Hello,
I have a list like the following:
tree<-list(); tree[[1]]$node<-list(); tree[[2]]$node<-list();
tree[[1]]$node$values <- 1:10
tree[[2]]$node$values <- 1:10
After building the list I have to generate the mean of all "values" elements
with equal indices.
Until now I use something like that
Thanks Bill!
Not asking for help with Stata at all, on the contrary: the article
mentioned using Stata to fit the model described earlier, and I wasn't
sure how to do the same in R (which is what I've used since college).
Thanks again, I'll play around a bit glmRob, see what happens (though
In R, the glm families poisson and quasipoisson will give you the same
estimates. Their standard errors will (usually) be different, though, and
family = quasipoisson does not give you an AIC (since it does not maximise a
true likelihood; it uses quasi-likelihood estimation).
I hope you are no
Hi,
I actually largely solved the problem, i am using this code:
var1 <- rnorm(5000,0,5)
var2 <- 0.90*var1+0.10*rnorm(5000,0,10)
plot(var1,var2)
Its somewhat simplistic to say the least but it works more or less, i will
definantly look into your suggestion though!
Thanks
Thor
Denmark
On Sun
Hello R-help,
According to a research article that covers the topic I'm analyzing,
in Stata, a Poisson pseudo-maximum-likelihood (PPML) estimation can be
obtained with the command
poisson depvar_ij ln(indepvar1_ij) ln(indepvar2_ij) ...
ln(indepvarN_ij), robust
I looked up Stata help fo
On Sep 11, 2010, at 9:00 PM, Peter Ehlers wrote:
On 2010-09-11 16:14, Dennis Murphy wrote:
Hi Baptiste,
You need to use the symbol("\nnn") concept, where nnn denotes the
octal
symbol number. For< it's 074 and for> it's 076. This little test
seemed to
work:
plot(1, 1, main = expression
On 2010-09-11 16:14, Dennis Murphy wrote:
Hi Baptiste,
You need to use the symbol("\nnn") concept, where nnn denotes the octal
symbol number. For< it's 074 and for> it's 076. This little test seemed to
work:
plot(1, 1, main = expression(symbol("\074")~'x, y'~symbol("\076")))
HTH,
Dennis
It
Is this the kind of thing you are talking about?
### 8< cut here 8< ###
A <- rep(NA, 100)
B <- sort(runif(25))
C <- sort(sample(1:100, 25))
A[C] <- B
B
C
A
### 8< cut here 8< ###
(The sorting is not necessary. It's only there to make checking what happened
easier.)
-Original Message---
Hi Greg,
I am sketchy on a few details of C, but does something like this work
for you? I just created C1 (renamed because C() is a function) with
two columns the first corresponding to A and the second to B. Then I
just used the first column to select elements of A and the second to
select elem
Oh,You actually want a mixture of two different normal random variables.
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Is this what you would expect to have. Definitely you can make this function
more elegant:
fn1 <- function(x = 10) {
cat("Please type the option number to get your Y value:\n\n")
cat(" 1. Y = 1.\n
2. Y = 2.\n
3. Use the default y.\n
4. Choose my own value for y.\n\n")
opt=scan
On Sat, Sep 11, 2010 at 7:17 PM, Christofer Bogaso
wrote:
> Dear all, I am looking for some procedure to send inputs to a function
> interactively. Here is an example:
>
> fn1 <- function(x = 10) {
> y <- 0
> # ask user whether he wants to put some other value for "y"
> # R will show 2 options: 1.
Have you tried scan()?:
> y=scan()
1: 2
2:
Read 1 item
> y
[1] 2
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Dear all, I am looking for some procedure to send inputs to a function
interactively. Here is an example:
fn1 <- function(x = 10) {
y <- 0
# ask user whether he wants to put some other value for "y"
# R will show 2 options: 1. y = 2
# 2. y = 3
# user will choose ei
Hi,
Suppose I have array A with 100 elements all filled in with "N/A". Array
B has 25 elements with actual values. Lastly, I have array C that
provides a map of where to put the elements from array A into array B.
Thus C would say put element 1 from B into element 3 from array A.
