similar to: pairwise scatterplot matrix

I've a vector of pairwise correlations in the order low-index element precedes the high-index element, say: corr(1,2)=0.1, corr(1,3)=0.2, corr(2,3)=0.3, corr(3,4)=0.4 How can I construct the corresponding correlation matrix? I tried using the "combn"-function in "combinat" package: library(combinat) combn(c(0.1,0.2,0.3,0.4),2) , but to no avail... Thank you for your

I'm applying a function (Cov.f) defined below to each element of a distance matrix. When I run the code below, I get a warning message (below) and elements of returned matrix [2,3] and [3,2] are not zero as I would expect. Clearly, there is an error... What am I doing wrong? Thanks. --Dale Warning message: In if (h <= phi) { : the condition has length > 1 and only the first element

Hi All I am having 2 very long character strings (550chars) and I want to put them as expressions together with c(). The problem is that I also get these double-quotes, as seen below in 'fct'. How can I remove these double-quotes? I tried as.name() but it did not work (because of size?). These are creating trouble with subsequent programs, which I tested with strings that for some

Hi I am using the following script to average a set of data 0f 62 columns into 31 colums. The data consists of values of ln(0.01) or -4.60517 instead of NA's. These need to be averaged for each row (i.e 2 values being averaged). What I would I need to change for me to meet the conditions: 1. If each run of the sample has a value, the average is given 2. If only one run of the sample has a

I am hoping someone can help translate some WinBUGS code into R code. I would like to use R to create the C[] matrix required for a car.proper model in WinBUGS, but I am having a difficult time negotiating the coding. The C matrix provides normalized weights for each pair of spatial areas. So the WinBUGS example is as follows: # of the weight matrix with elements Cij. The first J1 elements

Hi. My friend has a problem that smbtar cannot recognize space character in specified filenames. He tried to execute following command line to back file "long file name" up: $ smbtar -s SERVER -p PASSWORD -x SHARE -u username -t /tmp/backup.tar \ "long file name" but smbtar tried to include three files as 'long', 'file' and 'name'. :-( Here is the

Dear all: I tried to install R-1.1.0 im my Suse 6.0 linux box. My first step is: ./configure -with-f2c after i do > make command and i have the following error message: ############# error message make[5]: `Makedeps' is up to date. make[5]: Leaving directory `/root/R-1.1.0/src/unix/X11' make[5]: Entering directory `/root/R-1.1.0/src/unix/X11' gcc -I. -I../../../src/include

Dear all: I need to perform a Yates analysis of a 2-level full factorial experiment design. My question is: how i can do it with R, or best, what's the package who i need to use for this purpose. Thanks. ,__ __ /\ /| | | | | __ ,_ __, _ | | | | | / \_/ | / | |/ | | | \_|/\__/ |_/\_/|/|__/ |

I see that mksmbpasswd.sh (plus a little editing) would be a good way to set up users with "NO PASSWORD" in bulk. But suppose I don't want the risk of the "NO PASSWORD" approach. Now somebody with root access must run smbpasswd for each user they want to activate. But can't I just do this anyway (with the -a 'add user' option), without ever having run

Dear R Core, pairwise.wilcox.test does not handle "paired = TRUE" correctly; e.g. set.seed(13) x <- rnorm(20) g <- c(rep(1, 10), rep(2, 10)) wilcox.test(x ~ g)$p.value # 0.075 pairwise.wilcox.test(x, g)$p.value # 0.075, o.k wilcox.test(x ~ g, paired = TRUE)$p.value # 0.105 pairwise.wilcox.test(x, g, paired = TRUE)$p.value # 0.075, wrong The line wilcox.test(xi, xj,

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20081112/618073fe/attachment.pl>

Hello all, here's a collection of answers I got on my question concerning partial correlation coefficients: Some people gave a simple formula for the three-variable-case, as did Dave Lucy: pcor <- function(v1, v2, v3) { c12 <- cor(v1, v2) c23 <- cor(v2, v3) c13 <- cor(v1, v3) partial <- (c12-(c13*c23))/(sqrt(1-(c13^2)) * sqrt(1-(c23^2)))

I just thought of a useful metaphore for the problem I face. I am dealing with a problem in business finance, with two kinds of related events. However, imagine you have a known amount of carbon (so many kilograms), but you do not know what fraction is C14 (and thus radioactive). Only the C14 will give decay events (and once that event has occurred, the atom that decayed will never decay

Hello, after thorough searching of the R help files as well as S+-help, I'm coming to the list: Is there a possibility to compute partial correlation coefficients between multiple variables (correlation between two paired samples with the "effects of all other variables partialled out")? All I seem to find are the standard Pearson correlation coefficients (with cor()) and no clue

Hi, > The patch in PR is wrong since it will make both darwin and windows > unhappy. I will comment on gentoo PR about possible proper solution. I'll give it a try as you comment on the gentoo PR. But my ppc box is in my office, so I'll post a result on next Monday (JST). Thanks,

Hi, >> The patch in PR is wrong since it will make both darwin and windows >> unhappy. I will comment on gentoo PR about possible proper solution. > > I'll give it a try as you comment on the gentoo PR. > But my ppc box is in my office, so I'll post a result on next Monday (JST). I posted my result on the gentoo PR. https://bugs.gentoo.org/show_bug.cgi?id=403519#c12

Dear R-Friends, Sorry to bother everyone with my earlier question: "Do we have similar R function to work like Minitab's INDICATOR?" I found the way to make things work: e.g., > x <- c(2,2,5,3,6,5,NA) > model.matrix(~ factor(x) - 1) factor(x)2 factor(x)3 factor(x)5 factor(x)6 1 1 0 0 0 2 1 0 0 0

I'm here and alive. I submitted my problem somewhere and have been getting emails related to nouveau ever since. Yes there is a craigcgarner at gmail.com I am glad to be following along with these bugs as they directly affect me and my system :) This where I reported the problem nouveau at lists.freedesktop.org The other buglist just started ? Thank you, Craig On Wed, Dec 9, 2015 at

Dear friends: It's possible make a legend in the coplot output? My data are: "dados"<-structure (list ( Qu1 = c(1,2,3,4,5,6,7,8,3,2), Qu2 = c(3,4,5,6,7,8,9,9,6,3), Ku1 = structure (factor ( c(1,1,1,1,2,1,2,2,1,1),levels=1:2),.Label=c("Kmin","Kmax")), Ph1 = structure (factor (

Hi list, Using the script below, I have generated two lists (c and h) containing yearly matrices. Now I would like to divide the matrices in c into multiple matrices based on h. The number of matrices should be equal to: length(unique(DF1$B))*length(h). So each unique value in DF1$B get's a yearly matrix. Each matrix should contain all values from c where element cij is 1. An example for