Re: [R] manual R en español
http://cran.r-project.org/doc/contrib/R-intro-1.1.0-espanol.1.pdf ? __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Poor performance of "Optim"
I used to consider using R and "Optim" to replace my commercial packages: Gauss and Matlab. But it turns out that "Optim" does not converge completely. The same data for Gauss and Matlab are converged very well. I see that there are too many packages based on "optim" and really doubt if they can be trusted! -- View this message in context: http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3862229.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
-Original Message- From: r-help-boun...@r-project.org on behalf of yehengxin Sent: Sat 10/1/2011 8:12 AM To: r-help@r-project.org Subject: [R] Poor performance of "Optim" I used to consider using R and "Optim" to replace my commercial packages: Gauss and Matlab. But it turns out that "Optim" does not converge completely. The same data for Gauss and Matlab are converged very well. I see that there are too many packages based on "optim" and really doubt if they can be trusted! -- Considering that your post is pure whining without any evidence or reproducible example, considering that you speak of 'data' being converged, me think it's your fault, you cann't control optim well enough to get sensible results. There are many ways to use optim eh?, you can pass on the gradients, you can use a variety of methods, you can increase the number of iterations, et cetera, read optim's help, come back with a reproducible example, or quietly stick to your commercial sofware, leaving the whining to yourself. HTH Ruben -- Rubén H. Roa-Ureta, Ph. D. AZTI Tecnalia, Txatxarramendi Ugartea z/g, Sukarrieta, Bizkaia, SPAIN -- View this message in context: http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3862229.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
Is there a question or point to your message or did you simply feel the urge to inform the entire R-help list of the things that you consider? Josh On Fri, Sep 30, 2011 at 11:12 PM, yehengxin wrote: > I used to consider using R and "Optim" to replace my commercial packages: > Gauss and Matlab. But it turns out that "Optim" does not converge > completely. The same data for Gauss and Matlab are converged very well. I > see that there are too many packages based on "optim" and really doubt if > they can be trusted! > > -- > View this message in context: > http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3862229.html > Sent from the R help mailing list archive at Nabble.com. > > __ > 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.html > and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Covariance-Variance Matrix and For Loops
Hello again,
sapply works.
However it does not explicitly call a simplify function, but rather seems to
handle the case within its own body of code. I should be able to figure out
basically what simplify2array does from the code though.
function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE)
{
FUN <- match.fun(FUN)
answer <- lapply(X, FUN, ...)
if (USE.NAMES && is.character(X) && is.null(names(answer)))
names(answer) <- X
if (simplify && length(answer) && length(common.len <-
unique(unlist(lapply(answer,
length == 1L) {
if (common.len == 1L)
unlist(answer, recursive = FALSE)
else if (common.len > 1L) {
r <- as.vector(unlist(answer, recursive = FALSE))
if (prod(d <- c(common.len, length(X))) == length(r))
array(r, dim = d, dimnames = if (!(is.null(n1 <-
names(answer[[1L]])) &
is.null(n2 <- names(answer
list(n1, n2))
else answer
}
else answer
}
else answer
}
--
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Re: [R] Help with cast/reshape
df2<-melt(df1)
df3<-cast(df2,Index~Name)
df3
HTH,
Daniel
Dana Sevak wrote:
>
> I realize that this is terribly basic, but I just don't seem to see it at
> this moment, so I would very much appreciate your help.
>
>
> How shall I transform this dataframe:
>
>> df1
> Name Index Value
> 1 a 1 0.1
> 2 a 2 0.2
> 3 a 3 0.3
> 4 a 4 0.4
> 5 b 1 2.1
> 6 b 2 2.2
> 7 b 3 2.3
> 8 b 4 2.4
>
>
> into this dataframe:
>
>> df2
> Index a b
> 1 1 0.1 2.1
> 2 2 0.2 2.2
> 3 3 0.3 2.3
> 4 4 0.4 2.4
>
>
> df1 = data.frame(c("a", "a", "a", "a", "b", "b", "b", "b"),
> c(1,2,3,4,1,2,3,4), c(0.1, 0.2, 0.3, 0.4, 2.1, 2.2, 2.3, 2.4))
> colnames(df1) = c("Name", "Index", "Value")
>
> df2 = data.frame(c(1,2,3,4), c(0.1, 0.2, 0.3, 0.4), c(2.1, 2.2, 2.3, 2.4))
> colnames(df2) = c("Index", "a", "b")
>
>
> Thank you very much.
>
> Dana
>
>
> __
> R-help@ mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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[R] Problem with logarithmic nonlinear model using nls() from the `stats' package
Example:
> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
> y <- f(x <- c(1 : 10)); y
[1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
[9] 8.395662 8.599227
> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace =
> TRUE)
37.22954 : 1 2 3
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
In addition: Warning message:
In log(1 + c * x) : NaNs produced
What's wrong here? Am I handling this problem in the wrong way?
Any suggestions are welcome, thanks :)
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Re: [R] Poor performance of "Optim"
Le 01/10/11 08:12, yehengxin a écrit : I used to consider using R and "Optim" to replace my commercial packages: Gauss and Matlab. But it turns out that "Optim" does not converge completely. What it means "completely" ? The same data for Gauss and Matlab are converged very well. I see that there are too many packages based on "optim" and really doubt if they can be trusted! I don't understand the "too many". If a package needs an optimization, it is normal that it uses optim ! I use the same model in r, Excel solver (the new version is rather good) or Profit (a mac software, very powerful) and r is rather one of the best solution. But they are many different choices that can influence the optimization. You must give an example of the problem. I find some convergence problem when the criteria to be minimized is the result of a stochastic model (ie if the same set of parameters produce different objective value depending on the run). In this case the fit stops prematurely and the method SANN should be preferred. In conclusion, give us more information but take into account that non-linear optimization is a complex world ! Marc -- __ Marc Girondot, Pr Laboratoire Ecologie, Systématique et Evolution Equipe de Conservation des Populations et des Communautés CNRS, AgroParisTech et Université Paris-Sud 11 , UMR 8079 Bâtiment 362 91405 Orsay Cedex, France Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1.69.15.73.53 e-mail: marc.giron...@u-psud.fr Web: http://www.ese.u-psud.fr/epc/conservation/Marc.html Skype: girondot __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Returning vector of values shared across 3 vectors?
