Oops. Clicked "Send" too fast (don't shoot, Brian !). I forgot :
> sessionInfo()
R version 2.8.1 (2008-12-22)
i486-pc-linux-gnu
locale:
LC_CTYPE=fr_FR.UTF-8;LC_NUMERIC=C;LC_TIME=fr_FR.UTF-8;LC_COLLATE=fr_FR.UTF-8;
LC_MONETARY=C;LC_MESSAGES=fr_FR.UTF-8;LC_PAPER=fr_FR.UTF-8;LC_NAME=C;
LC_ADDRESS=
If you want to know how to calculate the CI, you can study the code of
this function. You can access the code by nlme:::intervals.lme.
Best
2009/3/17 WONG, Ka Yau :
> No standard deviation of variable in the random effect is
> provided. How to construct CI?
>
> Best Regards,
> Tommy
>
This is a surprisingly interesting problem that took a while to debug, because
the computations all seemed correct.
Your model hasn't converged yet. You can get the right answer either by
running longer:
summary(m1big_longer)
Large data regression model: bigglm(y ~ ttment, data = dat, fami
On Mon, 16 Mar 2009, David Winsemius wrote:
Use an offset with a non-intercept model?
Yes
dat$ones<-1
glm(y~0+x+offset(ones), data=dat, family=whatever)
I wonder if there could be problems depending on the choice of family.
Only in the sense that if the model fits badly it might run up a
try:
apply(data[,c(6:42)], 2, max)
(as stephen sefick Just mentioned...)
On Tue, Mar 17, 2009 at 2:56 AM, timbean wrote:
>
> I'd like to perform some calculations across a series of columns. Is there
> an
> easy way to do so? For example, I'd like to calculate the maximum value
> from
> c
Romain Francois wrote:
> Duncan Murdoch wrote:
>
> That is an issue. I guess I will fall back on what the parser says and
> infer on the scoping. Within the lines below, mean would be different
> each time
>
> mean( 1:10 )
> lapply( 1:10, mean)
> mean <- (1+4) / 2
> lapply( list( mean, median), fun
> "PaCo" == p connolly
> on Tue, 17 Mar 2009 10:42:49 +1300 writes:
PaCo> Quoting Mohan Singh :
>> Hii
>>
>> I am trying to plot lines at (0,0) axis where my scatterplot goes
>> between positive n negative values for x and y axis
>>
>> i can plot point us
Wacek Kusnierczyk wrote:
Romain Francois wrote:
Duncan Murdoch wrote:
That is an issue. I guess I will fall back on what the parser says and
infer on the scoping. Within the lines below, mean would be different
each time
mean( 1:10 )
lapply( 1:10, mean)
mean <- (1+4) / 2
lapply( list( mean,
rkevinbur...@charter.net wrote:
> I am having a hard time understanding just what 'sweep' does. The
> documentation states:
>
> Return an array obtained from an input array by sweeping out a summary
> statistic.
>
> So what does it mean "weeping out a summary statistic"?
>
from both the text
Hello,
I do loggin as root, and I tried three methods listed in INSTALL file in
RSQLite, but still did not work. The whole error list is:
** libs
WARNING: R include directory is empty -- perhaps need to install R-devel.rpm
or similar
gcc -std=gnu99 -I/usr/lib/R/include -DRSQLITE_USE_BUNDLED_SQLIT
Romain Francois wrote:
> Wacek Kusnierczyk wrote:
>>
>>
>>> Duncan Murdoch wrote:
>>>
>>>
So if you've got high standards, it's probably quite hard. On the
other hand, if you're willing to accept the usual sort of errors that
syntax highlighters make, it's not so bad, but not triv
On Tue, 2009-03-17 at 12:12 +0900, Ei-ji Nakama wrote:
> Hi
>
> > I have seen a lot of problems from people trying to compile R with
> > MKL. So I am writing my experience in case it helps and to ask one
> > question. I installed R-2.8.1.patched in Ubuntu 9.04 (gcc 4.3.3) using
> > MKL 10.1.1.019.
Hi, All.
I have a data frame with the part as :
.
