrett Grolemund wrote:
> I'm searching for an r command that will notify me if I create a time that
> does not exist due to Daylight Savings Time. For example, if I run the
On Mon, 10 May 2010, RATIARISON Eric wrote:
Hi,
I'm using maxlik with functions specified (L, his gradient & hessian).
Now I would like determine some robust standard errors of my estimators.
So I 'm try to use vcovHC, or hccm or robcov for example
but in use one of them with my result of ma
On May 10, 2010, at 6:17 PM, Chaudhari, Bimal wrote:
Are there functions/packages which support plots (bar and/or line)
where
I provide the point estimate and some error measure rather than the
raw
data?
I often have to summarize/present data from multiple sources where the
original data i
On May 10, 2010, at 4:29 PM, vjaneiro wrote:
I'm trying to use cex.axis or something like this to augment the size
labels' on levelplot axis, but it doesn't work.
Could someone help me?
Look at the thread "Re: [R] Increasing the font size on axes in
trellis" from two days ago.
--
D
On 10/05/2010 6:13 PM, Chaudhari, Bimal wrote:
Is there a predict method/syntax which I could use to generate
predictions (and other output from predict() methods) if I have the
model parameter estimates from a training dataset but not the data used
to generate the original model (the models were
Are there functions/packages which support plots (bar and/or line) where
I provide the point estimate and some error measure rather than the raw
data?
I often have to summarize/present data from multiple sources where the
original data is unavailable and do not yet have a good solution for
this.
Is there a predict method/syntax which I could use to generate
predictions (and other output from predict() methods) if I have the
model parameter estimates from a training dataset but not the data used
to generate the original model (the models were generated by a
collaborator using STATA and for
I'm trying to use cex.axis or something like this to augment the size
labels' on levelplot axis, but it doesn't work.
Could someone help me?
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Hi,
I'm using maxlik with functions specified (L, his gradient & hessian).
Now I would like determine some robust standard errors of my estimators.
So I 'm try to use vcovHC, or hccm or robcov for example
but in use one of them with my result of maxlik, I've a the following
error message :
under Unix/X11 (and IIRC under Windows), edit(some.data.frame) will
invoke a smallish spreadsheet-like data editor, which will allow you to
move around your data with no fuss. I probably wouldn't use it for any
serious data entry, but found damn useful in a lot of data-debugging
situations (you kno
Elisabeth, question to you: How is it that you recognise that it is
"a logaritmic axis with the base of 10", as opposed to any other base?
Ted.
On 10-May-10 17:15:04, Elisabeth Bjerke Rastad wrote:
> Hello!
> Thank you for answering!
> What I am trying to do is to plot my raw values (biomass of di
I realized that after I hit the send button...I was looking at the wrong
dataset,Brr
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
>
>From: jim holtman
>To: Felipe Carrillo
>Cc: r-h...@stat.math.ethz.ch
>Sent: M
It is working correctly, You have NAs in your data, so you get NAs in the
cumsum.
On Mon, May 10, 2010 at 5:09 PM, Felipe Carrillo
wrote:
> Hi: Thanks to Dennis and Fernando for your help reordering the levels.
> Now I have a different issue:
> I am trying to get the cumulative weekly values usi
On May 10, 2010, at 4:46 PM, Ralf B wrote:
I installed the lattice package, and got an error that R was not able
to remove the previous version of lattice. Now my installation seems
to be currupt, even affecting other packages. I am getting this error
when loading TTR:
library(TTR)
Loading r
Hi: Thanks to Dennis and Fernando for your help reordering the levels.
Now I have a different issue:
I am trying to get the cumulative weekly values using cumsum and it
appears to output the wrong values. Here's my dataset:
winter <- structure(list(week = c(26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L
Perhaps even ?View would be useful here. I've never used R Commander,
so don't know it would be useful in that environment.