try this: > vec1 <- > c(4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81) > vec2 <- c > (1,2,3,4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66) > vec3 <- c (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,52) > intersect(vec1,intersect(vec2, vec3)) [1] 4 5 6 7 8 9 10 11 12 52 > On Sat, Oct 1, 2011 at 2:00 AM, Chris Conner wrote: > Help-Rs, > > I've got three vectors representing participants: > > vec1 <- > c(4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81) > vec2 <- c > (1,2,3,4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66) > vec3 <- c (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,52) > > I'd like to return a vector that contains only the values that are shared > across ALL THREE vectors. So the statement would return a vector that looked > like this: > 4,5,6,7,8,9,10,11,12,52 > > For some reason I initially thought that a cbind and a unique() would handle > it, but then common sense sunk in. I think the sleep deprivation is starting > to take it's toll. I've got to believe that there is a simple solution to > this dilema. > > Thanks in adance for any help! > C > [[alternative HTML version deleted]] > > > __ > 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.html > and provide commented, minimal, self-contained, reproducible code. > > -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
Have you considered the "optimx" package? I haven't tried it, but it was produced by a team of leading researchers in nonlinear optimization, including those who wrote most of "optim" (http://user2010.org/tutorials/Nash.html) years ago. There is a team actively working on this. If you could provide specific examples where Gauss and Matlab outperformed the alternatives you've tried in R, especially if Gauss and Matlab outperformed optimx, I believe they would be interested. As previously noted, nonlinear optimization is a difficult problem. An overview of alternatives available in R, including optim and optimx, is available with the CRAN Task View on optimization (http://cran.fhcrc.org/web/views/Optimization.html). Hope this helps. Spencer On 10/1/2011 3:04 AM, Marc Girondot wrote: Le 01/10/11 08:12, yehengxin a écrit : I used to consider using R and "Optim" to replace my commercial packages: Gauss and Matlab. But it turns out that "Optim" does not converge completely. What it means "completely" ? The same data for Gauss and Matlab are converged very well. I see that there are too many packages based on "optim" and really doubt if they can be trusted! I don't understand the "too many". If a package needs an optimization, it is normal that it uses optim ! I use the same model in r, Excel solver (the new version is rather good) or Profit (a mac software, very powerful) and r is rather one of the best solution. But they are many different choices that can influence the optimization. You must give an example of the problem. I find some convergence problem when the criteria to be minimized is the result of a stochastic model (ie if the same set of parameters produce different objective value depending on the run). In this case the fit stops prematurely and the method SANN should be preferred. In conclusion, give us more information but take into account that non-linear optimization is a complex world ! Marc -- Spencer Graves, PE, PhD President and Chief Technology Officer Structure Inspection and Monitoring, Inc. 751 Emerson Ct. San José, CA 95126 ph: 408-655-4567 web: www.structuremonitoring.com __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding axis to an ellipse: "ellipse" package
Dear Rolf, I tryed to follow your advices but the results I am getting seems still strange to me. See below an example of a matrix: datamat <- matrix(c(2.2, 0.4, 0.4, 2.8), 2, 2) plot(ellipse(datamat),type='l') eigenval <- eigen(datamat)$values eigenvect <- eigen(datamat)$vectors eigenscl <- eigenvect * sqrt(eigenval) * (qchisq(0.95,2))# One solution to get rescale v1 <- (eigenvect[,1])*(sqrt(eigenval[1]))*(qchisq(0.95,2))#or directly rescale the vectors needed v2 <- (eigenvect[,2])*(sqrt(eigenval[2]))*(qchisq(0.95,2)) #Or v1 <- eigenscl[1,] v2 <- eigenscl[2,] segments(-v1[1],-v1[2],v1[1],v1[2]) segments(-v2[1],-v2[2],v2[1],v2[2]) The vectors don't seem to be scaled properly and I don't see what I am doing wrong. Any ideas? Thanks! Antoine -- View this message in context: http://r.789695.n4.nabble.com/Adding-axis-to-an-ellipse-ellipse-package-tp3847954p3862491.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Can I tell about someone's academic cheating
Hello, Can I tell about someone¡¦s academic cheating behavior in the mailing list? For I knew this person through this R mailing list. Thanks! Regards, Hong Yu [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] class definition
Hi everybody! I have a matrix of class "myClass", for example: myMat <- matrix(rnorm(30), nrow = 6) attr(myMat, "class") <- "myClass" class(myMat) When I extract part of ''myMat'', the corresponding class ''myClass'' unfortunately disappear: myMat.p <- myMat[,1:2] class(myMat.p) Please for any advice / suggestions, how define class, that during an operation does not disappear. Thanks, OV [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can I tell about someone's academic cheating
-Original Message- From: r-help-boun...@r-project.org on behalf of YuHong Sent: Sun 10/2/2011 3:27 AM To: r-help@r-project.org Subject: [R] Can I tell about someone's academic cheating Hello, Can I tell about someone¡¦s academic cheating behavior in the mailing list? For I knew this person through this R mailing list. Thanks! Regards, Hong Yu -- You have to provide a reproducible example ... Rubén H. Roa-Ureta, Ph. D. AZTI Tecnalia, Txatxarramendi Ugartea z/g, Sukarrieta, Bizkaia, SPAIN [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with logarithmic nonlinear model using nls() from the `stats' package
On Sat, Oct 1, 2011 at 5:28 AM, Casper Ti. Vector
wrote:
> Example:
>
>> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
>> y <- f(x <- c(1 : 10)); y
> [1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
> [9] 8.395662 8.599227
>> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace =
>> TRUE)
> 37.22954 : 1 2 3
> Error in numericDeriv(form[[3L]], names(ind), env) :
> Missing value or an infinity produced when evaluating the model
> In addition: Warning message:
> In log(1 + c * x) : NaNs produced
>
> What's wrong here? Am I handling this problem in the wrong way?
> Any suggestions are welcome, thanks :)
>
Its linear given c so calculate the residual sum of squares using lm
(or lm.fit which is faster) given c and optimize over c:
set.seed(123) # for reproducibility
# test data
x <- 1:10
y <- 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5)
# calculate residual sum of squares for best fit given c
fitc <- function(c) lm.fit(cbind(1, log(1 + c * x)), y)
rssvals <- function(c) sum(resid(fitc(c))^2)
out <- optimize(rssvals, c(0.01, 10))
which gives:
> setNames(c(coef(fitc(out$minimum)), out$minimum), letters[1:3])
a b c
0.7197666 2.007 2.899
--
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email: ggrothendieck at gmail.com
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Re: [R] Problem with logarithmic nonlinear model using nls() from the `stats' package
On Sat, Oct 1, 2011 at 9:27 AM, Gabor Grothendieck
wrote:
> On Sat, Oct 1, 2011 at 5:28 AM, Casper Ti. Vector
> wrote:
>> Example:
>>
>>> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
>>> y <- f(x <- c(1 : 10)); y
>> [1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
>> [9] 8.395662 8.599227
>>> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace =
>>> TRUE)
>> 37.22954 : 1 2 3
>> Error in numericDeriv(form[[3L]], names(ind), env) :
>> Missing value or an infinity produced when evaluating the model
>> In addition: Warning message:
>> In log(1 + c * x) : NaNs produced
>>
>> What's wrong here? Am I handling this problem in the wrong way?
>> Any suggestions are welcome, thanks :)
>>
>
> Its linear given c so calculate the residual sum of squares using lm
> (or lm.fit which is faster) given c and optimize over c:
>
> set.seed(123) # for reproducibility
>
> # test data
> x <- 1:10
> y <- 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5)
>
> # calculate residual sum of squares for best fit given c
> fitc <- function(c) lm.fit(cbind(1, log(1 + c * x)), y)
> rssvals <- function(c) sum(resid(fitc(c))^2)
>
> out <- optimize(rssvals, c(0.01, 10))
>
> which gives:
>
>> setNames(c(coef(fitc(out$minimum)), out$minimum), letters[1:3])
> a b c
> 0.7197666 2.007 2.899
Also you probably intended to write 10 instead of 1 as the arg to rnorm.