1422006-02-22 16:28:18 useractivity_act 8 0
1432006-02-22 16:28:26 42 0
1442006-02-22 16:28:28 33 0
1452006-02-22 16:28:31
On Mon, 2009-03-16 at 18:43 -0400, Stavros Macrakis wrote:
> Greg,
>
> I'm not bringing this up because I am bloody-minded, but because I
> think it's a serious issue with the R libraries -- there is no obvious
^ packages
> way to find relevant func
On Mon, 2009-03-16 at 18:33 -0400, Cristián wrote:
> Hi,
>
> I have seen a lot of problems from people trying to compile R with
> MKL. So I am writing my experience in case it helps and to ask one
> question. I installed R-2.8.1.patched in Ubuntu 9.04 (gcc 4.3.3) using
> MKL 10.1.1.019.
The R-adm
Hi, All.
I have a data frame with the part as :
.
1422006-02-22 16:28:18 useractivity_act 8 0
1432006-02-22 16:28:26 42 0
1442006-02-22 16:28:28 33 0
1452006-02-22 16:28:31
Thanks for your reply,
I couldn't use the method what the example gives. Because Data frame is quite
big, and it has so many rows which are repeated. I am not pretty sure about the
range of 1000. It turns out I should merge those row which repeated.
Tammy
> From: simon.pick...@bto.org
> To: me
Hi,
I guess I have a naive question. I use kde2d function in a standard way to
estimate kernel densities of x and y (x and y are vectors) and plot them
using image().
f1=kde2d(x,y)
image(f1)
But what if I want to see kernel estimates of three variables, x, y and z (a
vector) plotted together ? So
Hi,
I'm pretty sure you dont mean "merge" since merge() creates a new big data
frame from two small ones. It appears that you want to omit some rows.
As I said before, get used to selecting rows like this
data.frame<-data.frame[data.frame$conditional.rows>1000,]
(for example)
Although I don
Is this what you want it to do ('n' is the original data frame)?
> n
v V1 V2 V3 V4
1 k a 2 1 2
2 l b 1 2 1
3 m c 2 1 2
4 n d 1 2 1
5 o e 2 1 2
> n1 <- n[,3:5]
> n2 <- duplicated(n1)
> n3 <- n[!n2,]
> n3
v V1 V2 V3 V4
1 k a 2 1 2
2 l b 1 2 1
Pankaj
Tammy Ma wrote:
Hello,
I have a general help question that some of you might be able able to help.
I would like calculate sample size for a study (a two group comparison) based
on the outcome of reduction on (recurrent) events (say hospital admissions).
In a previous study, hospital admission rate of 140 a
Sorry, I didn't sent it twice, but there is some problem on my emailSorry
for this.
Tammy
> From: metal_lical...@live.com
> To: r-help@r-project.org
> Date: Tue, 17 Mar 2009 11:49:10 +0200
> Subject: [R] Merging
>
>
> Hi, All.
>
>
> I have a data frame with the part as :
>
> .
> 1
On Mar 17, 2009, at 4:59 AM, Wacek Kusnierczyk wrote:
rkevinbur...@charter.net wrote:
I am having a hard time understanding just what 'sweep' does. The
documentation states:
Return an array obtained from an input array by sweeping out a
summary statistic.
So what does it mean "weeping o
On 17/03/2009 6:44 AM, Josh wrote:
Hi,
I guess I have a naive question. I use kde2d function in a standard way to
estimate kernel densities of x and y (x and y are vectors) and plot them
using image().
f1=kde2d(x,y)
image(f1)
But what if I want to see kernel estimates of three variables, x, y a
Vijaykumar Muley wrote:
>
> Dear all,
>
> Myself Vijaykumar Muley working as senior research fellow. By training I
> am
> a computational biologist with not a strong knowledge of statistics. I
> have
> done some analysis which is explained as follows,
>
> I have 10340 (X) profiles of binary v
My aim is not only relect unduplicated column, I also want merge the last
column beside selection
Thanks.