I realize
Use scales = 'free' in facet_wrap() to vary both scales across panels. From
Ch. 7 of the ggplot2 book, the options are (in either type of faceting)
scales = 'fixed' => both x and y scales are fixed in all plots
scales = 'free_x'=> fix y scale, vary x scale across panels
scales = 'free_y'
Hi:
Do you mean multinomial or multivariate normal? If the latter, then in
addition to the previous response, there is package mvtnorm and a function
mvrnorm() in the MASS package to generate correlated multivariate normal
samples.
HTH,
Dennis
On Sat, Sep 11, 2010 at 12:52 PM, thedreamshaper
wro
Hi Baptiste,
You need to use the symbol("\nnn") concept, where nnn denotes the octal
symbol number. For < it's 074 and for > it's 076. This little test seemed to
work:
plot(1, 1, main = expression(symbol("\074")~'x, y'~symbol("\076")))
HTH,
Dennis
On Sat, Sep 11, 2010 at 10:01 AM, baptiste augu
Him I have a list of S4 objects. Each object has a field called name,
and what I'd like to do is to sort the list based on the value of
name. Currently I'm using the following code
tmp <- unlist(lapply(fps, function(x) as.integer(x...@name)))
tmp <- order(tmp, decreasing=FALSE)
fps <- fps[ tmp ]
Is this something you want to have (based on a simulated dataset)?
counts <- c(18,17,15,20,10,20,25,13,12)
#risk <- round(rexp(9,0.5),3)
risk<- c(2.242, 0.113, 1.480, 0.913, 5.795, 0.170, 0.846, 5.240, 0.648)
gm <- glm(counts ~ risk, family=quasipoisson)
summary(gm)
new.risk=seq(min(risk), max(r
This does not solve the problem, as I still do not know how to control
the y-range for individual facets. Data contains some outliers which
make the y-range too wide for me and I would explicitly set
the ylim = c(0,10) for facet "A" and ylim = c(42, 102) for facet "B".
How should I do it?
On 11 S
package: {mnormt}
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Swen,
facet_grid forces the scale for plots along an axis to be shared. Try
facet_wrap instead.
Jonathan
On Sat, Sep 11, 2010 at 2:21 PM, Sven Laur wrote:
> Faceting in ggplot2 seems to permit different scales for different facets,
> but I fail
> to see how one could control ylim and xlim ran
Faceting in ggplot2 seems to permit different scales for different
facets, but I fail
to see how one could control ylim and xlim ranges for each facet
separately.
For instance, I would like to set the ylim = c(0,10) for facet "A"
and ylim = c(42,102) for facet "B". Since the data is out of th
On Sep 11, 2010, at 3:15 PM, Maik Rehnus wrote:
Dear all,
I have a quasipoisson glm for which I need confidence bands in a
graphic:
gm6 <- glm(num_leaves ~ b_dist_min_new, family = quasipoisson, data
= beva)
summary(gm6)
library('VIM')
b_dist_min_new <- as.numeric(prepare(beva$dist_mi
I have had plenty of succes generating one dimensional variables and plotting
them, but what do i do for more (specifically 2) dimensional multinomial
variables?
I figure i have to create a vector consisting of two 1 dim normallly
distributed variables, that way i can also control the correlatio
Dear all,
I have a quasipoisson glm for which I need confidence bands in a graphic:
gm6 <- glm(num_leaves ~ b_dist_min_new, family = quasipoisson, data = beva)
summary(gm6)
library('VIM')
b_dist_min_new <- as.numeric(prepare(beva$dist_min, scaling="classical",
transformation="logarithm")).
My
Dear all,
Is it possible to extract cross-validation results from e1071's svm model?
For example, the following R code shows the result from the 10 fold
cross-validation.
model = svm(spam ~ ., data = spam, cross = 10)
summary(model)
But, I could not figure out how to get to the accuracy values
Another idea.
>From your formulation it seems that r does not depend on t and Xt so you
wouldn't need a solver.
You could solve for r explicitly so
r <- R0/sum(Xt)
/Berend
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You can also have a look at and try package nleqslv.
Berend
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What do people use to show angle brackets < > in R graphics? Have I
missed something obvious?
