search for: patrizio

...erate random coef. of linear combination bet = rnorm(M) # create matrix for contour-type plot Bx = apply(xgr2,1,btens) Bmat = matrix(t(Bx)%*%bet,ng) # plot the result contour(xgr,xgr,Bmat) persp(xgr,xgr,Bmat,theta=15) any of you have a better idea (ie more efficient)? Thanks in advance, Patrizio Frederic +------------------------------------------------- | Patrizio Frederic | Research associate in Statistics, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: patrizio.f...

...y condition of f, pk is unique, pk IS NOT calculated with least square. A quick google tour show a rich field of research and many algorithms proposed for computing such a task. I was wondered if some of you knows about some R implementations (packages) for computing BSA. Many thanks in advance, Patrizio as usual I apologize for my fragmented English -- +------------------------------------------------- | Patrizio Frederic, PhD | Assistant Professor, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax:...

Dear all, is there any way to transform a .Data directory created in S-plus 6.1 for windows in a .RData file? Thanks in advance, Patrizio Frederic

Hi all, I am running a bivariate regression with the following: p1=c(184,155,676,67,922,22,76,24,39) p2=c(1845,1483,2287,367,1693,488,435,1782,745) I1=c(1530,1505,2505,204,2285,269,1271,298,2023) I2=c(8238,6247,6150,2748,4361,5549,2657,3533,5415) R1=I1-p1 R2=I2-p2 x1=cbind(p1,R1) y1=cbind(p2,R2) fit1=lm(y1~-1+x1) summary(fit1) Response 2: Coefficients: Estimate Std. Error t value

...i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 7.0 year 2008 month 04 day 22 svn rev 45424 language R version.string R version 2.7.0 (2008-04-22) thank you in advance, Patrizio Frederic ------------------------------------------------- | Patrizio Frederic | Research associate in Statistics, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: patrizio.fr...

...V2 V3 1 "1" "red" "0.4" 2 "2" "blue" NA 3 "3" "blue" "1.1" # and if I sapply, I loose factors labels > sapply(data,choose.last) V1 V2 V3 1 1 3 0.4 2 2 1 NA 3 3 1 1.1 any hint? Thanks in advance, Patrizio +------------------------------------------------- | Patrizio Frederic, PhD | Research associate in Statistics, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: patrizio.frede...

...i386 os mingw32 system i386, mingw32 status major 2 minor 7.1 year 2008 month 06 day 23 svn rev 45970 language R version.string R version 2.7.1 (2008-06-23) what's going wrong? Thanks in advance, Regards, Patrizio Frederic

...a is a large country I could be interested in cutting it in two peaces eastern and western. Can I do it using wrld_simpl or should I search for other data? If so where can I find free maps data? Some documentation exists but it seems sparse to me. Any hints will be appreciated. Thanks in advance, Patrizio

...I remember the specific package was mentioned. 2) .RData created with pc2 is not loaded in pc1. I got this message: Error in load("K:\\.RData") : ReadItem: unknown type 136, perhaps written by later version of R I've no idea of what type 136 is. Any hints? Thanks in advance, Patrizio -- +------------------------------------------------- | Patrizio Frederic, PhD | Assistant Professor, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: patrizio.frederic at un...

Dear all, I'm stuck in a seemingly trivial task that I need to perform for many datasets. Basically, I want to replace NA with 0 in a specified range of columns in a dataframe. I know the first and last column to be recoded only by its name. I can select the columns starting like this a[match('first',names(a)): match('last',names(a))] The question is how can replace all NA

Hi, In order to use the mle2-function, one has to define the likelihood function itself. As we know, the likelihood function is a sum of the logarithm of probability density functions (pdf). I have implemented myself the pdfs that I am using. My problem is, that the pdfs values are negative and I cann't take the logarithm of them in the log-likelihood function. So how can one take the

Hi R-users, I have this code to uniformise the data using gamma: > length(dp1) [1] 696 > dim(dp1) [1] 58 12 > dim(ahall) [1]  1 12 > dim(bhall) [1]  1 12 > trans_dt <- function(dt,a,b) + { n1 <- ncol(dt) +   n2 <- length(dt) +   trans  <- vector(mode='numeric', length=n2) +   dim(trans) <- dim(dt) +   for (i in 1:n1) +   {  dt[,i] <- as.vector(dt[,i])

Dear R Users, I am executing the following command to produce a line graph: matplot(aggregate_1986[,1], aggregate_1986[,2:3], type="l", col=2:3) On the x-axis I have values of Latitude (in column 1) ranging from -60 to +80 (left to right on the x-axis). However, I wish to have these values shown in reverse on the x-axis, going from +80 to -60 (ie. North to South in terms of Latitude).

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) ,

Hi, I would like to generate two correlated variables. I found that funktion for doing that: a <- rmvnorm(n=10000,mean=c(20,20),sigma=matrix(c(5,0.8*sqrt(50), 0.8*sqrt(50),10),2,2)) (using library(mvtnorm)) Now I also want to generate two correlated variables where the error variance vary over the variable-correlation. And I want to plot this for showing heteroscedasticity. Like shown

Hi, At each iteration in my program,I need to generate tree vectors,X1,X2,X3, from exponential distribution with parameters a1,a2,a3. Can you help me please how can I do it such that it take a little time? thank you khazaei

Dear group, Here are my objects in my environment: > ls() [1] "Pos100415" "Pos100416" "posA" "pose15" "pose16" "pose16t" "position" "trade" "x" I need to pass the object "Pos100415" to a function. This element is a data.frame, obtained through a function: Pos(x)<-myfun(x)

...nfig [1]) and the idea itself. Sincerly Nico P.S.: Please CC, I am not subscribed. [0]: http://unix.schottelius.org/ccollect/ [1]: http://nico.schotteli.us/papers/linux/cconfig/ -- ``...if there's one thing about Linux users, they're do-ers, not whiners.'' (A quotation of Andy Patrizio I completely agree with) -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 189 bytes Desc: Digital signature Url : http://lists.samba.org/archive/rsync/attachments/20061020/e9c0d691/attachment.bin

Hi Readers, I have a question. I have a large dataset and want to throw away columns that have the same value in the column itself and I want to know which column this was. For example > x<-data.frame(id=c(1,2,3), snp1=c("A","G", "G"),snp2=c("G","G","G"),snp3=c("G","G","A"))

Hi again! Suppose I have the following: > xy <- round(rexp(20),1) > xy [1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9 2.5 0.1 1.5 0.4 > table(xy) xy 0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4 1 4 2 1 2 1 2 1 1 2 1 1 1 > Is there a way to set things up to have 0 - 0.4 0.5 - 0.9 etc. please? I know there is the cut