Tammy
> Date: Tue, 17 Mar 2009 06:59:20 -0400
> From: pcho...@ncsu.edu
> To: metal_lical...@live.com
> CC: r-help@r-project.org
> Subject: Re: [R] Merging
>
> Is this what you want it t
I am assuming that for each species and station_id that you want
the value in that temp record whose date/time is closest to the
date/time in the species record along with the identifying information
(species, station_id) and date/time of the species. That interpretation
does give the same answer
David Winsemius wrote:
>
>>A <- array(1:16, dim = c(4, 4))
>># sweep by * 1:4, with recycling
>>sweep(A, 1, 1:4, '*')
>>
>> be careful to note that here '1' means that the operation is performed
>> *columnwise*, unlike in the case of apply, where '1' means *rowwise*:
>
> The sweep opera
Is this what you want it to do? (aDF is the original data frame)
> aDF
V1V2 V3 V4
1 2006-02-22 16:28:18 useractivity_act 8 0
2 2006-02-22 16:28:26 4 2 0
3 2006-02-22 16:28:28 3 3 0
4 2006-02-22 16:28:31
Dear R user,
I am looking for a code on double cross validation in
LASSO , one for optimizing the parameter and other one is for MSEP. If any
one have it, please foroward to me. I am using different package like LARS,
chemometric etc.
Thanks in advance
Alex
[[altern
Does anyone know the solution`?
Thanks,
Tammy
> From: metal_lical...@live.com
> To: r-help@r-project.org
> Date: Tue, 17 Mar 2009 11:49:10 +0200
> Subject: [R] Merging
>
>
> Hi, All.
>
>
> I have a data frame with the part as :
>
> .
> 1422006-02-22 16:28:18 useractivity_act
# How do I make from a SpatialGridDataframe a normal data frame? If I do
this in package gstat:
library(gstat)
g.dummy <- gstat(formula = z~1, locations = ~x+y, dummy = TRUE, beta = 0,
model = vgm(1,"Exp",15), nmax = 20)
set.seed(1)
iso <- predict(g.dummy, newdata = xy, nsim = 10)
gridded(iso) =
I have two numerical vectors; hun2 is a subset of hun1:
> hun1
[1] 147 198 141 170 184 136 150 134 144 161 124 103 107 110 121 79 106
51
[19] 102 109 52 71 52 80 95 78 75 96 101 107 107 80 71 49 71
64
[37] 58 68 75 85 55 52 70 90 98 90 100 98 74 78 60 69 106
103
Hello,
I have a general help question that some of you might be able able to help.
I would like to design a study (a two group comparison) based on reduction on
events(say hospital admissions). In a previous study hospital admission rate of
140 admissions per 72 patients (over a 4 month period)
Alex,
> I am looking for a code on double cross validation in
> LASSO , one for optimizing the parameter and other one is for MSEP. If any
> one have it, please foroward to me. I am using different package like LARS,
> chemometric etc.
It won't do double cross-validation, but the caret package
In thinking about this some more an SQL solution would be a
bit easier. Try this where species and temp are from
the prior post (i.e. after processing with chron).
The first statement finds the minimum distances for
each species and station_id combination. The
second finds all combinations of a
I still couldn't solve itany help would be preciated!
Tammy
> From: metal_lical...@live.com
> To: r-help@r-project.org
> Date: Tue, 17 Mar 2009 11:49:10 +0200
> Subject: [R] Merging
>
>
> Hi, All.
>
>
> I have a data frame with the part as :
>
> .
> 1422006-02-22 16:28
Hi Lars,
The approach suggested by Paul Smith is one way to try and solve your problem.
Another aproach is to try and eliminate, if possible, one of the variables
using the equality constraint, so that the problem now become unconstrained in
(p-1) dimensions (where p is the original dimension
Tammy,
You won't get more help if you post this again every half an hour... I don't quite understand what you exactly want to do, but it seemed at
least to me that Keith's suggestion (see again below) pretty much does what you want. If not, could you please specify *EXACTLY* what you
want to do
Dear R list members,
I'd like to make a graph of coefficients of the intercept, variable 1, and
variable 2 (and possibly the interaction between variable 1 and variable
2). When I use the lmList function as attached below, it shows a nice
coefficient graph.
> PACRP.lis <- lmList(PAffect ~ CRPC +
Dear Francisco,
I was able to duplicate the problem that you reported, and in addition
discovered that the problem seems to be peculiar to the poisson family.
lm(y~ttment, data=dat) and biglm(y~ttment, data=dat) produce identical
results, as do glm(y~ttment, data=dat) and bigglm(y~ttment, data=dat
Thanks.
