Hello,
I frequently have to export a large quantity of data from some
source (for example, a database, or a hand-written perl script) and then
read it into R. This occasionally takes a lot of time; I'm usually using
read.table("filename",comment.char="",quote="") to read the data once it
Dear R communities
May I seek your advices on how to change color the default in levelplot(), e.g.
from the default of pink and light blue, to e.g. red and green ?
The levelplot function has 1 of the arguments being panel (which is actually
panel.levelplot), but I am not sure where the commands
Hi all,
I'm using the mle.stepwise function in the wle package to run forward
selection. My statistical background tells me that the first variable
selected should be the one with the highest correlation with the response,
however that's not the case. The two highest correlations with the respons
On Nov 8, 2009, at 7:23 PM, rusers.sh wrote:
for (i in num) {
c_num<-c[c$b==num,]
write.csv(c_num,file="c:/c_num.csv")
}
Warning messages:
1: In c$b == num :
longer object length is not a multiple of shorter object length
This is because you're comparing column b to the entire vector of
Hi all,
I think the code to calculate the default value of the symbreaks
argument in the gplots::heatmap.2 function is wrong.
The documentation says symbreaks defaults to true if any negative
values are detected in the data passed in.
The relevant code in the parameter list of this functi
Dear Peng,
I'm tempted to try to get an entry in the fortunes package but will instead
try to answer your questions directly:
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Peng Yu
> Sent: November-08-09 7:41 PM
> To: r-h...@
Emmanuel Charpentier wrote:
Le dimanche 08 novembre 2009 à 17:07 -0200, Iuri Gavronski a écrit :
Hi,
I would like to fit Logit models for ordered data, such as those
suggested by Greene (2003), p. 736.
Does anyone suggests any package in R for that?
look up the polr function in package MASS
Have you considered using split?
-Ista
On Sun, Nov 8, 2009 at 7:23 PM, rusers.sh wrote:
> Hi Rusers,
> I hope to divide the original dataset into several subsets and output
> these multilple datasets. But errors appeared in my loops. See example.
> ##
> a<-c(1:10)
> b<-c(rep(1,3),rep(2,3),r
> x
Year Depth Biomass1 Biomass2
1 199910 14.3 14.7
2 199915 14.7 15.6
> require(reshape)
> melt(x, id=c('Year','Depth'))
Year Depth variable value
1 199910 Biomass1 14.3
2 199915 Biomass1 14.7
3 199910 Biomass2 14.7
4 199915 Biomass2 15.6
On Sun, N
Dear John,
I did read Section 9.1.2 and various other textbooks before posting my
questions. But each reference uses slightly different notations and
terminology. I get confused and would like a description that
summaries everything so that I don't have to refer to many different
resources. May I
Ah, I believe I constructed my *.csv wrong in that I only had 1 observation
within groups whereas I needed at least 2.
Originally I had:
Year Depth Biomass1 Biomass2
1999 10 14.3 14.7
1999 15 14.7 15.6
etc.
but I switched this to:
Year
Hi Rusers,
I hope to divide the original dataset into several subsets and output
these multilple datasets. But errors appeared in my loops. See example.
##
a<-c(1:10)
b<-c(rep(1,3),rep(2,3),rep(3,4))
c<-data.frame(a,b) #c is the example data
num<-c(unique(b))
# I hope to get the subsets c_1.
Dear Peng Yu,
Perhaps you're referring to my text, Applied Linear Regression Analysis and
Generalized Linear Models, since I seem to recall that you sent me a number
of questions about it. See Section 9.1.2 on linear contrasts for the answer
to your question.
I hope this helps,
John
---
exp(1)
#[1] 2.718282
On Nov 8, 2009, at 5:03 PM, Crab wrote:
How do you express "e" the base of the natural log in R.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
On 08/11/2009 5:03 PM, Crab wrote:
How do you express "e" the base of the natural log in R.
e <- exp(1)
__
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/postin
How do you express "e" the base of the natural log in R.
--
View this message in context:
http://old.nabble.com/Express-%22e%22-in-R--tp26258503p26258503.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing li
Normal distribution check within R can be done with functions available in
nortest package. This package consists of several normality tests. In order
to install package type install.packages('nortest'). Afterwards, you should
consider running ks.test() only if mu and sigma parameters are known (t
> Sys.Date()
[1] "2009-11-08"
> as.Date("05/29/1971", format = "%m/%d/%Y")
[1] "1971-05-29"
> as.numeric((Sys.Date() - as.Date("05/29/1971", format = "%m/%d/
%Y")) / 365.25)
[1] 38.44764
or perhaps more clearly:
EndDate <- Sys.Date()
StartDate <- as.Date("05/29/1971", format = "%m/%d/%Y")
If I've understood correctly, you have cell sizes of 1. This is not enough.