Michael H wrote:
R experts,
I am working with large multivariable data frames (> 50 variables)
and I would like to scroll horizontally across my output
to view each d
Dear all:
I would like to create a landscape of environmental values that follow a
uniform frequency distribution and also have spatial autocorrelation in the
landscape. I was wondering if there is an algorithm and/or package out there
that creates autocorrelation of values that are distributed ac
I installed the lattice package, and got an error that R was not able
to remove the previous version of lattice. Now my installation seems
to be currupt, even affecting other packages. I am getting this error
when loading TTR:
> library(TTR)
Loading required package: xts
Loading required package:
Hi Mike,
You can set options(width = x), where x equals the number of columns.
-Ista
On Monday 10 May 2010 16:06:54 Michael H wrote:
> R experts,
>
> I am working with large multivariable data frames (> 50 variables)
> and I would like to scroll horizontally across my output
> to view each data
Hi All
I need some help with plotting a step function, currently I'm using
sfun <- stepfun(c(1, 2, 5,10, 20), c(0, 11, 22, 33, 44, 0), f=0)
plot(sfun, pch=NA, main="", xlim=c(1,20))
which I working fine, but my data is in the following format:
Min Max Value
1 2 11
2 5 22
510 33
10 20 44
1. To
I tried writing that function before, but the expression() command is not
preserved when we incorporate it into a function(). The first example you gave
me, if run in R, produced a title that reads "title" instead of "stuff". It
seems that in this case the expression() can't read the character s
R experts,
I am working with large multivariable data frames (> 50 variables)
and I would like to scroll horizontally across my output
to view each data frame rather than having to scroll down vertically-
wrapped data frames.I have been using R Commander as a programming
interface. If I assign
I am writing to ask if R has a build- in function to calculate this
polylogarithm Li_n(z) function , also known as the Jonquière's function
defined as
Li_n(z)=sum_(k=1)^infty(z^k)/(k^n)
Thanks
Andy
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https://stat.et
Many thanks for this suggestion. Indeed, "labels" turned out to be a
factor, and after reordering the levels I got the plot I wanted.
Alexey
David Winsemius wrote:
On May 10, 2010, at 10:00 AM, Alexey Bessudnov wrote:
Dear all,
I'm trying to create a dot plot with error bands with
Dotplo
Hi,
I use R with a screen reader to do statistical and financial analysis.
There are a couple of things that your student can do.
R was created on Linux, and then ported to Windows. So she can use R under
Linux where the CLI is perfectly accessible.
There are 2 main screen readers for Linux:
Sigh; that's because I forget to include it in the 'substitute' list.
Here is a version that works (and has been tested...).
ourtitle <- function(title, footnotes=NULL) {
if (length(footnotes) > 0) {
fn <- paste(footnotes, collapse=' ')
title <- eval(substitute(expression(bold(paste(titl
When running DEMA(data, 5) on a vector 'data' of length 5, my R engine
stops. Is this function or the R environment facing a bug here or am I
doing something wrong? DEMA should work if the smoothing window size
is the same size as the the data length, right?
(I am working with Eclipse 3.5. and the
Hi,
Lattice and ggplot2 are both ideally suited for this task. Consider
this example,
library(ggplot2)
d = data.frame(x=1:10, a1=rnorm(10), b1=rnorm(10))
m = melt(d, id ="x") # reshape into long format
qplot(x, value, data=m, geom="path", colour=variable)
library(lattice)
xyplot(value~x, data=m
Lets say I have a generated data frame with variables that follow a
naming convention:
title,a1,a2,b1,b2,b3,c1,c2,c3,c4...
I am plotting every column (starting from a1) as a line in a plot.
That works. However my diagram becomes very unorganized. Creating
legends is nice, but trying out different
I think your problem is that HL07 is not a data frame but a list. read.spss()
produces a list rather than a data frame by default, and svydesign() requires a
data frame. You can use as.data.frame() to turn HL07 into a data.frame.