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Re: [R] Odd gridding pattern when plotting
I think you found a bug introduced in R-2.13.x that has been fixed in
R-2.13.2 which has been released yesterday.
Best,
Uwe Ligges
On 30.09.2011 21:36, Balko, Justin wrote:
Thanks, that kind of helps. However, some of my previous code uses functions
like heatmap.2 which has multiple images (legend/color key) as well as the
actual heatmap. Employing useRaster=TRUE here only applies to the heatmap and
not the legend. Not a huge deal. Is there anyway to set an option in R to
always use rastering when drawing in the interface?
Thanks again,
Justin
-Original Message-
From: David L Carlson [mailto:dcarl...@tamu.edu]
Sent: Friday, September 30, 2011 1:54 PM
To: Balko, Justin; r-help@r-project.org
Subject: RE: [R] Odd gridding pattern when plotting
From ?image
" Images for large z on a regular grid are more efficient with useRaster enabled and
can prevent rare anti-aliasing artifacts, but may not be supported by all graphics
devices."
Adding useRaster=TRUE to the two image() calls gets rid of the white grid lines.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Balko, Justin
Sent: Friday, September 30, 2011 10:43 AM
To: r-help@r-project.org
Subject: [R] Odd gridding pattern when plotting
Hi, I'm no longer on the subscribing list, but was hoping to get my question
posted. Please inform if this is ok, although I am guessing you wont post with
the image below. If so, let me know and I will resend without the image.
Thanks
Hi,
I just upgraded my system and my version of R all at once. Upon running old
code for heatmaps etc, I suddenly notice that there is an odd grid pattern
appearing in all of my plots. An example is below:
#example from ?image
require(grDevices) # for colours
x<- y<- seq(-4*pi, 4*pi, len=27)
r<- sqrt(outer(x^2, y^2, "+"))
image(z = z<- cos(r^2)*exp(-r/6), col=gray((0:32)/32)) image(z, axes = FALSE, main =
"Math can be beautiful ...",
xlab = expression(cos(r^2) * e^{-r/6})) contour(z, add = TRUE,
drawlabels = FALSE)
Any ideas what is causing this? I can't seem to figure it out. I'm not sure
the bmp image can/will be posted, so maybe you can just take my word for it.
It is a gridding pattern in white, that appears over the plot area only.
Vertical lines are every 4 units, evenly spaced. Horizontal lines appear at
every unit, then stop for a while (6-7 units, then appear every unit for 4-5
units). Simple plots like plot(x,y) do not seem to produce it, or at least I
can't see it. Any ideas are helpful.
Thanks!
Justin M. Balko, Pharm.D., Ph.D.
Research Fellow, Arteaga Lab
Department of Medicine
Division of Hematology/Oncology
Vanderbilt University
777 Preston Research Building
Nashville TN, 37232-6307
Ph: 615-936-1495
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Re: [R] Problem with logarithmic nonlinear model using nls() from the `stats' package
Ah, now I see... Thanks very much :) On Sat, Oct 01, 2011 at 09:27:34AM -0400, Gabor Grothendieck wrote: > On Sat, Oct 1, 2011 at 5:28 AM, Casper Ti. Vector > wrote: > Its linear given c so calculate the residual sum of squares using lm > (or lm.fit which is faster) given c and optimize over c: > > set.seed(123) # for reproducibility > x <- 1:10 > y <- 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) > fitc <- function(c) lm.fit(cbind(1, log(1 + c * x)), y) > rssvals <- function(c) sum(resid(fitc(c))^2) > out <- optimize(rssvals, c(0.01, 10)) > > which gives: > 0.7197666 2.007 2.899 -- Using GPG/PGP? Please get my current public key (ID: 0xAEF6A134, valid from 2010 to 2013) from a key server. signature.asc Description: Digital signature __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] class definition
On 01.10.2011 13:21, Omphalodes Verna wrote: Hi everybody! I have a matrix of class "myClass", for example: myMat<- matrix(rnorm(30), nrow = 6) attr(myMat, "class")<- "myClass" class(myMat) When I extract part of ''myMat'', the corresponding class ''myClass'' unfortunately disappear: myMat.p<- myMat[,1:2] class(myMat.p) Please for any advice / suggestions, how define class, that during an operation does not disappear. You will need a "[" method for your class. Uwe Ligges Thanks, OV [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Covariance-Variance Matrix and For Loops
Surprising: must be newer update than I realizedanyways, here's
the code if you want to add it manually:
simplify2array <-
function (x, higher = TRUE)
{
if (length(common.len <- unique(unlist(lapply(x, length >
1L)
return(x)
if (common.len == 1L)
unlist(x, recursive = FALSE)
else if (common.len > 1L) {
n <- length(x)
r <- as.vector(unlist(x, recursive = FALSE))
if (higher && length(c.dim <- unique(lapply(x, dim))) ==
1 && is.numeric(c.dim <- c.dim[[1L]]) && prod(d <- c(c.dim,
n)) == length(r)) {
iN1 <- is.null(n1 <- dimnames(x[[1L]]))
n2 <- names(x)
dnam <- if (!(iN1 && is.null(n2)))
c(if (iN1) rep.int(list(n1), length(c.dim)) else n1,
list(n2))
array(r, dim = d, dimnames = dnam)
}
else if (prod(d <- c(common.len, n)) == length(r))
array(r, dim = d, dimnames = if (!(is.null(n1 <- names(x[[1L]])) &
is.null(n2 <- names(x
list(n1, n2))
else x
}
else x
}
On Sat, Oct 1, 2011 at 4:22 AM, sf1979 wrote:
> Hello again,
>
> sapply works.
>
> However it does not explicitly call a simplify function, but rather seems to
> handle the case within its own body of code. I should be able to figure out
> basically what simplify2array does from the code though.
>
> function (X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE)
> {
> FUN <- match.fun(FUN)
> answer <- lapply(X, FUN, ...)
> if (USE.NAMES && is.character(X) && is.null(names(answer)))
> names(answer) <- X
> if (simplify && length(answer) && length(common.len <-
> unique(unlist(lapply(answer,
> length == 1L) {
> if (common.len == 1L)
> unlist(answer, recursive = FALSE)
> else if (common.len > 1L) {
> r <- as.vector(unlist(answer, recursive = FALSE))
> if (prod(d <- c(common.len, length(X))) == length(r))
> array(r, dim = d, dimnames = if (!(is.null(n1 <-
> names(answer[[1L]])) &
> is.null(n2 <- names(answer
> list(n1, n2))
> else answer
> }
> else answer
> }
> else answer
> }
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Covariance-Variance-Matrix-and-For-Loops-tp3859441p3862347.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.html
> and provide commented, minimal, self-contained, reproducible code.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Nearest neighbour in a matrix
Hi, sorry for the late reply. I just wanted to thank both of you for your answers. They were helpful and also thank you for mentioning the website that has the tutorials which is a most helpful resource. Cheers, Léa -- View this message in context: http://r.789695.n4.nabble.com/Nearest-neighbour-in-a-matrix-tp3845747p3862973.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is the output of survfit.coxph survival or baseline survival?