Thanks,
baptiste
On 9 September 2010 17:57, baptiste auguie
wrote:
> Dear list,
>
> I read in ?plotmath that I can use bgroup to draw scalable delimiters
> such as [ ] and ( ). The same technique fails
thanks a lot!
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Hi:
This doesn't directly answer your question, but you may want to investigate
the tikzDevice package. From its DESCRIPTION file:
The TikZ device enables LaTeX-ready output from R graphics functions. This
is done by producing code that can be understood by the TikZ graphics
language. All text in
Use BBsolve in the BB package. It'll do darn near anything.
Carl
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and provide commented, mi
Hi Martin,
indeed, as mentioned in the bug-report, the results are inconsistent,
and each time I rerun, I get different results... Sometimes, the range
is correct even on my machine, but mostly I get values >1 back:
Here is another run:
> ### load the data: a list called approx.data
> load(file=
> Duncan Murdoch
> on Sat, 11 Sep 2010 11:23:02 -0400 writes:
> On 11/09/2010 11:13 AM, Martin Maechler wrote:
>>> Duncan Murdoch
>>> on Sat, 11 Sep 2010 10:32:38 -0400 writes:
>>
>> > On 11/09/2010 10:04 AM, Martin Maechler wrote:
>> >>> "SW" ==
On 11/09/2010 11:13 AM, Martin Maechler wrote:
Duncan Murdoch
on Sat, 11 Sep 2010 10:32:38 -0400 writes:
> On 11/09/2010 10:04 AM, Martin Maechler wrote:
>>> "SW" == Samuel Wuest
>>> on Thu, 26 Aug 2010 14:34:26 +0100 writes:
>>
SW> Hi Greg,
SW> thanks fo
On Sep 11, 2010, at 11:07 AM, threshold wrote:
Dear R users, say
plot(rnorm(1)~rnorm(1), xlab=paste('abc', expression(x>=1)),
plot(rnorm(1)~rnorm(1), xlab=expression(abc~x >= 1))
# or if you are uncertain whether abc is a plotmath-special then this
also works
plot(rnorm(1)~rnorm(1), xla
> Duncan Murdoch
> on Sat, 11 Sep 2010 10:32:38 -0400 writes:
> On 11/09/2010 10:04 AM, Martin Maechler wrote:
>>> "SW" == Samuel Wuest
>>> on Thu, 26 Aug 2010 14:34:26 +0100 writes:
>>
SW> Hi Greg,
SW> thanks for the suggestion:
>>
SW> I hav
On 11/09/2010 10:53 AM, Martin Maechler wrote:
"MM" == Martin Maechler
on Sat, 11 Sep 2010 16:04:37 +0200 writes:
"SW" == Samuel Wuest
on Thu, 26 Aug 2010 14:34:26 +0100 writes:
SW> Hi Greg,
SW> thanks for the suggestion:
SW> I have attached some small dataset that can
Dear R users, say
plot(rnorm(1)~rnorm(1), xlab=paste('abc', expression(x>=1)),
I want proper sign of weak inequality not just '>='
will appreciate!
robert
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> "MM" == Martin Maechler
> on Sat, 11 Sep 2010 16:04:37 +0200 writes:
> "SW" == Samuel Wuest
> on Thu, 26 Aug 2010 14:34:26 +0100 writes:
SW> Hi Greg,
SW> thanks for the suggestion:
SW> I have attached some small dataset that can be used to reproduce the
Hi all,
Does one of you know if there is any way to combine a "nls" method in
the stat_smooth of ggplot?
