similar to: Weighting data when running regressions

Hi, I'm looking for some way to pick up the numbers which are contained and buried in a long character. For example, outtree.new="(((B:1204.25,E:1204.25):7581.11,F:8785.36):8353.85,C:17139.21);" num.char =

I have a data frame named "database" with panel data, a little piece of which looks like this: Symbol Name Trial Factor1 Factor2 External 1 548140 A 1 -3.87 -0.32 0.01 2 547400 B 1 12.11 -0.68 0.40 3 547173 C 1

hello all sorry for the following "none" R related question. does anyone know of a reference to calculate the following identity: |I + ABC| where I is an identity matrix and A, B,C may not have to be square matrices? you help will be greatly appreciated. H. V. Henderson; S. R. Searle SIAM Review, Vol. 23, No. 1. (Jan., 1981), pp. 53-60. provides a result to

Hello, I just discovered that I cannot close the Excel application and task manager shows numerous copies of Excel.exe I tried both x$Quit() # shown in the rcom archive and x$Exit() and Excel refuses to die. Thank you very much. S. "You can't kill me, I will not die" Mojo Nixon I also have a problem with saving. It produces a pop-up dialog and does not take my second

After many hours of debugging code, I came to the conclusion that I have a fundamental misunderstanding regarding eval, and hope that someone here can explain to me, why the following code acts as it does: foo <- function(expr) { eval(substitute(expr), envir=list(a=5), enclos=parent.frame()) } bar <- function(er) { foo(er) } > foo(a) [1] 5 > bar(a) Error in eval(expr,

Hi there, I am stuck trying to solve what should be a fairly easy problem. I have a data frame that essentially consists of (ID, time as seqMonth, variable, value) and i want to find the regression coefficient of value vs time for each combination of ID and Variable. I have tried several approaches and none of them seems to work as i expected. For example, i have tried:

Since we're on the topic of book reviews, I just received Phil Spector's new R book called "Data Manipulation with R" and it is also quite a nice book. I haven't gone through it all and I won't give a detailed review but I have gotten a lot out of the first 100 pages that I have read. Note that I've been using R for almost 1.5 years so , for me, it's a

Greetings, I have a problem that I am sure is very straightforward, but I just can't wrap my head around it. I've read the help pages on text, plotmath, expression, substitute, but somehow I can't find the answer to this simple question. Basically consider the following example: plot( NULL, xlim = c(0,2), ylim = c(0,2) ) expressions <- expression( -infinity, infinity )

Dear All, I am a beginner creating R packages. I followed the Leisch (2009) tutorial and the document ?Writing R Extensions? to write an example. I installed R 2.12.2 (I also tried R2.13.2), the last version of Rtools and the recommended packages in a PC with Windows 7 Home Premium. I can run R CMD INSTALL linmod in the command prompt and the R CMD check linmod. The following outputs are

Dear useRs, I have some trouble with the calculation of Cook's distance in R. The formula for Cook's distance can be found for example here: http://en.wikipedia.org/wiki/Cook%27s_distance I tried to apply it in R: > y <- (1:400)^2 > x <- 1:100 > lm(y~x) -> linmod # just for the sake of a simple example >

Hello all, I ran into the following, to me unexpected, behavior. I have (for reasons that don't necessarily pertain to the question at hand, hence I won't go into them) the need/desire to use an empty string for the name of a vector entry. Perhaps I did not read ?"[" very carefully, but it seems to me that he following lines should return "1" at the end:

I am reading Section 5 and 6 of http://cran.r-project.org/doc/contrib/Leisch-CreatingPackages.pdf It seems that I have to do the following two steps in order to make an R package. But when I am testing these package, these two steps will run many times, which may take a lot of time. So when I still develop the package, shall I always source('linmod.R') to test it. Once the code in

I just acquired a copy of "Statistical Models in S", I guess most commonly known as the "white book", and realized to my dismay that most of the code is not directly executable in R, and I was wondering if there was a source discussing the things that are different and what the new ways of calling things are. For instance, the first obstacle was the solder.balance data

I am getting some unexpected results from some functions of igraph and it is possible that I am misinterpreting the vertex numbers. Eg., the max betweenness measure seems to be from a vertex that is not connected to a single other vertex. Below if my code snippet: require(igraph) my.graph <- graph.adjacency(adjmatrix = my.adj.matrix, mode=c("undirected")) most.between.vert <-

I'm trying to do a linear regression between the columns of matrices. In example below I want to regress column 1 of matrix xdat with column1 of ydat and do a separate regression between the column 2s of each matrix. But the output I get seems to give correct slopes but incorrect intercepts and another set of slopes with value NA. How do I do this correctly? I'm after the slope and

Hi, folks, As seen in the following codes: x1=rlnorm(10) x2=rlnorm(10,mean=2) y=rlnorm(10,mean=10)### Fake dataset linmod=lm(log(y)~log(x1)+log(x2)) After the regression, I would like to know the mean of y. Since log(y) is normal and y is lognormal, I need to know the mean and variance of log(y) first. I tried mean (y) and mean(linmod), but either one is what I want. Any tips? Thanks in

Dear R helpers, I have a problem with exporting a chart (to any format). The graphic device becomes inactive and I get the 'Error: invalid graphics state' error message. I searched the help, web and FAQ but couldn't find the solution. This is my code: I chart a histogram for differences in R2 by sample size (an extract from the data is below). Altogether I have n=2500 observations

Greetings, I am trying to add a boxplot to the bottom of a histogram, right between the histogram bars and the x axis. Here is the code I am using at the moment (the par line is probably not relevant for our discussion): hs <- hist(x, breaks = 20, plot = F) par(mar = c(3,3,2,1)) hist(x, breaks = 20, main = NULL, ylim = c(-2, max(hs$counts))) boxplot(x, horizontal = T, axes = T, add =

I have the following relatively simple problem. Say we have three factors, and we want to create a cross-tabulation against each of the other two: x <- factor(rbinom(5, 1, 1/2)) y <- factor(rbinom(5, 1, 1/2)) z <- factor(rbinom(5, 1, 1/2)) table(x,y) table(x,z) This looks like: y x 0 1 0 2 0 1 1 2 z x 0 1 0 1 1 1 2 1 I would like to get (surely this will

Hi, Below I have a function mlogl_k, later it's called with "nlm" . __BEGIN__ vsamples<- c(14.7, 18.8, 14, 15.9, 9.7, 12.8) mlogl_k <- function( k_func, x_func, theta_func, samp) { tot_mll <- 0 for (comp in 1:k_func) { curr_mll <- (- sum(dgamma(samp, shape = x_func, scale=theta_func, log = TRUE))) tot_mll <- tot_mll + curr_mll }