On Sep 30, 2011, at 9:31 PM, koshihaku wrote:
Dear all,
I am confused with the output of survfit.coxph.
Someone said that the survival given by summary(survfit.coxph) is the
baseline survival S_0, but some said that is the survival
S=S_0^exp{beta*x}.
Which one is correct?
It may depend on who _some_ and _someone_ mean by S_0 and who they
are. I have in the past posted erroneous answers, but the name on
which to search the archives is 'Terry Therneau'. My current
understanding is that the survival S_0 is the estimated survival for a
hypothetical subject whose continuous and discrete covariates are all
at their means. (But I have been wrong before.) Here is some of what
Therneau has said about it:
http://finzi.psych.upenn.edu/Rhelp10/2010-October/257941.html
http://finzi.psych.upenn.edu/Rhelp10/2009-March/190341.html
http://finzi.psych.upenn.edu/Rhelp10/2009-February/189768.html
By the way, if I use "newdata=" in the survfit, does that mean the
survival
is estimated by the value of covariates in the new data frame?
In one sense yes, but in another sense, no. If you have a cox fit and
you supply newdata, the beta estimates and the baseline survival come
from in the original data. If you just give it a formula, then there
is no newdata argument, only a data argument.
Try this:
fit <- coxph( Surv(futime, fustat)~rx, data=ovarian)
plot( survfit(fit, newdata=data.frame(rx=1) ) )
plot( survfit( Surv(futime, fustat)~rx, data=ovarian) )
Then flipping back and forth between those curves might clarify, at
least to the extent that I understand this question.
And here's a pathological extrapolation:
plot(survfit(fit, newdata=data.frame(rx=1:3)))
# There is no rx=3 in the original data but it wasn't defined as a
factor when given to coxph.
# Just checked to see if you could extrapolate past the end of a range
of factors and very sensibly you cannot.
> fit <- coxph( Surv(futime, fustat)~factor(rx), data=ovarian)
> plot(survfit(fit, newdata=data.frame(rx=1:3)))
Error in model.frame.default(data = data.frame(rx = 1:3), formula =
~factor(rx), :
factor 'factor(rx)' has new level(s) 3
--
David.
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[R] error using ddply to generate means
Dear list,
I encounter an error when I try to use ddply to generate means as follows:
fun3<-structure(list(sector = structure(list(gics_sector_name = c("Financials",
"Financials", "Materials", "Materials")), .Names = "gics_sector_name",
row.names = structure(c("UBSN VX Equity",
"LLOY LN Equity", "AI FP Equity", "AKE FP Equity"), .Dim = 4L), class
= "data.frame"),
bebitpcchg = c(-0.567449058550428, 0.99600643852127, NA,
-42.7587478692081), ticker = c("UBSN VX Equity", "LLOY LN Equity",
"AI FP Equity", "AKE FP Equity")), .Names = c("sector", "bebitpcchg",
"ticker"), row.names = c(12L, 24L, 36L, 48L), class = "data.frame")
fun3
gics_sector_name bebitpcchg ticker
12 Financials -0.5674491 UBSN VX Equity
24 Financials 0.9960064 LLOY LN Equity
36Materials NA AI FP Equity
48Materials -42.7587479 AKE FP Equity
fun4<-ddply(fun3,c("sector"),summarise,avgbebitchg=mean(bebitpcchg,na.rm=TRUE))
Error in `[.data.frame`(x, order(x, na.last = na.last, decreasing =
decreasing)) :
undefined columns selected
This is a small sample of my data. I’m probably overlooking some
problem in my syntax, but would be very grateful if someone could
point it out.
Thanks in advance,
Aidan.
sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_Ireland.1252 LC_CTYPE=English_Ireland.1252
LC_MONETARY=English_Ireland.1252
[4] LC_NUMERIC=C LC_TIME=English_Ireland.1252
attached base packages:
[1] grid stats graphics grDevices utils datasets
methods base
other attached packages:
[1] plm_1.2-7 sandwich_2.2-7 MASS_7.3-12
Formula_1.0-1 nlme_3.1-100
[6] bdsmatrix_1.0 RBloomberg_0.4-149 rJava_0.8-8
gtools_2.6.2 gdata_2.8.2
[11] ggplot2_0.8.9 proto_0.3-9.2 zoo_1.7-4
reshape_0.8.4 plyr_1.6
loaded via a namespace (and not attached):
[1] lattice_0.19-23 tools_2.13.0
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Re: [R] error using ddply to generate means
Hi:
Here's the problem:
> str(fun3)
'data.frame': 4 obs. of 3 variables:
$ sector:'data.frame': 4 obs. of 1 variable:
..$ gics_sector_name: chr "Financials" "Financials" "Materials" "Materials"
$ bebitpcchg: num -0.567 0.996 NA -42.759
$ ticker: chr "UBSN VX Equity" "LLOY LN Equity" "AI FP Equity"
"AKE FP Equity"
Notice that fun3$sector is a data frame, not a variable. By leaving
fun3 intact, the summary is gotten with
ddply(fun3, .(sector$gics_sector_name), summarise,
avgbebitchg=mean(bebitpcchg,na.rm=TRUE))
sector$gics_sector_name avgbebitchg
1 Financials 0.2142787
2 Materials -42.7587479
You might consider reframing fun3, pardon the pun.
HTH,
Dennis
On Sat, Oct 1, 2011 at 7:58 AM, Aidan Corcoran
wrote:
> Dear list,
>
> I encounter an error when I try to use ddply to generate means as follows:
>
> fun3<-structure(list(sector = structure(list(gics_sector_name =
> c("Financials",
> "Financials", "Materials", "Materials")), .Names = "gics_sector_name",
> row.names = structure(c("UBSN VX Equity",
> "LLOY LN Equity", "AI FP Equity", "AKE FP Equity"), .Dim = 4L), class
> = "data.frame"),
> bebitpcchg = c(-0.567449058550428, 0.99600643852127, NA,
> -42.7587478692081), ticker = c("UBSN VX Equity", "LLOY LN Equity",
> "AI FP Equity", "AKE FP Equity")), .Names = c("sector", "bebitpcchg",
> "ticker"), row.names = c(12L, 24L, 36L, 48L), class = "data.frame")
>
> fun3
>
> gics_sector_name bebitpcchg ticker
> 12 Financials -0.5674491 UBSN VX Equity
> 24 Financials 0.9960064 LLOY LN Equity
> 36 Materials NA AI FP Equity
> 48 Materials -42.7587479 AKE FP Equity
>
>
> fun4<-ddply(fun3,c("sector"),summarise,avgbebitchg=mean(bebitpcchg,na.rm=TRUE))
>
> Error in `[.data.frame`(x, order(x, na.last = na.last, decreasing =
> decreasing)) :
> undefined columns selected
>
> This is a small sample of my data. I’m probably overlooking some
> problem in my syntax, but would be very grateful if someone could
> point it out.