Regards
--
-
Benoit Boulinguiez
Ph.D student
Ecole de Chimie de Rennes (ENSCR) Bureau 1.20
Equipe CIP UMR CNRS 6226 "Sciences Chimiques de Rennes"
Avenue du Général Leclerc
CS
sseq <- c(1, seq(5, 120, by = 5))
for(i in 1:length(sseq)){
assign(paste("arima", i, sep=""), arima0(data.ts[sseq[i]:(sseq[i]+115)],
order=c(1,1,1)))
}
pred1 = predict(arima1, n.ahead = 5, se.fit = TRUE)$pred
how do I traverse the arima models so I repeat the above prediction
procedure(bold) on
On Sep 11, 2010, at 10:04 AM, Martin Maechler wrote:
"SW" == Samuel Wuest
on Thu, 26 Aug 2010 14:34:26 +0100 writes:
SW> Hi Greg,
SW> thanks for the suggestion:
SW> I have attached some small dataset that can be used to
reproduce the
SW> odd behavior of the approxfun-funct
On 11/09/2010 10:04 AM, Martin Maechler wrote:
"SW" == Samuel Wuest
on Thu, 26 Aug 2010 14:34:26 +0100 writes:
SW> Hi Greg,
SW> thanks for the suggestion:
SW> I have attached some small dataset that can be used to reproduce the
SW> odd behavior of the approxfun-function.
> "SW" == Samuel Wuest
> on Thu, 26 Aug 2010 14:34:26 +0100 writes:
SW> Hi Greg,
SW> thanks for the suggestion:
SW> I have attached some small dataset that can be used to reproduce the
SW> odd behavior of the approxfun-function.
SW> If it gets stripped off my ema
On Sep 11, 2010, at 9:50 AM, Gregory Ryslik wrote:
Hi Everyone,
I am implementing a special case of Random forests. At one point, I
have a list of which I then sample for replacement. So if the list
is 100 elements, I get 100 elements some of them duplicates. How can
I easily get the ele
Hi Everyone,
I am implementing a special case of Random forests. At one point, I have a list
of which I then sample for replacement. So if the list is 100 elements, I get
100 elements some of them duplicates. How can I easily get the elements that
were not included in the list? I realize i can
There are currently 2 packages that can be of help for solving the type of
problems that you are describing:
1. Rsolnp
2. alabama
For optimizing nonlinear objective functions subject to linear
equality/inequality constraints, you also have the `spg' function in the "BB"
package. In contras
On Sep 11, 2010, at 8:02 AM, David Winsemius wrote:
On Sep 11, 2010, at 6:29 AM, khush wrote:
Dear all,
I have installed R using yum install R-2.9. I am able to use R for
general
functions but when I installed some library say gplots I am getting
the
below error.
install.pa
On Sep 11, 2010, at 8:39 AM, Peng, C wrote:
or:
k=0
for (i in 1:k) if(k>0) print(i)
Because of the way the ":" operator works, I would have tested k >=1
> k=0.5
> for (i in 1:k) if (k>0){print(i)}
[1] 1
But Gabor's suggestion to use seq_len(k) is cleaner, anyway.
--
David Winsemius, MD
Cheers Duncan, for your FAST answers and patience. (fast patience?!)
I was close. But closer reading of ?save would also have worked i see
now :$. LETTERS[] = :)
thanks again,
Karl
On 9/11/2010 2:10 PM, Duncan Murdoch wrote:
On 11/09/2010 8:00 AM, Karl Brand wrote:
Esteemed R users and dev
or:
k=0
for (i in 1:k) if(k>0) print(i)
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On Sat, Sep 11, 2010 at 4:39 AM, Yuan Jian wrote:
> Hello,
> I have a simple question: I want to list numbers 1:k, but if k <1, I hope
> nothing listed.
> how should we do?
> k=2
> for (i in 1:k) print(i)
> [1] 1 # <-correct
> [1] 2
> k=0
> for (i in 1:k) print(i)
> [1] 1 #< wrong
>
I've been caught out by this more times than I care to admit -
forgetting that an R for loop isn't a C for loop.
Here's one solution...
k <- start.value
while (k <= end.value) {
# do stuff
k <- k + 1
}
Michael
On 11 September 2010 18:39, Yuan Jian wrote:
> Hello,
> I have a simple question
Hello,
I have a simple question: I want to list numbers 1:k, but if k <1, I hope
nothing listed.
how should we do?
k=2
for (i in 1:k) print(i)
[1] 1 # <-correct
[1] 2
k=0
for (i in 1:k) print(i)
[1] 1 #< wrong
[1] 0
thanks
jian
[[alternative HTML version delete
On 11/09/2010 8:00 AM, Karl Brand wrote:
Esteemed R users and developers,
How does one 'programatically' list or call objects for use in a function?
It depends on the function.