The weights also look a bit strange -- you have sampling prob
On May 10, 2010, at 2:55 PM, Sergio Andrés Estay Cabrera wrote:
Dear R users and specially Albyn and Giovanni,
thanks for your answers, but in fact I supposed the same at the
beginning of my problem. However, when I generate the data seldom I
obtain the expected correlation. For example us
You probably want to write your own titling function to add
superscripts. The following code will serve as a reasonable starting point:
# make the title expression
ourtitle <- function(title, footnotes=NULL) {
if (length(footnotes) > 0) {
fn <- paste(footnotes, collapse=' ')
title <- eva
Dear R users and specially Albyn and Giovanni,
thanks for your answers, but in fact I supposed the same at the
beginning of my problem. However, when I generate the data seldom I
obtain the expected correlation. For example using this code:
fz<-function(n,t,rho){
f<-NULL
for(i in 1:n){
s<-rmv
Dear Elisabeth,
log(X) shall return logarithm of X, with the base of e.
If you are not sure, you can specify e to be the base, e.g.:
log(X,exp(1))
.
Snow in Tromsø is melting, so will your problem.
King regards,
I'm searching for an r command that will notify me if I create a time that
does not exist due to Daylight Savings Time. For example, if I run the
following command on a windows machine
> ISOdatetime(2010,03,14,2,10,0, tz = "") # My system time is set to the
United States Central Time Zone
[1] NA
Dear R users,
I recently developed a plotting function in R and introduced it to my
coworkers. The function is designed to make plotting easier and more efficient,
which will in turn be more cost-effective for the company. The reviews for the
function have been positive thus far, except for one
Thank you Andy. I'm aware of that but my question is about the way the locfit
package use to automatically determine the bandwidth/smoothing parameter if
Gauss kernel density function is used as the weight function. Does the
generalized cross validation criteria is imposed here or does the valu
Hello:
I am using the ggplot2 package on R 2.10.1. I am plotting points using
geom_pointrange. Is there a way to overlay hashmarks on the points,
specifically the median and the min and max of the range?
Cheers,
Michael
[[alternative HTML version deleted]]
Sums of correlated increments have the same correlation as the original
variables...
library(mvtnorm)
X<- matrix(0,nrow=1000,ncol=2)
for(i in 1:1000){
Y <- rmvnorm(1000,mean=mu,sigma=S)
X[i,] <- apply(Y,2,sum)
}
cor(Y)
[,1] [,2]
[1,] 1.000 0.4909281
[2,] 0.4909281 1.0
Hello!
Thank you for answering!
What I am trying to do is to plot my raw values (biomass of different
species) on a logaritmic y-axis with the base of e. When I type "log="y"",
the axis transforms into a logaritmic axis with the base of 10.
Best regards,
Elisabeth
> Dear Elisabeth,
>
> I'm
Loader's book is the ultimate reference (other than the source code...). I
don't have it with me right now, so can't tell you for sure. However, it seems
like an odd thing to do to use a different kernel at the selection step than
the final estimation step. I should think if you specify the k
no one?
any pointers would really be greatly appreciated!
thanks,
kay
-
Kay Cichini
Postgraduate student
Institute of Botany
Univ. of Innsbruck
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(Near) non-identifiability (especially in nonlinear models, which include
linear mixed effects models, Bayesian hierarchical models, etc.) is
typically a strong clue; usually indicated by software complaints (e.g.
convergence failures, running up against iteration limits, etc.).
However this is s
Dear Elisabeth,
I'm not sure if I have understood your question -- are you trying
to use a different logarithmic base?