>
> Thanks in advance,
> Aidan.
>
> sessionInfo()
>
> R version 2.13.0 (2011-04-13)
> Platform: i386-pc-mingw32/i386 (32-bit)
>
> locale:
> [1] LC_COLLATE=English_Ireland.1252 LC_CTYPE=English_Ireland.1252
> LC_MONETARY=English_Ireland.1252
> [4] LC_NUMERIC=C LC_TIME=English_Ireland.1252
>
> attached base packages:
> [1] grid stats graphics grDevices utils datasets
> methods base
>
> other attached packages:
> [1] plm_1.2-7 sandwich_2.2-7 MASS_7.3-12
> Formula_1.0-1 nlme_3.1-100
> [6] bdsmatrix_1.0 RBloomberg_0.4-149 rJava_0.8-8
> gtools_2.6.2 gdata_2.8.2
> [11] ggplot2_0.8.9 proto_0.3-9.2 zoo_1.7-4
> reshape_0.8.4 plyr_1.6
>
> loaded via a namespace (and not attached):
> [1] lattice_0.19-23 tools_2.13.0
>
> __
> 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.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Covariance-Variance Matrix and For Loops
That's very helpful Michael, thank you. I will add it to the arsenal. -- View this message in context: http://r.789695.n4.nabble.com/Covariance-Variance-Matrix-and-For-Loops-tp3859441p3863098.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error while using shapiro.test()
Thank you very much! your response solved my issue. I needed to determine the probability of normality for word types per page. -- View this message in context: http://r.789695.n4.nabble.com/error-while-using-shapiro-test-tp3861535p3863205.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Gstat - Installation Fail _ download source and compile help ...
Hello I have been trying to install gstat on university's unix based system ( i am not familiar with many technical aspects of installation) but i am getting a particular error which i could not find a solution to online. Here is what the technical support guy mailed me back, i am sure someone who understands the technicalities can explain me this procedure in a more lucid way. * **Technical Assistant's reply* * Unfortunately, the error is due to a type being used in one of the source files which has not yet been defined in an include file. The "u_int" type is defined in /usr/include/sys/types.h: typedef __u_int u_int; And, the "__u_int" type is defined in /usr/include/bits/types.h: typedef unsigned int __u_int; Note that is included at the top of , so only the would need to be included. Without including , the program won't recognize "u_int" as a valid type. So, this is an issue with the configuration or perhaps source for the given program being compiled by the package installation function of R. My suggestion would be to search for the given error message on any support/help/discussion boards/websites related to the R program. Or, do a google search to see if anyone else has encountered the same error and find their suggested solution. Otherwise, you can manually download the source to your directory and attempt to tweak the "configure" command, which would generate a more correct Makefile. Or, in the least desirable scenario, insert the needed "#include " in the given *.c file yourself and compile. * Can anyone make out anything from this , i want to tweak the configure command but do not know how to proceed. [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Create web applications to run R scripts
Hello, is there anything similar to the Rwui package to create web applications to run R scripts? Many thanks, syrvn -- View this message in context: http://r.789695.n4.nabble.com/Create-web-applications-to-run-R-scripts-tp3863457p3863457.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create web applications to run R scripts
On Sat, Oct 01, 2011 at 11:34:47AM -0700, syrvn wrote: > Hello, > > is there anything similar to the Rwui package to create web applications to > run R scripts? There is an entire section of the R FAQ devoted to this. Dirk -- Three out of two people have difficulties with fractions. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Printing an xtable with type = html
Maybe some of the comments in this post may be informative to you: http://r.789695.n4.nabble.com/improve-formatting-of-HTML-table-td3736299.html On Wed, Sep 28, 2011 at 6:21 AM, David Scott wrote: > > I have been playing around with producing tables using xtable and the type = > "html" argument when printing. For example, if xtbl is the output of a > dataframe which has been run through xtable, using the command: > > print(xtbl, type = "html", > html.table.attributes = "border = '1', align = 'center'") > > I would be interested to see other examples of the use of xtable to produce > html. There is a whole vignette on using xtable to produce all sorts of > tables for incorporation into a TeX document but I have found no examples of > producing html with any table attributes. > > Ideally xtable should be able to access a css file but I don't see any > mechanism for doing that. Perhaps someone can enlighten me. > > David Scott > > -- > _ > David Scott Department of Statistics > The University of Auckland, PB 92019 > Auckland 1142, NEW ZEALAND > Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 > Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 > > __ > 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.html > and provide commented, minimal, self-contained, reproducible code. > __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Understanding the workflow between sweave, R and Latex
On 9/30/2011 9:08 AM, syrvn wrote:
Hi Duncan,
I use Eclipse and StatET plus TexClipse and Sweave which comes with the
StatET package.
So fore me it is basically one click as well to produce the pdf from the
.Rnw file.
I installed the MacTex live 2011 version on my computer and thought it might
actually be
easy to find out how and where latex searches for packages. But I did not
find the place
where all this is coded...
First, since this is Mac-related, you would probably get better answers
on the R-sig-mac list.
Second, most latex distributions support both a system 'texmf' tree and
one or more local/user texmf trees, that you can configure with
something like Preferences somewhere in MacTex. On my linux system, I use
~/texmf/ and simply copied Sweave.sty to
~/texmf/tex/latex/misc/Sweave.sty (if my path-memory serves)
No more worries (unless Sweave.sty is changed in a new R distro)
Finally, it does help to RTFM, where you can find other options
under ?RweaveLatex in the Details section.
The LaTeX file generated needs to contain the line \usepackage{Sweave},
and if this is not present in the Sweave source file (possibly in a
comment), it is inserted by the RweaveLatex driver. If stylepath = TRUE,
a hard-coded path to the file ‘Sweave.sty’ in the R installation is set
in place of Sweave. The hard-coded path makes the LaTeX file less
portable, but avoids the problem of installing the current version of
‘Sweave.sty’ to some place in your TeX input path. However, TeX may not
be able to process the hard-coded path if it contains spaces (as it
often will under Windows) or TeX special characters.
The default for stylepath is now taken from the environment variable
SWEAVE_STYLEPATH_DEFAULT, or is FALSE it that is unset or empty. If set,
it should be exactly TRUE or FALSE: any other values are taken as FALSE.
As from R 2.12.0, the simplest way for frequent Sweave users to ensure
that ‘Sweave.sty’ is in the TeX input path is to add
‘R_HOME/share/texmf’ as a ‘texmf tree’ (‘root directory’ in the parlance
of the ‘MiKTeX settings’ utility).
By default, ‘Sweave.sty’ sets the width of all included graphics to:
\setkeys{Gin}{width=0.8\textwidth}.
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Re: [R] Sum of Probabilities in a matrix...
