similar to: merge(join) problem

Hi, I created the model below, which returns me the following warning message: In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : Could not compute QR decomposition of Hessian. Optimization probably did not converge. ######### Model ######## mDPDF = data.frame(mj1,mj2,mj3,mj4,mj5,eL1,eL2,eL3,eL4,eL5,aC1,aC2,aC3,aC4,disR1,disR2,disR3,disR4,disR5,

Hi, I need this plot: given: x,y - numerical vectors of length N plot xi vs mean(yj such that |xj - xi|<epsilon) (running mean?) alternatively, discretize X as if for histogram plotting and plot mean y over the center of the histogram group. is there a simple way? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://thereligionofpeace.com

I have a data.frame object like: > data.frame(x=c('A','A','B','B'), y=c('Ab','Ac','Ba','Bd')) x y 1 A Ab 2 A Ac 3 B Ba 4 B Bd how could I create a tree structure object like this: |---Ab A---| _| |---Ac | | |---Ba B---| |---Bb Thanks, Zech [[alternative HTML version deleted]]

Hi, I'm working on the codes below however every time I run them when they get to OpenBUGS I keep getting the error message: array index is greater than array upper bound for hab. Any help would be greatly appreciated, Suzie Codes: ungulate <- read.csv(file.choose ()) #ungulate ungulate <- as.matrix(ungulate);colnames(ungulate)<-NULL;rownames(ungulate)<-NULL

Hello, This is Elaine. I am trying a path analysis using lavaan Package. There are three explanatory variables: X, Z, and M. The response variable is Y. A, b, and c have direct effects on Y. On the other hand, X and Z also have direct effects on M. In other words, X and Z have indirect effects on Y. I found the code example of lavaan package describes only one indirect effect as below. Please

I see this: --8<---------------cut here---------------start------------->8--- > length(which(is.na(z$language))) [1] 0 > locals <- z[z$country == mycountry,] > length(which(is.na(locals$language))) [1] 229 --8<---------------cut here---------------end--------------->8--- where are those locals without the language coming from?! -- Sam Steingold (http://sds.podval.org/) on

Is there a way for an apply-type function to return a data frame? the closest thing I think of is foo <- as.data.frame(sapply(...)) names(foo) <- c(....) is there a more "elegant" way? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://honestreporting.com

Hi, I have a data frame df and a list of names of columns that I want to turn into factors: df.names <- attr(df,"names") sapply(factors, function (name) { pos <- match(name,df.names) if (is.na(pos)) stop(paste(name,": no such column\n")) df[[pos]] <- factor(df[[pos]]) cat(name,"(",pos,"):",is.factor(df[[pos]]),"\n")

I am trying to get stock metadata from Yahoo finance (or maybe there is a better source?) here is what I did so far: yahoo.url <- "http://finance.yahoo.com/d/quotes.csv?f=j1jka2&s="; stocks <- c("IBM","NOIZ","MSFT","LNN","C","BODY","F"); # just some samples socket <-

When I do > all(all$X.Time == all$Y.Time); [1] TRUE as expected, but > all.equal(all$X.Time,all$Y.Time); Error in target[[i]] : subscript out of bounds why? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://mideasttruth.com http://honestreporting.com http://dhimmi.com http://jihadwatch.org http://pmw.org.il http://ffii.org The dark past once was the

Hi, It appears that deal does not support missing values (NA), so I need to remove them (NAs) from my data frame. how do I do this? (I am very new to R, so a detailed step-by-step explanation with code samples would be nice). Some columns (variables) have quite a few NAs, so I would rather drop the whole column than sacrifice all the rows (observations) which have NA in that column. How do I

I have a data frame with several columns. I want to select the rows with no NAs (as with complete.cases) and all columns identical. E.g., for --8<---------------cut here---------------start------------->8--- > f <- data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40)) > f a b c 1 1 1 1 2 NA NA NA 3 NA 3 5 4 4 40 40 --8<---------------cut

Hi I wonder if there's a smarter way to do this procedure. I have a vector of filenames wher I only am interested in the first part of the filename. I would use the following method of extracting the first part. But is there a more simple way of doing this? Names <-

Can someone recommend a good quality 24 or greater port channel bank? Steve -------------- next part -------------- An HTML attachment was scrubbed... URL: http://lists.digium.com/pipermail/asterisk-users/attachments/20070506/e5896021/attachment.htm

Hi, I am confused by the way the indexing works. I read a table from a csv file like this: ysmd <- read.csv("ysmd.csv",header=TRUE); ysmd.table <- hash(); for (i in 1:length(ysmd$X.stock)) ysmd.table[ysmd$X.stock[i]] <- ysmd[i,]; the first column ("X.stock") is a string (factor): > ysmd$X.stock[[100]] [1] FLO 7757 Levels: A AA AA- AAAAA AAC AACC AACOU AACOW AADR

I find myself doing --8<---------------cut here---------------start------------->8--- tab <- table(...) tab <- tab[tab > 0] tab <- sort(tab,decreasing=TRUE) --8<---------------cut here---------------end--------------->8--- all the time. I am wondering if the "drop 0" (and maybe even sort?) can be effected by some magic argument to table() which I fail to discover

when do I need to use which()? > a <- c(1,2,3,4,5,6) > a [1] 1 2 3 4 5 6 > a[a==4] [1] 4 > a[which(a==4)] [1] 4 > which(a==4) [1] 4 > a[which(a>2)] [1] 3 4 5 6 > a[a>2] [1] 3 4 5 6 > seems unnecessary... -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://jihadwatch.org http://palestinefacts.org http://mideasttruth.com

I get this error from read.table(): Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 234 did not have 8 elements The error is genuine (an extra field separator between 1st and 2nd element). 1. is there a way to see this bad line 234 from R without diving into the file? 2. is there a way to ignore the bad lines and get the data from the good lines only (I do

Hi, to count vector elements with some property, the standard idiom seems to be length(which): --8<---------------cut here---------------start------------->8--- x <- c(1,1,0,0,0) count.0 <- length(which(x == 0)) --8<---------------cut here---------------end--------------->8--- however, this approach allocates and discards 2 vectors: a logical vector of length=length(x) and an

Is there an analogue of common lisp "*" variable which contains the value of the last expression? E.g., in lisp: > (+ 1 2) 3 > * 3 I wish I could recover the value of the last expression without re-evaluating it. thanks -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://camera.org http://ffii.org