ANOVA compares within group variance to between group variance, and
your within group variances are zero.
You need more data, or to collapse some cells.
Jeremy
2009/11/8 znd :
>
> Hello, I'm new to R and have been f
I don't understand under what situation ordered factor rather than
unordered factor should be used. Could somebody give me some examples?
What are the implications of order vs. unordered factors? Could
somebody recommend a textbook to me?
__
R-help@r-pro
To clarify.
Lets turn a date into an age. Given 05/29/1971 in mm/dd/
format. What is the year difference between then and today?
This would be the "age" requested that starts 05/29/1971 as
one.
Thanks,
Jim
David Winsemius wrote:
>
> On Nov 8, 2009, at 3:11 PM, frenchcr wrote:
>
>>
>>
>> w
On Nov 8, 2009, at 3:11 PM, frenchcr wrote:
why do you use 365.25?
As opposed to what?
--
David
dates<-as.character(data[,"date_commissioned"]); # convert dates to
characters
#dates[1:10]
#[1] "19910101" "19860101" "19910101" "19860101" "19910101" "19910101"
"19910101" "19910101" "199
Hello, I'm new to R and have been following many guides including the two-way
anova (http://www.personality-project.org/r/r.anova.html). Using that
walkthrough including the supplied data I do str(data.ex2) and receive the
appropriate types of data as follows:
> str(data.ex2)
'data.frame': 16 o
Patrick Giraudoux a écrit :
Duncan Murdoch a écrit :
On 08/11/2009 12:07 PM, Patrick Giraudoux wrote:
Hi,
I am routinely compiling a package and since I have moved to R
2.10.0, it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function ca
On Sun, Nov 8, 2009 at 11:28 AM, Peter Dalgaard
wrote:
> Gabor Grothendieck wrote:
>>
>> On Sun, Nov 8, 2009 at 11:59 AM, Peng Yu wrote:
>>>
>>> On Sun, Nov 8, 2009 at 10:11 AM, Duncan Murdoch
>>> wrote:
On 08/11/2009 11:03 AM, Peng Yu wrote:
>
> I'm wondering which textbook di
Try this:
library(gsubfn)
demo("gsubfn-cut")
and note the strapply call that converts the levels from cut to a matrix.
On Sun, Nov 8, 2009 at 4:08 PM, jose romero wrote:
> Hello list:
>
> I am using "cut" and "table" to obtain a frequency table from a numeric
> sample vector. The idea is to c
jose romero wrote:
Hello list:
I am using "cut" and "table" to obtain a frequency table from a
numeric sample vector. The idea is to calculate mean and standard
deviation on grouped data. However, I can't extract the midpoints of the
class intervals, which seem to be strings treated as factors.
Hi,
Maybe something like this (inspired by ?cut),
cut2num <- function(f){
labs <- levels(f)
d <- data.frame(lower = as.numeric( sub("\\((.+),.*", "\\1", labs) ),
upper = as.numeric( sub("[^,]*,([^]]*)\\]", "\\1", labs) ))
d$midpoints <- rowMeans(d)
d
}
a <- c(1, 2, 3, 4
Hello list:
I am using "cut" and "table" to obtain a frequency table from a numeric sample
vector. The idea is to calculate mean and standard deviation on grouped data.
However, I can't extract the midpoints of the class intervals, which seem to be
strings treated as factors. How do i extrac
why do you use 365.25?
dates<-as.character(data[,"date_commissioned"]); # convert dates to
characters
#dates[1:10]
#[1] "19910101" "19860101" "19910101" "19860101" "19910101" "19910101"
"19910101" "19910101" "19910101" "19910101"
dateObs <- as.Date(dates,format="%Y%m%d")
#dateObs[1:10]
#[1] "1
it sure does thank you!