Hi all, I have 2 columns in a mtrix, one of which is a column of probabilities and the other is simply a vector of integers. I want to sum all the probabilities with the same integer value and put it in a new column. For example, If my matrix is: 0.98 2 0.2 1 0.01 2 0.5 1 0.6 6 Then I should get: 0.98 20.99 0.2 10.70 0.01 20.99 0.5 10.70 0.6 60.60 Any help is greatly appreciated. -- Thanks, Jim. [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
With respect, your statement that R's optim does not give you a reliable
estimator is bogus. As pointed out before, this would depend on when optim
believes it's good enough and stops optimizing. In particular if you stretch
out x, then it is plausible that the likelihood function will become flat
enough "earlier," so that the numerical optimization will stop earlier
(i.e., optim will "think" that the slope of the likelihood function is flat
enough to be considered zero and stop earlier than it will for more
condensed data). After all, maximum likelihood is a numerical method and
thus an approximation. I would venture to say that what you describe lies in
the nature of this method. You could also follow the good advice given
earlier, by increasing the number of iterations or decreasing the tolerance.
However, check the example below: for all purposes it's really close enough
and has nothing to do with optim being "unreliable."
n<-1000
x<-rnorm(n)
y<-0.5*x+rnorm(n)
z<-ifelse(y>0,1,0)
X<-cbind(1,x)
b<-matrix(c(0,0),nrow=2)
#Probit
reg<-glm(z~x,family=binomial("probit"))
#Optim reproducing probit (with minor deviations due to difference in
method)
LL<-function(b){-sum(z*log(pnorm(X%*%b))+(1-z)*log(1-pnorm(X%*%b)))}
optim(c(0,0),LL)
#Multiply x by 2 and repeat optim
X[,2]=2*X[,2]
optim(c(0,0),LL)
HTH,
Daniel
yehengxin wrote:
>
> What I tried is just a simple binary probit model. Create a random data
> and use "optim" to maximize the log-likelihood function to estimate the
> coefficients. (e.g. u = 0.1+0.2*x + e, e is standard normal. And y = (u
> > 0), y indicating a binary choice variable)
>
> If I estimate coefficient of "x", I should be able to get a value close to
> 0.2 if sample is large enough. Say I got 0.18.
>
> If I expand x by twice and reestimate the model, which coefficient should
> I get? 0.09, right?
>
> But with "optim", I got something different. When I do the same thing in
> both Gauss and Matlab, I can exactly get 0.09, evidencing that the
> coefficient estimator is reliable. But R's "optim" does not give me a
> reliable estimator.
>
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Re: [R] Sum of Probabilities in a matrix...
Let's make it a data frame instead:
# Read the data from your post into a data frame named d:
d <- read.table(textConnection("
0.98 2
0.2 1
0.01 2
0.5 1
0.6 6"))
closeAllConnections()
# Use the ave() function and append the result to d:
d$sumprob <- with(d, ave(V1, V2, FUN = sum))
> d
V1 V2 sumprob
1 0.98 20.99
2 0.20 10.70
3 0.01 20.99
4 0.50 10.70
5 0.60 60.60
HTH,
Dennis
On Sat, Oct 1, 2011 at 6:06 PM, Jim Silverton wrote:
> Hi all,
> I have 2 columns in a mtrix, one of which is a column of probabilities and
> the other is simply a vector of integers. I want to sum all the
> probabilities with the same integer value and put it in a new column.
> For example,
> If my matrix is:
>
> 0.98 2
> 0.2 1
> 0.01 2
> 0.5 1
> 0.6 6
>
>
> Then I should get:
> 0.98 2 0.99
> 0.2 1 0.70
> 0.01 2 0.99
> 0.5 1 0.70
> 0.6 6 0.60
>
> Any help is greatly appreciated.
>
>
>
> --
> Thanks,
> Jim.
>
> [[alternative HTML version deleted]]
>
> __
> 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.html
> and provide commented, minimal, self-contained, reproducible code.
>
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[R] R Studio and Rcmdr/RcmdrPlugins
Dear R People: Hope you're having a great weekend! Anyhow, I'm currently experimenting with R Studio on a web server, which is the best thing since sliced bread, Coca Cola, etc. My one question: there is a way to show plots. is there a way to show Rcmdr or its Plugins, please? I tried, but it doesn't seem to work. Thanks so much, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding axis to an ellipse: "ellipse" package
See comments in-line: On 01/10/11 23:26, Antoine wrote: Dear Rolf, I tryed to follow your advices but the results I am getting seems still strange to me. See below an example of a matrix: datamat<- matrix(c(2.2, 0.4, 0.4, 2.8), 2, 2) plot(ellipse(datamat),type='l') eigenval<- eigen(datamat)$values eigenvect<- eigen(datamat)$vectors eigenscl<- eigenvect * sqrt(eigenval) * (qchisq(0.95,2))# One solution to get rescale This is wrong because you are multiplying the i-th row of ``eigenvect'' the square root of the i-th eigenvalue. The *columns* of ``eigenvect'' are the eigenvectors. So you need to multiply the j-th column by the square root of the j-th eigenvalue. v1<- (eigenvect[,1])*(sqrt(eigenval[1]))*(qchisq(0.95,2))#or directly rescale the vectors needed v2<- (eigenvect[,2])*(sqrt(eigenval[2]))*(qchisq(0.95,2)) The foregoing is correct except that you need to take the square root of the chi-squared quantile. #Or v1<- eigenscl[1,] v2<- eigenscl[2,] segments(-v1[1],-v1[2],v1[1],v1[2]) segments(-v2[1],-v2[2],v2[1],v2[2]) The vectors don't seem to be scaled properly and I don't see what I am doing wrong. Any ideas? Here is correct code: require(ellipse) S <- matrix(c(2.2, 0.4, 0.4, 2.8), 2, 2) # Note the ``asp=1'' which makes orthogonal lines # look orthogonal: plot(ellipse(S),type='l',asp=1) E <- eigen(S) Val <- E$values Vec <- E$vectors v1 <- sqrt(Val[1]*qchisq(0.95,2))*Vec[,1] v2 <- sqrt(Val[2]*qchisq(0.95,2))*Vec[,2] segments(-v1[1],-v1[2],v1[1],v1[2]) segments(-v2[1],-v2[2],v2[1],v2[2]) cheers, Rolf Turner __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum of Probabilities in a matrix...