Actually, I have been trying many time using method from KJ, There is always
the following information coming up or Rgui exit without reason.. I have no
idea about this?
gh<-with(act_data12, aggregate(V4, by=list(V1,V2,V3), sum))
Error in unlist(y, recursive = FALSE) :
promise alrea
R-help,
I have an image plot and I wish to set a legend to it
but I don't have an idea how to do it. Below it's the code line.
image(yyy, col = rev(heat.colors(10)), axes = T)
Thanks in advance.
__
R-help@r-project.org mailing list
https://stat.ethz.c
Looks as if you copied my code without modifying to suit your data frame.
My data frame (aDF) had column names V1, V2, V3, V4
> str(aDF)
'data.frame': 12 obs. of 4 variables:
$ V1: chr "2006-02-22 16:28:18" "2006-02-22 16:28:26" "2006-02-22
16:28:28" "2006-02-22 16:28:31" ...
$ V2: Factor w
I have the following data frames, avGain and retGain. They have the same
dimensions.
The following line of code replaces row j of avGain as desired:
avGain[j, ] <- mean( retGain[jStart:j, ] )
However, the following line does not work:
avGain[j, ] <- ( avGain[j-1, ] * ( DAYS - 1 ) +
R-help,
I'm sorry that I said " I don't have idea ".
What I mean t is that I just don't know how to set a legend
when "image" is called.
Thanks in advance
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
I guess no reply means there is not an existing package to produce the plot?
I will post the results of my script to hopefully help others who are trying to
formulate the same plot.
Thanks again.
--- On Mon, 3/16/09, Jason Rupert wrote:
> From: Jason Rupert
> Subject: [R] R package to aut
Luis, take a look at the ?legend function, it can produce almost any kind of
legend you may need!
Hope It helps.
Rodrigo.
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de Luis Ridao Cruz
Enviada em: terça-feira, 17 de março de 2009 11:
In case it helps, this refers to the columns by position rather than name...
> aggregate(aDF[,4], by=aDF[,-4], sum)
It sums the fourth column (I'm guessing this is what Tammy meant by "merge")
for unique combinations of all columns except the fourth.
It doesn't need the column names inside the ag
You can use something like this to make sure that h2 is a subset of h1
in consecutive order:
> h1 <- 1:50
> h2 <- 25:30
> hm <- match(h2, h1)
> all(diff(hm) == 1)
[1] TRUE
>
On Tue, Mar 17, 2009 at 6:42 AM, emj83 wrote:
>
> I have two numerical vectors; hun2 is a subset of hun1:
>
>> hun1
> [
I would like to build a package for the HT HG-U133+ PM arrays from affy,
but I can't find any good documentation on how to go about it. Naively
using makecdfenv's make.cdf.package() causes R to seg-fault.
I'm unfamiliar with the CDF format as such, but I'm guessing that it's
changed somewhat be
My data frame is:
> str(act_data12)
'data.frame': 1687 obs. of 5 variables:
$ Date: chr "2006-02-22" "2006-02-22" "2006-02-22" "2006-02-22" ...
$ Dtime : chr "14:36:10" "14:36:11" "14:36:12" "14:37:38" ...
$ FocusApp_seq: chr "useractivity_act" "4" "3" "0" ...
$ App_duratio
Here is what I have so far:
> test_data<-rnorm(100)
> par(mfrow=c(1,3)) # 3 rows by 1 columns layout of plots
> hist(test_data)
> boxplot(test_data)
> qqnorm(test_data)
I noticed that I can rotate a boxplot via "horizontal", but apparently "hist"
does not have that functionality.
I tried stack
I find Sweave very useful and I was trying to combine it with the
latex package fancyvrb. I was trying to get line numbering and labels
in order to reference the lines where particular commands occur.
Unfortunately, I haven't been able to figure out how to do it. Maybe
somebody can help me.
The
I believe that hist will return a vector that could be passed to
barplot:
h.islands <- hist(islands)
> barplot(h.islands$intensities, horiz=TRUE) # or
> barplot(h.islands$counts, horiz=TRUE)
David Winsemius
On Mar 17, 2009, at 11:38 AM, Jason Rupert wrote:
Here is what I have so far:
Does anyone have a sample file with an example/template for Rdoc (as
implemented in R.oo)? The command I am using is Rdoc$compile(), but I
am having difficulties figuring out what the Macros look like.