> will this work for you
>
> x <- c('19910101', '19950302', '20010502')
> today <- Sys.Date()
> x.date <- as.Date(x, format="%Y%m%d")
> round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
>
On Sun, Nov 8, 2009 at 2:44 PM, wrote:
> Hi Jim,
>
Thanks a lot, this works.
jim holtman wrote:
> First of all, allocate 'theta' to be the final size you need. Every
> time through your loop you are extending it by one, meaning you are
> spending a lot of time copying the data each time. Do something like:
>
> theta <- numeric(n)
>
> and then
Le dimanche 08 novembre 2009 à 17:07 -0200, Iuri Gavronski a écrit :
> Hi,
>
> I would like to fit Logit models for ordered data, such as those
> suggested by Greene (2003), p. 736.
>
> Does anyone suggests any package in R for that?
look up the polr function in package MASS (and read the releva
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jens Malmros
> Sent: Sunday, November 08, 2009 11:11 AM
> To: r-help@r-project.org
> Subject: [R] MCMC gradually slows down
>
> Hello,
>
> I have written a simple Metropolis-Has
Hi there,
I tried to fit a penalized spline for a continuous risk factor in
recurrent events data. I found that I can include both pspline and
frailty terms in coxph. So I use code like
fit1 <- coxph(Surv(start, end, event)~pspline(age, df=0) + male +
white +frailty(id), data.all)
It yie
As Jim has noted, if the dates you have below are an 'end date', you
need to define the time0 or start date for each to calculate the
intervals. On the other hand, are the dates you have below the start
dates and you need to calculate the time to today? In the latter case,
see ?Sys.Date to
strapply in the gsubfn package can do that. It applies the indicated
function, here just c, to the back references from the pattern match
and then simplifies the result using simplify. (If you omit simplify
here it would give a one element list like strsplit does.)
library(gsubfn)
pat <- "(.*?) (.
Try this:
Instead of:
theta = c()
use:
theta <- rep(NA, 50)
or however many iterations you want the algorithm to run.
Best,
--
Wolfgang Viechtbauerhttp://www.wvbauer.com/
Department of Methodology and StatisticsTel: +31 (0)43 388-2277
School for Public Health
First of all, allocate 'theta' to be the final size you need. Every
time through your loop you are extending it by one, meaning you are
spending a lot of time copying the data each time. Do something like:
theta <- numeric(n)
and then see how fast it works.
On Sun, Nov 8, 2009 at 2:11 PM, Jens
Is this what you want:
> x <- ' one two three '
> y <-
> sub(".*?([^[:space:]]+)[[:space:]]+([^[:space:]]+)[[:space:]]+([ehrt]{5}).*",
+ "\\1 \\2 \\3", x, perl=TRUE)
> unlist(strsplit(y, ' '))
[1] "one" "two" "three"
On Sun, Nov 8, 2009 at 1:51 PM, Hadley Wickham wrot
To answer my own question:
The package mvtBinaryEP was helpful!
Christian Lerch schrieb:
Dear All,
I try to simulate correlated binary data for a clinical research project.
Unfortunately, I do not come to grips with bindata().
Consider
corr<-data.frame(ID=as.factor(rep(c(1:10), each=5)),
What is the frame of reference to determine the age? Check out 'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr wrote:
>
> Ive got a big column of dates (also some fields dont have a date so they have
> NA instead),
> that i have converted into date format as so...
>
>
> dates<-as.character
Many thanks, Bill and Emmanuel!
Christian
Emmanuel Charpentier schrieb:
Le dimanche 08 novembre 2009 à 00:05 +0100, Christian Lerch a écrit :
Dear list members,
I try to simulate an incomplete block design in which every participants
receives 3 out of 4 possible treatment. The outcome in bina
Ive got a big column of dates (also some fields dont have a date so they
have NA instead),
that i have converted into date format as so...
dates<-as.character(data[,"date_commissioned"]); # converted dates to
characters
dates[1:10]
[1] "19910101" "19860101" "19910101" "19860101" "19910101" "199
I have two related variables, each with 16 points (x and Y). I am given
variance and the y-intercept. I know how to create a regression line and
find the residuals, but here is my problem. I have to make a loop that uses
the seq() function, so that it changes the slope value of the y=mx + B
equati
Hello,
I have written a simple Metropolis-Hastings MCMC algorithm for a
binomial parameter:
MHastings = function(n,p0,d){
theta = c()
theta[1] = p0
t =1
while(t<=n){
phi = log(theta[t]/(1-theta[t]))
phisim = phi + rnorm(1,0,d)
Hi,
I would like to fit Logit models for ordered data, such as those
suggested by Greene (2003), p. 736.
Does anyone suggests any package in R for that?
By the way, my dependent variable is ordinal and my independent
variables are ratio/intervalar.
Thanks,
Iuri.