On 02/10/11 14:06, Jim Silverton wrote: Hi all, I have 2 columns in a mtrix, one of which is a column of probabilities and the other is simply a vector of integers. I want to sum all the probabilities with the same integer value and put it in a new column. For example, If my matrix is: 0.98 2 0.2 1 0.01 2 0.5 1 0.6 6 Then I should get: 0.98 20.99 0.2 10.70 0.01 20.99 0.5 10.70 0.6 60.60 Any help is greatly appreciated. Suppose your matrix is called "m". Execute: > ttt <- tapply(m[,1],m[,2],sum) > m <- cbind(m,ttt[match(m[,2],names(ttt))]) > dimnames(m) <- NULL # To tidy up a bit. You get: > m [,1] [,2] [,3] [1,] 0.982 0.99 [2,] 0.201 0.70 [3,] 0.012 0.99 [4,] 0.501 0.70 [5,] 0.606 0.60 Easy-peasy. cheers, Rolf Turner __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tobit Fixed Effects
Any good news Arne? *Felipe Nunes* CAPES/Fulbright Fellow PhD Student Political Science - UCLA Web: felipenunes.bol.ucla.edu On Thu, Sep 29, 2011 at 5:10 AM, Arne Henningsen < arne.henning...@googlemail.com> wrote: > Hi Felipe > > On 25 September 2011 00:16, Felipe Nunes wrote: > > Hi Arne, > > my problem persists. I am still using censReg [version - 0.5-7] to run a > > random effects model in my data (>50,000 cases), but I always get the > > message. > > tob7 <- censReg(transfers.cap ~ pt.pt + psdb.pt + pt.opp + pt.coa + > psdb.coa > > + pib.cap + transfers.cap.lag + pib.cap + ifdm + log(populat) + > > mayor.vot.per + log(bol.fam.per+0.01) + factor(uf.name) + factor(year) - > 1, > > left=0, right=Inf, method="BHHH", nGHQ=8, iterlim=1, data = pdata2) > > Error in maxNRCompute(fn = logLikAttr, fnOrig = fn, gradOrig = grad, > > hessOrig = hess, : > > NA in the initial gradient > > If I sent you my data set, could you try to help me? I have been > struggling > > with that for two months now. > > Thanks for sending me your data set. With it, I was able to figure > out, where the NAs in the (initial) gradients come from: when > calculating the derivatives of the standard normal density function [d > dnorm(x) / d x = - dnorm(x) * x], values for x that are larger than > approximately 40 (in absolute terms) result in so small values (in > absolute terms) that R rounds them to zero. Later, these derivatives > are multiplied by some other values and then the logarithm is taken > ... and multiplying any number by zero and taking the logarithms gives > not a finite number :-( > > When *densities* of the standard normal distribution become too small, > one can use dnorm(x,log=TRUE) and store the logarithm of the small > number, which is much larger (in absolute terms) than the density and > hence, is not rounded to zero. However, in the case of the > *derivative* of the standard normal density function, this is more > complicated as log( d dnorm(x) / d x ) = log( dnorm(x) ) + log( - x ) > is not defined if x is positive. I will try to solve this problem by > case distinction (x>0 vs. x<0). Or does anybody know a better > solution? > > /Arne > > -- > Arne Henningsen > http://www.arne-henningsen.name > [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] getting list of data.frame names
Dear R People: This is probably a very simple question. I know that if I want to get a list of the classes of the objects in the workspace, I can do this: > sapply(ls(), function(x)class(get(x))) aa1.dfbd "list" "data.frame""integer""numeric" Now I want to get just the data frames. > sapply(ls(), function(x)class(get(x))=="data.frame") a a1.df b d FALSE TRUE FALSE FALSE However, I would like the names of the data frames, rather than the True/False for the objects. I've been trying all sorts of combinations/permutations with no success. Any suggestions would be much appreciated. Thanks, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] getting list of data.frame names
Hi Erin, Try this: names(which(sapply(.GlobalEnv, is.data.frame))) Cheers, Josh On Sat, Oct 1, 2011 at 8:37 PM, Erin Hodgess wrote: > Dear R People: > > This is probably a very simple question. I know that if I want to get > a list of the classes of the objects in the workspace, I can do this: > >> sapply(ls(), function(x)class(get(x))) > a a1.df b d > "list" "data.frame" "integer" "numeric" > > Now I want to get just the data frames. >> sapply(ls(), function(x)class(get(x))=="data.frame") > a a1.df b d > FALSE TRUE FALSE FALSE > > However, I would like the names of the data frames, rather than the > True/False for the objects. > > I've been trying all sorts of combinations/permutations with no success. > > Any suggestions would be much appreciated. > > Thanks, > Sincerely, > Erin > > > > -- > Erin Hodgess > Associate Professor > Department of Computer and Mathematical Sciences > University of Houston - Downtown > mailto: erinm.hodg...@gmail.com > > __ > 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.html > and provide commented, minimal, self-contained, reproducible code. > -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Fitting 3 beta distributions
Hi, I want to fit 3 beta distributions to my data which ranges between 0 and 1. What are the functions that I can easily call and specify that 3 beta distributions should be fitted? I have already looked at normalmixEM and fitdistr but they dont seem to be applicable (normalmixEM is only for fitting normal dist and fitdistr will only fit 1 distribution, not 3). Is that right? Also, my data has 26 million data points. What can I do to reduce the computation time with the suggested function? thanks a lot in advance, eagerly waiting for any input. Best Nitin -- ÎI+IÐ [[alternative HTML version deleted]] __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] deSolve - Function daspk on DAE system - Error
I'm getting this error on the attached code and breaking my head but can't
figure it out. Any help is much appreciated. Thanks, Vince
CODE:
library(deSolve)
Res_DAE=function(t, y, dy, pars) {
with(as.list(c(y, dy, pars)), {
res1 = -dS -dES-k2*ES
res2 = -dP + k2*ES
eq1 = Eo-E -ES
eq2 = So-S -ES -P
return(list(c(res1, res2, eq1, eq2)))
})
}
pars <- c(Eo=0.02, So=0.02, k2=250, E=0.01); pars
yini <- c(S=0.01, ES = 0.01, P=0.0, E=0.01); yini
times <- seq(0, 0.01, by = 0.0001); times
dyini = c(dS=0.0, dES=0.0, dP=0.0)
## Tabular output check of matrix output
DAE <- daspk(y = yini, dy = dyini, times = times, res = Res_DAE, parms =
pars, atol = 1e-10, rtol = 1e-10)
ERROR:
daspk-- warning.. At T(=R1) and stepsize H (=R2) the nonlinear solver
f
nonlinear solver failed to converge repeatedly of with abs (H) =
H
repeatedly of with abs (H) = HMIN preconditioner had repeated
failur
0.0D+00 0.5960464477539D-14
Warning messages:
1: In daspk(y = yini, dy = dyini, times = times, res = Res_DAE, parms =
pars, :
repeated convergence test failures on a step - inaccurate Jacobian or
preconditioner?
2: In daspk(y = yini, dy = dyini, times = times, res = Res_DAE, parms =
pars, :
Returning early. Results are accurate, as far as they go
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Re: [R] Entering data into a multi-way array?