For example, i thought i could do something better than this:
save(A.cwb, B.cwb, C.cwb, D.cwb, E.cwb, F.cwb, file
On Sep 11, 2010, at 7:18 AM, KARAVASILIS GEORGE wrote:
Hello, R users.
I am trying to embed Computer modern fonts to an R plot and I get
the following error.
CM <- Type1Font("CM",
+ c(paste("cm-lgc/fonts/afm/public/cm-lgc/",
+ c("fcmr8a.afm", "fcmb8a.afm", "fcmri8a.afm", "f
On Sep 11, 2010, at 6:29 AM, khush wrote:
Dear all,
I have installed R using yum install R-2.9. I am able to use R for
general
functions but when I installed some library say gplots I am getting
the
below error.
install.packages("gplots")
Warning in install.packages("gplots")
Esteemed R users and developers,
How does one 'programatically' list or call objects for use in a function?
For example, i thought i could do something better than this:
save(A.cwb, B.cwb, C.cwb, D.cwb, E.cwb, F.cwb, file="afile.RData")
with something like these-
prfxs <- c("A", "B", "C", "D"
On 11/09/2010 4:42 AM, Werner W. wrote:
Hi,
I am wondering if there is any convenient way to comment out an entire region of
a Sweave file which comprises R and Latex code. Currently I'm doing it for the R
and Latex parts separately or transfer the unwanted part into a different file.
But bot
On 11/09/2010 12:25 AM, Stephen Liu wrote:
Hi folk,
How to chcek whether a package is available on R repo? What command shall I
run? TIA
?available.packages
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Hello, R users.
I am trying to embed Computer modern fonts to an R plot and I get the
following error.
CM <- Type1Font("CM",
+ c(paste("cm-lgc/fonts/afm/public/cm-lgc/",
+ c("fcmr8a.afm", "fcmb8a.afm", "fcmri8a.afm", "fcmbi8a.afm"), sep=""),
+ "./cmsyase.afm"))
> pdf("cm.pdf",
OK
thanks to all, I don't need, more
Up to this moment the fastest working code is Jonathan Chang's
http://stackoverflow.com/questions/3686982/r-adding-zeroes-after-old-zeroes-in-a-vector/3689360#3689360
rr <- rle(tmp)
## Pad so that it always begins with 1 and ends with 1
if (rr$values[1] == 0)
Dear all,
I have installed R using yum install R-2.9. I am able to use R for general
functions but when I installed some library say gplots I am getting the
below error.
> install.packages("gplots")
Warning in install.packages("gplots") :
argument 'lib' is missing: using
'/home/fedora/R/i386-r
On 09/10/2010 10:43 PM, mamunbabu2001 wrote:
Hi Josh,
Thanks for your reply. I gave a reply yesterday but found that it was not
posted.
I managed to plot the bar pot and overlay points.
The problem I am facing now is the spread of Y scale. The values I am
plotting
in Y scale are very close. so
Would using the
?save
function on the rpart object do what you want ?
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statist
Hi all,
I'm trying to traverse a dendrogram object (cast from an hclust clustering).
What's the simplest way to obtain the height per leaf node? I suppose I can
print the dendrogram text and parse the height values, etc... But is there a
way to query for/access a leaf's branch height?
Thanks,
Pa
On 09/11/2010 05:00 AM, Peng, C wrote:
>
> Thanks David.
> func() simply prints out the 0010 as a text value. It is still not numeric.
> I am just curious about it.
>
>> is.numeric(func4(0100))
> 00100[1] FALSE
>
>
Well, you can look at the kind of things as.octmode & friends do.
The only th
Hi,
I am wondering if there is any convenient way to comment out an entire region
of
a Sweave file which comprises R and Latex code. Currently I'm doing it for the
R
and Latex parts separately or transfer the unwanted part into a different file.
But both are not great solutions. (I am doing t
Sorry my bad, example too simple
try that one out.
ggplot(diamonds, aes(clarity, fill=color,colour = cut)) +
geom_bar(position = "dodge")
I want change the filling in the "colour" legend, not the filling of the
bars.
Regards
Le 10/09/2010 20:41, Ista Zahn a écrit :
ggplot(diamonds, aes(cl
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