Apparently, in the headers, I see that I should use one of:
# @RdocClass className
# @RdocMeth
Dear experts,
I use R to conduct multilevel modeling. However, I have a problem
about the interpretation of random effect. Unlike the variables in fixed
effects, the variables in random effects have not shown the standard error
(s.e.) and p-value, so I don't know whether the
UNIX only (At least it works under Linux and Mac OS X):
(user <- system("users", intern=T))
Paolo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide
Dear all,
I read a file with all numbers with readLines function, as below,
> f <- file("data.txt")
> a <- readLines(f)
but all the values in a are in format "", and I cannot do the
calculation with them since they are not numeric. I wonder how should I skip
those quotes, thank you for help!
not off the top of my head, but if you provide a reproducible example
(cut and paste into R) then I may be able to help you.
On Tue, Mar 17, 2009 at 10:44 AM, Bill Cunliffe wrote:
> I have the following data frames, avGain and retGain. They have the same
> dimensions.
>
>
>
> The following line
That's by intent, by the way. The standard errors of the variance
components are only useful if the distribution is symmetric, and this is
not always true. If you were using lmer, and not lme, then you could use
the mcmcsamp function to look at the distribution of the random effects
to see if it is
Your first example that works would just repeat the atomic vector to the
length of the row. To replace a row with another data frame, you could use
rbind:
avGain <- rbind(avGain[-j,],b)
Or if the positioning is important,
2) avGain <- rbind(avGain[1:(j-1),],b,avGain[(j+1):nrow(avGain),])
I didn
I don't know if this will help, but look at the R graph gallery.
There may be something there, with code, that will work.
On Tue, Mar 17, 2009 at 10:39 AM, Jason Rupert wrote:
>
> I guess no reply means there is not an existing package to produce the plot?
>
> I will post the results of my script
Jason,
be carefully to the order of intensities (counts or densities):
x=rnorm(1000)
par(mfrow=c(2,2))
h=hist(x,breaks=bk<-c(-5,-3,-2,-1,-.5,0,1,3,5))
barplot(rev(h$intensities),rev(bk[2:9]-bk[1:8]),space=0,horiz=T) # compare to
axis(2)
barplot(h$intensities,bk[2:9]-bk[1:8],space=0,horiz=T)
axis(2
try this:
create a file called mySweave.sty, containing the following code:
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{mySweave}{}
\RequirePackage{ifthen}
\newboolean{swe...@gin}
\setboolean{swe...@gin}{true}
\newboolean{swe...@ae}
\setboolean{swe...@ae}{true}
\RequirePackage{ifthen}
\RequirePack
OK, I can't see anything wrong with that.
a) post the results of ...
> sessionInfo()
... so we can check you're using a reasonably up to date system
b) send a copy of your data set to me at...
k.jewell at campden.co.uk
... and I'll see if I can reproduce your error. I can take most common da
No, I meant the Combinations package, it is apparently an Omegahat package
(http://www.omegahat.org/Combinations/). It looks similar to the permn
function as far as the usage goes, but the documentation includes additional
information on other resources.
--
Gregory (Greg) L. Snow Ph.D.
Statis
The order() function allows you to specify multiple vectors, which are used
successively to break ties. If I want to use many vectors to break ties
(say, 25 or more), that are columns of a matrix or elements of a list, does
anyone know a shortcut to do this without passing 25 arguments to order()?
Thanks for your help. Part of the problem was that I thought I was dealing
with a data frame when it was actually an array. I forced all to be
matrices for consistency. However, I see that
avGain <- rbind(avGain[1:(j-1),],b,avGain[(j+1):nrow(avGain),])
would be a great solution as posit
On Tue, Mar 17, 2009 at 12:56 PM, Patrick Burns wrote:
> Use 'do.call' as discussed in 'The R Inferno'
> and elsewhere.
Perfect, thanks.
--
Levi Waldron
post-doctoral fellow
Jurisica Lab, Ontario Cancer Institute
Division of Signaling Biology
IBM Life Sciences Discovery Centre
TMDT 9-304D
101
Hello,
I would be pleased if you could help me in determining the standard error of
the estimate for place2L1.N2 from the following table:
Call:
glm(formula = y ~ places2 + shellheight, family = quasibinomial)
Deviance Residuals:
Min1QMedian3Q Max
-3.41415 -
This question is better asked on the Bioconductor list
(bioconduc...@stat.math.ethz.ch).