Greene, W. H. Econometric Anal
Hi all,
Is there a tool in base R to extract matched expressions from a
regular expression? i.e. given the regular expression "(.*?) (.*?)
([ehtr]{5})" is there a way to extract the character vector c("one",
"two", "three") from the string "one two three" ?
Thanks,
Hadley
--
http://had.co.nz/
Thanks a lot for your explanation. That actually makes sense.
On Sat, Nov 7, 2009 at 3:54 PM, Duncan Murdoch wrote:
> On 07/11/2009 6:31 PM, Ivan wrote:
>
>> Well, the problem is that I want those console to be from the same session
>> (i.e., they share same objects etc.). I do not think multipl
Duncan Murdoch a écrit :
On 08/11/2009 12:07 PM, Patrick Giraudoux wrote:
Hi,
I am routinely compiling a package and since I have moved to R
2.10.0, it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function calls gpsbabel via the system.
Eric Fail wrote:
Dear list users
How is it possible to visualise both a linear trend line and a quadratic trend
line on a plot
of two variables?
Here my almost working exsample.
data(Duncan)
attach(Duncan)
plot(prestige ~ income)
abline(lm(prestige ~ income), col=2, lwd=2)
Now I
On 08/11/2009 12:07 PM, Patrick Giraudoux wrote:
Hi,
I am routinely compiling a package and since I have moved to R 2.10.0,
it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function calls gpsbabel via the system. The gpsbabel program must
Gabor Grothendieck wrote:
On Sun, Nov 8, 2009 at 11:59 AM, Peng Yu wrote:
On Sun, Nov 8, 2009 at 10:11 AM, Duncan Murdoch wrote:
On 08/11/2009 11:03 AM, Peng Yu wrote:
I'm wondering which textbook discussed the various contrast matrices
mentioned in the help page of 'contr.helmert'. Could so
Dear list users
How is it possible to visualise both a linear trend line and a quadratic trend
line on a plot
of two variables?
Here my almost working exsample.
data(Duncan)
attach(Duncan)
plot(prestige ~ income)
abline(lm(prestige ~ income), col=2, lwd=2)
Now I would like to add
On Sun, Nov 8, 2009 at 11:59 AM, Peng Yu wrote:
> On Sun, Nov 8, 2009 at 10:11 AM, Duncan Murdoch wrote:
>> On 08/11/2009 11:03 AM, Peng Yu wrote:
>>>
>>> I'm wondering which textbook discussed the various contrast matrices
>>> mentioned in the help page of 'contr.helmert'. Could somebody let me
With xx as your sample data will this work? See ?addmargins
jj <- table(xx)
addmargins(jj, 2)
# or for both margins
addmargins(jj, c(1,2))
or
apply(jj, 1, sum)
--- On Sun, 11/8/09, sylvain willart wrote:
> From: sylvain willart
> Subject: [R] Counting non-empty levels of a factor
> To:
Hi,
I am routinely compiling a package and since I have moved to R 2.10.0,
it troncates some section texts in the doc:
With the following section in the rd file:
\details{
The function calls gpsbabel via the system. The gpsbabel program must
be present and on the user's PATH for the function
On Sun, Nov 8, 2009 at 10:11 AM, Duncan Murdoch wrote:
> On 08/11/2009 11:03 AM, Peng Yu wrote:
>>
>> I'm wondering which textbook discussed the various contrast matrices
>> mentioned in the help page of 'contr.helmert'. Could somebody let me
>> know?
>
> Doesn't the reference on that page discuss
On Sun, Nov 08, 2009 at 10:31:12AM -0600, Dirk Eddelbuettel wrote:
>
> On 8 November 2009 at 17:05, Lorenzo Isella wrote:
> | I am experiencing some really strange problems in updating my system
> | whenever I have both the R repository and the multimedia repository
> | available.
> [...]
> | deb
I'm not quite sure I understood the second queston but does this work?
subset(temp, xx$v2==-9)
subset(temp, xx$v2!= -9)
--- On Sun, 11/8/09, Ashta wrote:
> From: Ashta
> Subject: [R] look up and Missing
> To: r-h...@stat.math.ethz.ch
> Received: Sunday, November 8, 2009, 10:23 AM
> HI R-User
On Nov 8, 2009, at 11:08 AM, David Winsemius wrote:
On Nov 8, 2009, at 10:23 AM, Ashta wrote:
HI R-Users
Assume that I have a data frame 'temp' with several variables
(v1,v2,v3,v4,v5.).
v1 v2 v3 v4 v5
1 2 3 36
5 2 420
2 -9 5 43
6 2 1 34
1, I wa
Dear Ashta,
Is this what you want?
x <- read.table(textConnection("v1 v2 v3 v4 v5
1 2 3 36
5 2 420
2 -9 5 43
6 2 1 34"), header = TRUE)
closeAllConnections()
x
# Option 1
x1 <- x # copy of x just for this example
x1$v2[which(x1$v2 == -9)] <- NA
x
try this:
temp.new <- temp[temp$v2 != -9, ]
temp.new
I hope it helps.