Kind regards,
Xianwen
On Sun, May 9, 2010 at 8:05 PM, Elis
R experts -
I'm using John Fox's sem package to analyze a simple path model (two correlated
predictor variables directly influencing a single criterion variable):
Predictor1 -> Criterion
Predictor2 -> Criterion
Predictor1 <-> Predictor2
I'm giving a presentation on this material next week, and
Hi,
I used
#--
# Análise do PESO -
#--
pesom1 = cbind(PESO[MÊS==1])
pesom3 = cbind(PESO[MÊS==3])
#- Grupo Placebo -
pesom1p = pesom1[GRUPO=='P']
pesom3p = pesom3[GRUPO=='P']
t.test(pesom1p, pesom3p, paired=T)
#- Grupo Cogumelo -
p
See the "kern" argument in ?locfit.raw.
Andy
From: FMH
> Hi,
>
> In the locfit package, could someone please let me know the
> automatic selection of smoothing parameter if Gauss kernel
> density function is used as weight function?
>
> thanks
> Fir
>
>
>
> [[alternative HTML
On May 10, 2010, at 10:00 AM, Alexey Bessudnov wrote:
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is
a character vector that contains labels.
I think matrix(nrow=0, ncol=0) will do it.
-Ista
On Mon, May 10, 2010 at 7:15 AM, anderson nuel wrote:
> Dear r-help,
>
> Could you help me to find the function which create an empty matrix.
>
> I use matrix(), but it gives *a single value that is NA and length of
> this
> matrix is 1.*
> **
>
On 10/05/2010 7:15 AM, anderson nuel wrote:
Dear r-help,
Could you help me to find the function which create an empty matrix.
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
Not sure what you mean by an empty matrix, but here's one interpretation:
Hi,
My first suggestion would be to supply a sample data (maybe using the
function dput) showing what you have, what you want to do, and what
you've tried (you say that subset() didn't work but we don't know how
you've typed it).
Then, we'll see!
Ivan
Le 5/10/2010 14:35, Silvano a écrit :
On 10.05.2010 13:15, anderson nuel wrote:
Dear r-help,
Could you help me to find the function which create an empty matrix.
Help is:
?matrix
Uwe Ligges
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
**
*Best Regards*
[[alternative HT
On 10.05.2010 14:35, Silvano wrote:
Hi,
I have a dataset with many variables and observations.
The variable Group has two levels: C and P,
the Month variable has four levels: 0, 1, 2 and 3.
I want to extract a subset of the variable Weight, considering only 1
and 3 levels for Months of the Gr
Hi:
This is more a statistical question than an R question (apologies!).
I have some income data as follows:
<$5000 : 598
$5000-$1 : 2586
$65001-$7 : 202
$70001+ : 446
I.e an open ended income class for incomes > $70k.
What would be the best way to estimate mean income?
Somethin
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is a
character vector that contains labels.
The problem is that labels are automatically sorted alp
Dear r-help,
Could you help me to find the function which create an empty matrix.
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
**
*Best Regards*
[[alternative HTML version deleted]]
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now. it is ok. I uninstall R2.11.0, then delete an packages in the library,
and install again R2.11.0. ok, it does works.
thank you!
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Sent from th
Hi,
I have a dataset with many variables and observations.
The variable Group has two levels: C and P,
the Month variable has four levels: 0, 1, 2 and 3.
I want to extract a subset of the variable Weight,
considering only 1 and 3 levels for Months of the Group
variable.
I tried the command s
Hi,
In the locfit package, could someone please let me know the automatic selection
of smoothing parameter if Gauss kernel density function is used as weight
function?
thanks
Fir
[[alternative HTML version deleted]]
__
R-help@r-proj
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is a
character vector that contains labels.
The problem is that labels are automatically sorted alp
The Mantel-Byar test is a simple modification to the logrank test. Can
R perform this test or any other similar test?
Raphael
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PLEASE do read the posting guide http://www.R
Steve,
The list r-sig-debian is a more appropriate forum for this question as it is
dedicated to R on Debian / Ubuntu.