James Mcininch wrote:
I would like to build a package for the HT HG-U133+ PM arrays from affy,
but I can't find any good documentation on how to go about it. Naively
using makecdfenv's make.cdf.package()
In the case of 1:1 merging with distinct sets of non-ID variables in two
or more datasets, would the following code, which doesn't need to form
the larger merged data frame, be useful or faster? [A generalization of
with() would make this even better. I've often wondered about the
utility of
Dear r helpers,
I'm trying to run the models of the BIOMOD package and got this message:
*
Error in nnet.default (x, y, w,...): **initial value in 'vmmin' is not finite*
What does this mean? and how can I correct it?
Thank you
*Andreia*
[[alternative HTML version deleted]]
Dear List,
I recently got the chance to interview Jon Peck of SPSS Inc, a pioneering
technical statistician working since 1983 (when there were only two
substantial statistical software companies as per him ;) (not anymore ;)
and currently he is a Principal Software Engineer and Technical Advisor
Yes, the slow convergence is easier to get with the log link. Overshooting the
correct coefficient vector has more dramatic effects on the fitted values and
weights, and in this example the starting value of (0,0) is a long way from the
truth. The same sort of thing happens in the Cox model
Is it possible to render a volume like the one attached, in R using packages
misc3d or rgl if I have the data for the individual surfaces (top and bottom
for each layer).
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
On 3/17/2009 1:57 PM, kapo coulibaly wrote:
Is it possible to render a volume like the one attached, in R using packages
misc3d or rgl if I have the data for the individual surfaces (top and bottom
for each layer).
Your figure wasn't attached, but I'd guess the answer is yes. rgl
allows you
hello,
i use the vars package and would like to save the plots plotted with the
plot method for objects with class attribute varest etc.
so, if
library(vars)
data(Canada)
var.canada <- VAR(Canada)
then
plot(var.canada)
produces a plot for every of the four variable
On Tue, Mar 17, 2009 at 12:54 PM, Bill Cunliffe wrote:
> Thanks for your help. Part of the problem was that I thought I was
> dealing with a data frame when it was actually an array. I forced all to be
> matrices for consistency. However, I see that
>
>
>
> avGain <- rbind(avGain[1:(j-1),],b,a
Hi Lars,
Consider the following problem:
max x + y
subject to
x^2 + y^2 =1.
The solution is obviously (x,y) = (sqrt(2) / 2, sqrt(2) / 2).
Now, consider the unconstrained maximization problem on the variables
x, y and lambda:
max x + y + lambda * (x^2 + y^2 - 1)
(Notice that the objective fu
You do not mention your OS, so further details may be needed.
Graphical output can be constrained by machine details. As a first
step, why not try a format that has multiple page possibilities such
as pdf ... this gives me 4 pages:
pdf(file ="outtest.pdf")
?lm
ctl <- c(4.17,5.58,5.18,6
Awesome.
This seems to produce the vertical histogram as needed, but is there then a way
to come back and add on a probability distribution line?
That is, add something like the following to the barplot:
points(density(x), type='l', lwd=3, col='red')
Thanks again.
--- On Tue, 3/17/09, Patr
I've searched the help archives of both lists and apologize if I missed the
answer to my question:
Is there an update on developing mcmcsamp for glmer?
I'm using R v. 2.7.2 (on our Unix server - will hopefully be updated soon) and
2.8.1 on my PC and get the message for both:
gm1 <- glmer(cbind
Hi,
In the package survival, using the function survreg for AFT model, I
only see 4 distributions for the response y: weibull, gaussian,
logistic, lognormal and log-logistic, which correspond to certain
distributions for the error terms. I'm wondering if there is a package
or how to obta
Can you reverse the x and y columns of the density object?
plot(density(c(-20,rep(0,98),20))$y, density(c(-20,rep(0,98),20))$x,
xlim = c(0,1), ylim=c(-4,4), type='l', lwd=3, col='red')
--
David Winsemius
On Mar 17, 2009, at 2:38 PM, Jason Rupert wrote:
Awesome.