Best,
Dimitris
Ashta wrote:
HI R-Users
Assume that I have a data frame 'temp' with several variables (v1,v2,v3,v4,v5.).
v1 v2 v3 v4 v5
1 2 3 36
5 2 420
2 -9 5 43
6 2 1 3
On Nov 8, 2009, at 10:23 AM, Ashta wrote:
HI R-Users
Assume that I have a data frame 'temp' with several variables
(v1,v2,v3,v4,v5.).
v1 v2 v3 v4 v5
1 2 3 36
5 2 420
2 -9 5 43
6 2 1 34
1, I want to look at the entire row values of when v2 =-9
On 8 November 2009 at 17:05, Lorenzo Isella wrote:
| I am experiencing some really strange problems in updating my system
| whenever I have both the R repository and the multimedia repository
| available.
[...]
| deb http://cran.ch.r-project.org/bin/linux/debian lenny-cran/
This URLs doesn't exis
On Nov 8, 2009, at 11:03 AM, Peng Yu wrote:
I'm wondering which textbook discussed the various contrast matrices
mentioned in the help page of 'contr.helmert'. Could somebody let me
know?
My version of "Modern Applied Statistics in S" (aka MASS) deals with
it in enough detail for a person
Here is how to find out which rows contain -9 and then you can do with
it as you please:
> x
v1 v2 v3 v4 v5
1 1 2 3 3 6
2 5 2 4 2 0
3 2 -9 5 4 3
4 6 2 1 3 4
> which(apply(x, 1, function(.row) any(.row == -9)))
[1] 3
>
>
On Sun, Nov 8, 2009 at 10:23 AM, Ashta wrote:
> HI R-
Mangalani Peter Makananisa would like to recall the message, "Scheffe test".
Please Note: This email and its contents are subject to our email legal notice
which can be viewed at http://www.sars.gov.za/Email_Disclaimer.pdf
[[alternative HTML version deleted]]
__
Dear Prof,
Please help me with the R code which compute SCHEFFE TEST
Thanking you in advance
Kind regards
Mangalani Peter Makananisa
Statistical Analyst
South African Revenue Service (SARS)
Segmentation and Research : Data Modelling
Tel: +2712 422 7357
Cell: +2782 456 4669
Dear all,
Please help me with the R code which compute SCHEFFE TEST
Thanking you in advance
Kind regards
Mangalani Peter Makananisa
Statistical Analyst
South African Revenue Service (SARS)
Segmentation and Research : Data Modelling
Tel: +2712 422 7357
Cell: +2782 456 4669
F
Dear all,
Please help me with the R code which compute SCHEFFE TEST
Thanking you in advance
Kind regards
Mangalani Peter Makananisa
Statistical Analyst
South African Revenue Service (SARS)
Segmentation and Research : Data Modelling
Tel: +2712 422 7357
Cell: +2782 456 4669
F
Dear All,
I have just installed a fresh Debian testing (squeeze) on my system
(amd64 architecture).
I am experiencing some really strange problems in updating my system
whenever I have both the R repository and the multimedia repository
available.
This is my source.list (when I disable the multimed
I'm wondering which textbook discussed the various contrast matrices
mentioned in the help page of 'contr.helmert'. Could somebody let me
know?
BTW, in R version 2.9.1, there is a typo on the help page of
'contr.helmert' ('cont.helmert' should be 'contr.helmert').