On 10 May 2010 at 08:24, steve_fried...@nps.gov wrote:
|
| Hello,
|
| I've tried to install randomForest on a Ubuntu 8.04 Hardy Heron system.
|
| I've repeatedly rec'd the er
On 05/10/2010 12:32 AM, bbslover wrote:
many thanks . I can try to use test set with 100 samples.
anther question is that how can I rationally split my data to training set
and test set? (training set with 108 samples, and test set with 100 samples)
as I know, the test set should the same di
On 9 May 2010 15:25, Ky Mathews wrote:
> Hi,
>
>
>
> I want to customise the segments on an xyplot. Below is a simple example
> of what I'm trying to do...
>
>
>
> #Example dataset
>
> x <- c(-0.25, 0.25, 0.8)
>
> y <- c(-0.5, 0, 0.75)
>
> gp <- c("A", "I", "C")
>
> my.data <- cbind.data.frame
>> Would you please confirm that it is a bootstrap sample with
>> replacement?
Someone should note that the definition of a bootstrap sample is a
sample with replacement (usually of size n).
I've read quite a few papers where they claim to be using the
bootstrap. Upon further review (sometimes to
On 10/05/2010 7:43 AM, Assa Yeroslaviz wrote:
Hello everybody,
I'm trying to install a package I have built. This package contains three
scripts with various functions (S3 as well as S4 classes)
I run at first the package.skeleton command with:
> package.skeleton("affyAnalysis", namespace=TRUE
I don't know much about how permissions are managed under ubuntu, but
you can try a couple of things:
- Dirk had worked very hard at automating building of CRAN packages for
Debian (from which Ubuntu is based). I believe randomForest is among
one of those available. I'm not sure if you need to s
Hello,
I've tried to install randomForest on a Ubuntu 8.04 Hardy Heron system.
I've repeatedly rec'd the error:
> install.packages("randomForest", dependencies = TRUE)
ERROR: compiliation failed for package 'randomForest'
** Removing '/home/admuser/R/i486-pc-linux-gnu-library/2.6/randomForest
I've not seen the book myself, but Graham Williams (author of the rattle
package) has been working on a book that perhaps may fit your need.
http://datamining.togaware.com/survivor/index.html
Andy
From: Wensui Liu
>
> good question!
> if there is such a book, i'd also like to read as well.
>
>
See the "replace" argument in ?randomForest.
Andy
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri
> Liakhovitski
> Sent: Friday, May 07, 2010 12:21 PM
> To: r-help
> Subject: [R] randomForest: sampling with replacem
Hello everybody,
I'm trying to install a package I have built. This package contains three
scripts with various functions (S3 as well as S4 classes)
I run at first the package.skeleton command with:
> package.skeleton("affyAnalysis", namespace=TRUE, code_files =c("defS3.R",
"defS4.R", "qc.R"))
On 10/05/2010 7:36 AM, Ruben Garcia Berasategui wrote:
Dear all,
when trying to replicate John M. Quick's example for correlations between
multiple variables posted on:
http://rtutorialseries.blogspot.com/2009/11/r-tutorial-series-zero-order.html
with R 2.11.0 (GUI 1.33) using my MacBook Pro
Dear all,
when trying to replicate John M. Quick's example for correlations between
multiple variables posted on:
http://rtutorialseries.blogspot.com/2009/11/r-tutorial-series-zero-order.html
with R 2.11.0 (GUI 1.33) using my MacBook Pro with OX X 10.5.8 I got the
following error message
> d
When you call a function R passes a "promise" to it for each argument.
A promise consists of the unevaluated variable together with the
environment in which it should evaluate the variable when time comes
to evaluate it. Thus tmp[[1]] contains function(y) y and in the
environment of function(y) y
sayan dasgupta wrote:
Hey guys,
I have a doubt here , It is something simple I guess, what am I missing out
here ??
f <- function(y) function() y
tmp <- vector("list", 5)
for (i in 1:5) tmp[[i]] <- f(i)
tmp[[1]]() # returns 5;
z <- f(6)
tmp[[1]]() # still returns 5; it should return 6 "ideall
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