This seems to
Lars,
Here is how you can solve the example given by Paul using spg() in "BB"
package.
f <- function(x, lambda) x[1] + x[2] - lambda * (x[1]^2 + x[2]^2 - 1)^2 #
look at how Penalized function is formulated
eps <- Inf
tol <- 1.e-05
p0 <- c(1, 1)
lambda <- 0.1 # start with a small value for
Here is how you can implement the Lagrangian multiplier approach and solve
the first-order KKT conditions to obtain the solution for Paul Smith's
example:
require(BB)
f2 <- function(x) {
f <- rep(NA, length(x))
f[1] <- 1 + 2 * x[1] * x[3] # x[3] is the Lagrangian multiplier
f[2] <- 1 +
Folks,
As reproduced in the code below, our data is being transformed during the
sqlFetch() operation. This has apparently been an issue since 2002. Are
there any ways for correctly reading data from MS SQL Server 2005? Why
does RODBC ignore the type information supplied by SQL Server?
req
On Tue, Mar 17, 2009 at 7:10 PM, Ravi Varadhan wrote:
> Here is how you can implement the Lagrangian multiplier approach and solve
> the first-order KKT conditions to obtain the solution for Paul Smith's
> example:
>
> require(BB)
>
> f2 <- function(x) {
> f <- rep(NA, length(x))
> f[1] <-
The code I'm using is shown below.
I would like to have a larger median at the top of the plot so that I can show
the entirity of "title_text".
Several times I tried messing with "par(mar", but that seemed to make matters
worse.
By any chance can anyone provide any insight as to the best
When trying to remember what did I do in the session, especially after
coming back to it after a few days, I'd like to mimic Unix's ls -ltrh
-- does R retain the timing a certain variable is created? If not,
would it make a useful addition, to have ls with an option to sort by
creation tim
Ravi Varadhan wrote:
>
>
> require(BB)
>
> f2 <- function(x) {
> f <- rep(NA, length(x))
> f[1] <- 1 + 2 * x[1] * x[3] # x[3] is the Lagrangian multiplier
> f[2] <- 1 + 2 * x[2] * x[3]
> f[3] <- x[1]^2 + x[2]^2 - 1 # the equality constraint
> f
> }
>
>
You can also try my packag
Hi
Jason Rupert wrote:
> I guess no reply means there is not an existing package to produce
> the plot?
>
> I will post the results of my script to hopefully help others who are
> trying to formulate the same plot.
>
> Thanks again.
I don't know of an existing function that does that particul
In most biometric applications, those variances are treated as
nuisance parameters. They only need to be controlled for, while the
main purpose is to get the right point estimates and standard errors
for the fixed effects. In social science multilevel modeling (of which
education is probably the he
Paul,
This is great! Like you said it is really close. I made a few changes and for
some reason the y-axis label magically came back. I tried to remove it but it
wouldn't stay away.
Also, for some reason the title exceeds the margins of the layout. I am going
ot mess around with this a
I'm moving my R applications to a Redhat OS and want to install ESS.
My sys admin has downloaded the rpm (emacs-common-ess-5.3.8-1.fc8.src.rpm),
but when he tried to do the install he rec'd a number of warnings. We are
not sure if the warnings are telling us that the installation did not occur
On Mar 17, 2009, at 3:29 PM, steve_fried...@nps.gov wrote:
I'm moving my R applications to a Redhat OS and want to install ESS.
My sys admin has downloaded the rpm (emacs-common-
ess-5.3.8-1.fc8.src.rpm),
but when he tried to do the install he rec'd a number of warnings.
We are
not sure if
Using ggplot2:
library(ggplot2)
test_data<-rnorm(100)
a=data.frame(obs=test_data,condition='None')
p1=qplot(
data=a
,x=obs
,geom='histogram'
)+coord_flip()
p2=qplot(
data=a
,y=obs
,x=condition
,geom='boxplot'
)+opts(
R doesn't keep track of when objects were created, so that's not possible.
If this is important to you, you could look at the 'trackObjs' package, which
does this and also stores individual objects in individual files (and writes
them to the file when they are changed in R).
-- Tony Plate
Ale
1 - 100 of 135 matches
Mail list logo