Thanks a lot for those solutions,
Both are working great, and they do slightly different (but both very
interesting) things,
Moreover, I learned about the length() function ... one more to add to
my personal cheat sheet
King Regards
2009/11/8 David Winsemius :
>
> On Nov 8, 2009, at 9:11 AM, David
Hi Henrik,
I am using your saveObject/loadObject to handle over 1000 matrices. It
worked beautifully. Because I need to load those matrices often for
evaluating a few functions on them and those matrices do not fit all in
memory at once, is there a way to speed up the loading part? I tried save
al
my code is not right below:
rm(list=ls())
#define data.frame
a=c(1,2,3,5,6); b=c(1,2,3,4,7); c=c(1,2,3,4,8); d=c(1,2,3,5,1);
e=c(1,2,3,5,7)
data.f=data.frame(a,b,c,d,e)
#backup data.f
origin.data<-data.f
#get correlation matrix
cor.matrix<-cor(origin.data)
#backup corre
Aha, what is that blog post and what does not work for you?
I haven't got any report so far and do not have Windows 7 easily
available yet.
Best,
Uwe Ligges
Peter Flom wrote:
Good morning
I just got a new computer with Windows 7. R works fine, but the editor I am used to
using "RWinEdt"
HI R-Users
Assume that I have a data frame 'temp' with several variables (v1,v2,v3,v4,v5.).
v1 v2 v3 v4 v5
1 2 3 36
5 2 420
2 -9 5 43
6 2 1 34
1, I want to look at the entire row values of when v2 =-9
like
2 -9 5 43
I wrote
Good morning
I just got a new computer with Windows 7. R works fine, but the editor I am
used to using "RWinEdt" does not. I did find one blog post on how to get
RWinEdt to work in Windows 7, but I could not get those instructions to work
either.
Is there a patch for RWinEdt?
If not, is the
On Nov 8, 2009, at 9:11 AM, David Winsemius wrote:
On Nov 8, 2009, at 8:38 AM, sylvain willart wrote:
Hi everyone,
I'm struggling with a little problem for a while, and I'm wondering
if
anyone could help...
I have a dataset (from retailing industry) that indicates which
brands
are pr
Dan, with base graphics, the> plot command determines the ylim, >lines can
only add lines/points to an exisiting graph, it cannot modify the existing
ylim.
You have two choices. 1) Before proceeding to the graph, determine which of
the two data sets has the greater range and call >plot with that
On Nov 8, 2009, at 8:38 AM, sylvain willart wrote:
Hi everyone,
I'm struggling with a little problem for a while, and I'm wondering if
anyone could help...
I have a dataset (from retailing industry) that indicates which brands
are present in a panel of 500 stores,
store , brand
1 , B1
1 , B2
Not sure what you mean.
yi <- c(2,3,2,4,3,6)
xi <- c(1,4,3,2,4,5)
res <- lm(yi ~ xi)
hatvalues(res)
X <- cbind(1, xi)
diag( X%*%solve(t(X)%*%X)%*%t(X) )
Same result.
Best,
--
Wolfgang Viechtbauerhttp://www.wvbauer.com/
Department of Methodology and StatisticsTel: +
Hi everyone,
I'm struggling with a little problem for a while, and I'm wondering if
anyone could help...
I have a dataset (from retailing industry) that indicates which brands
are present in a panel of 500 stores,
store , brand
1 , B1
1 , B2
1 , B3
2 , B1
2 , B3
3 , B2
3 , B3
3 , B4
I would lik
Sigmund Freud wrote:
Hello:
I am trying to understand the method 'hatvalues(...)', which returns something similar to the diagonals of the plain vanilla hat matrix [X(X'X)^(-1)X'], but not quite.
A Fortran programmer I am not, but tracing through the code it looks like perhaps some sort of co
Le dimanche 08 novembre 2009 à 00:05 +0100, Christian Lerch a écrit :
> Dear list members,
>
> I try to simulate an incomplete block design in which every participants
> receives 3 out of 4 possible treatment. The outcome in binary.
>
> Assigning a binary outcome to the BIB or PBIB dataset of th
Hi,
Why, when I run the script below, is my y-axis range command being ignored?
I.e., the y axis graphed consistently fits the line specified in the plot
command, but never fits the line I'm trying to add in the subsequent line
command.
This is my first R script, so if I'm doing anything else wac
ok,I understand your means, maybe PLS is better for my aim. but I have done
that, also bad. the most questions for me is how to select less variables
from the independent to fit dependent. GA maybe is good way, but I do not
learn it well.
Ben Bolker wrote:
>
> bbslover yeah.net> writes:
>
>>
Thanks Paul.
I'm still struggeling with some beginners issues on the ps-import
(windows troubles with installing ghostscript), but when I resolved
them I'm sure that I can use your example code which loos great to me.
Thanks a lot,
Thomas
2009/11/4 Paul Murrell :
> Hi
>
>
> Thomas Steiner wrote:
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