I am trying to replicate the script, appended below. My data is in OOCalc files. The script (below) synthesizes a dataset (it serves as a "tutorial"), but I will need to get my data from OOCalc into R for use in that script (which uses arrays). I've worked my way through the script, and understand how most of it works (except the first bit - Step 1 - which is irrelevant to me, anyway). [begin script] ### Supplementary material with the paper ### Interpretation of ANOVA models for microarray data using PCA ### J.R. de Haan et al. Bioinformatics (2006) ### Please cite this paper when you use this code in a publication. ### Written by J.R. de Haan, December 18, 2006 ### Step1: a synthetic dataset of 500 genes is generated with 5 classes ### 1 unresponsive genes (300 genes) ### 2 constant genes (50 genes) ### 3 profile 1 (50 genes) ### 4 profile 2 (50 genes) ### 5 profile 3 (50 genes) #generate synthetic dataset with similar dimensions: # 500 genes, 3 replicates, 10 timepoints, 4 treatments X <- array(0, c(500, 3, 10, 4)) labs.synth <- c(rep(1, 300), rep(2, 50), rep(3, 50), rep(4, 50), rep(5, 50)) gnames <- cbind(labs.synth, labs.synth) #print(dim(gnames)) gnames[1:300,2] <- "A" gnames[301:350,2] <- "B" gnames[351:400,2] <- "C" gnames[401:450,2] <- "D" gnames[451:500,2] <- "E" ### generate 300 "noise" genes with expressions slightly larger than ### the detection limit (class 1) X[labs.synth==1,1,,] <- rnorm(length(X[labs.synth==1,1,,]), mean=50, sd=40) X[labs.synth==1,2,,] <- X[labs.synth==1,1,,] + rnorm(length(X[labs.synth==1,1,,]), mean=0, sd=10) X[labs.synth==1,3,,] <- X[labs.synth==1,1,,] + rnorm(length(X[labs.synth==1,1,,]), mean=0, sd=10) # generate 50 stable genes at two levels (class 2) X[301:325,1,,] <- rnorm(length(X[301:325,1,,]), mean=1500, sd=40) X[301:325,2,,] <- X[301:325,1,,] + rnorm(length(X[301:325,1,,]), mean=0, sd=10) X[301:325,3,,] <- X[301:325,1,,] + rnorm(length(X[301:325,1,,]), mean=0, sd=10) X[326:350,1,,] <- rnorm(length(X[326:350,1,,]), mean=3000, sd=40) X[326:350,2,,] <- X[326:350,1,,] + rnorm(length(X[326:350,1,,]), mean=0, sd=10) X[326:350,3,,] <- X[326:350,1,,] + rnorm(length(X[326:350,1,,]), mean=0, sd=10) # generate50 genes with profile 1 (class 3) increase.range <- matrix(rep(1:50, 10), ncol=10, byrow=FALSE) profA3 <- matrix(rep(c(10, 60, 110, 150, 150, 150, 150, 150, 150, 150) , 50), ncol=10, byrow=TRUE) * increase.range X[351:400,1,,1] <- profA3 + rnorm(length(profA3), mean=0, sd=40) profB3 <- matrix(rep(c(10, 100, 220, 280, 280, 280, 280, 280, 280, 280), 50), ncol=10, byrow=TRUE) * increase.range X[351:400,1,1:10,2] <- profB3 + rnorm(length(profA3), mean=0, sd=40) profC3 <- matrix(rep(c(10, 120, 300, 300, 280, 280, 280, 280, 280, 280), 50), ncol=10, byrow=TRUE) * increase.range X[351:400,1,1:10,3] <- profC3 + rnorm(length(profA3), mean=0, sd=40) profD3 <- matrix(rep(c(100, 75, 50, 50, 50, 50, 50, 50, 75, 100), 50), ncol=10, byrow=TRUE) X[351:400,1,1:10,4] <- profD3 + rnorm(length(profA3), mean=0, sd=40) #again replicates X[351:400,2,,] <- X[351:400,1,,] + rnorm(length(X[351:400,2,,]), mean=0, sd=10) X[351:400,3,,] <- X[351:400,1,,] + rnorm(length(X[351:400,3,,]), mean=0, sd=10) # generate50 genes with profile 2 (class 4) increase.range <- matrix(rep(1:50, 10), ncol=10, byrow=FALSE) profA4 <- matrix(rep(c(10, 60, 110, 150, 125, 100, 75, 50, 50, 50) , 50), ncol=10, byrow=TRUE) * increase.range X[401:450,1,,1] <- profA4 + rnorm(length(profA4), mean=0, sd=40) profB4 <- matrix(rep(c(10, 100, 220, 280, 200, 150, 100, 50, 50, 50), 50), ncol=10, byrow=TRUE) * increase.range X[401:450,1,1:10,2] <- profB4 + rnorm(length(profA4), mean=0, sd=40) profC4 <- matrix(rep(c(10, 150, 300, 220, 150, 100, 50, 50, 50, 50), 50), ncol=10, byrow=TRUE) * increase.range X[401:450,1,1:10,3] <- profC4 + rnorm(length(profA4), mean=0, sd=40) profD4 <- matrix(rep(c(150, 100, 50, 50, 75, 75, 75, 100, 100, 100), 50), ncol=10, byrow=TRUE) X[401:450,1,1:10,4] <- profD4 + rnorm(length(profA4), mean=0, sd=40) #again replicates X[401:450,2,,] <- X[401:450,1,,] + rnorm(length(X[401:450,2,,]), mean=0, sd=10) X[401:450,3,,] <- X[401:450,1,,] + rnorm(length(X[401:450,3,,]), mean=0, sd=10) # generate50 genes with profile 3 (class 5) increase.range <- matrix(rep(1:25, 20), ncol=10, byrow=FALSE) profA4 <- matrix(rep((200 - c(10, 60, 110, 150, 125, 100, 75, 50, 50, 50)), 50), ncol=10, byrow=TRUE) * increase.range X[451:500,1,,1] <- profA4 + rnorm(length(profA4), mean=0, sd=40) profB4 <- matrix(rep((200 - c(10, 100, 180, 200, 200, 150, 100, 50, 50, 50)), 50), ncol=10, byrow=TRUE) * increase.range X[451:500,1,1:10,2] <- profB4 + rnorm(length(profA4), mean=0, sd=40) profC4 <- matrix(rep((200 - c(10, 150, 200, 180, 150, 100, 50, 50, 50, 50)), 50), ncol=10, byrow=TRUE) * increase.range X[451:500,1,1:10,3]
Re: [R] Poor performance of "Optim"
Thank you for your response! But the problem is when I estimate a model without knowing the true coefficients, how can I know which "reltol" is good enough? "1e-8" or "1e-10"? Why can commercial packages automatically determine the right "reltol" but R cannot? -- View this message in context: http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3864243.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
What I tried is just a simple binary probit model. Create a random data and use "optim" to maximize the log-likelihood function to estimate the coefficients. (e.g. u = 0.1+0.2*x + e, e is standard normal. And y = (u > 0), y indicating a binary choice variable) If I estimate coefficient of "x", I should be able to get a value close to 0.2 if sample is large enough. Say I got 0.18. If I expand x by twice and reestimate the model, which coefficient should I get? 0.09, right? But with "optim", I got something different. When I do the same thing in both Gauss and Matlab, I can exactly get 0.09, evidencing that the coefficient estimator is reliable. But R's "optim" does not give me a reliable estimator. -- View this message in context: http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3863969.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
[R] Multivariate Laplace density
Can anyone show how to calculate a multivariate Laplace density? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Multivariate-Laplace-density-tp3864072p3864072.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poor performance of "Optim"
Oh, I think I got it. Commercial packages limit the number of decimals shown. -- View this message in context: http://r.789695.n4.nabble.com/Poor-performance-of-Optim-tp3862229p3864271.html Sent from the R help mailing list archive at Nabble.com. __ 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.html and provide commented, minimal, self-contained, reproducible code.
