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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

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

Suppose I have a vector of strings: c("A1B2","A3C4","B5","C6A7B8") [1] "A1B2" "A3C4" "B5" "C6A7B8" where each string is a sequence of <column><value> pairs (fixed width, in this example both value and name are 1 character, in reality the column name is 6 chars and value is 2 digits). I need to

...;xfsdump" (if you use an XFS file system). Some versions of Arcserve can only backup local files... hope this helps -------FROM------------------------------- Patrick Niessen IT-Manager / Planning & Controlling Gerling Allgemeine Versicherungs-AG, Japan Branch fQ?[fSf"fO?EfAf<fQf}fCfl*?OE???Z? "?-{Zx"X TEL. +81-3-5214-1361 / FAX. +81-3-5214-1365 URL: http://www.gerling.co.jp > > Message: 18 > From: "Norman Zhang" <nzhang@arkon-group.com> > To: <samba@lists.samba.org> > Date: Mon, 18 Nov 2002 16:40:40 -0800 > Subject: [S...

suppose I have two factor vectors: x <- as.factor(c("a","b","a","c","b","c")) y <- as.factor(c("b","a","a","c","c","b")) I can compute their entropies: entropy(table(x)) [1] 1.098612 using library(entropy) but it is not clear how to compute their mutual information

I just got this error: > library(igraph) > comp <- decompose.graph(gr) Error: protect(): protection stack overflow Error: protect(): protection stack overflow > what can I do? the digraph is, indeed, large (300,000 vertexes), but there are very many very small components (which I would rather not discard). PS. the doc for decompose.graph does not say which mode is the default. --

How do I read/write liblinear models to files? E.g., if I train a model using the command line interface, I might want to load it into R to look the histogram of the weights. Or I might want to train a model in R and then apply it using a command line interface. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/

When a sparse matrix is multiplied by a regular one, the result is usually not sparse. However, when matrix.csr is multiplied by a regular matrix in R, a matrix.csr is produced. Is there a way to avoid this? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://truepeace.org

Dear Samba Team, We're running Samba version 2.2.4 on a Solaris 8 file server. Recently all PC clients were upgraded from NT4 to WinXP and although all drives are mapping (following the EnablePlainTextPassword fix) the Samba drives are not being reconnected correctly during a reboot of the client. We are used to the NT clients prompting for a password during boot to authenticate any

all of a sudden, after a SparseM upgrade(?) I get this error: > str(z) Formal class 'matrix.csr' [package "SparseM"] with 4 slots ..@ ra : num [1:85372672] -0.4288 0.0397 0.0104 -0.1843 -0.1203 ... ..@ ja : int [1:85372672] 1 2 3 4 5 6 7 8 9 10 ... ..@ ia : int [1:699777] 1 123 245 367 489 611 733 855 977 1099 ... ..@ dimension: int [1:2] 699776 122

I need an analogue of "uniq -c" for a data frame. xtabs(), although dog slow, would have footed the bill nicely: --8<---------------cut here---------------start------------->8--- > x <- data.frame(a=1:32,b=1:32,c=1:32,d=1:32,e=1:32) > system.time(subset(as.data.frame(xtabs( ~. , x )), Freq != 0 )) user system elapsed 12.788 4.288 17.224 --8<---------------cut

I have a graph with edge and vertex weights, stored in two data frames: --8<---------------cut here---------------start------------->8--- vertices <- data.frame(vertex=c("a","b","c","d"),weight=c(1,2,1,3)) edges <-

I did this: nb <- naiveBayes(users, platform) pl <- predict(nb,users) nrow(users) ==> 314781 ncol(users) ==> 109 1. naiveBayes() was quite fast (~20 seconds), while predict() was slow (tens of minutes). why? 2. the predict results were completely off the mark (quite the opposite of the expected overfitting). suffice it to show the tables: pl: android blackberry ipad

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 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 am trying to aggregate some data and I am confused by the results. I load a data frame "all" from a csv file, and then I do: (FOO,BAR,X,Y come from the header line in the csv file, BTW, how do I rename a column?) byFOO <- aggregate(list(all$BAR,all$QUUX,all$X/all$Y), by = list(FOO=all$FOO), FUN = mean); I expect a data frame with 4