similar to: matrix.csr %*% matrix --> matrix

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

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

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/

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. --

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

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

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 know it does not look very good - using the same column names to mean different things in different data frames, but here you go: --8<---------------cut here---------------start------------->8--- > x <- data.frame(a=c(1,2,3),b=c(4,5,6)) > y <- data.frame(b=c(1,2),a=c("a","b")) >

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 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

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

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

How do I concatenate two vectors of factors? --8<---------------cut here---------------start------------->8--- > a <- factor(5:1,levels=1:9) > b <- factor(9:1,levels=1:9) > str(c(a,b)) int [1:14] 5 4 3 2 1 9 8 7 6 5 ... > str(unlist(list(a,b),use.names=FALSE)) Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...

> write.matrix.csr(mx, y = y, file = file) > table(y) 0 1 5194394 23487 $ cut -d' ' -f1 f | sort | uniq -c 23487 2 5194394 1 i.e., 0 is written as 1 and 1 is written as 2. why? is there a way to disable this? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org

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

Hi, I find this behavior unexpected: --8<---------------cut here---------------start------------->8--- > strsplit(c("a,b;c","d;e,f"),c(",",";")) [[1]] [1] "a" "b;c" [[2]] [1] "d" "e,f" --8<---------------cut here---------------end--------------->8--- I thought that it should be identical to this:

I read a dataset with times in them, e.g., "09:31:29.18761". I then parse them: > all$X.Time <- strptime(all$X.Time, format = "%H:%M:%OS6"); and get a vector of NAs (how do I check that except for a visual inspection?) then I do > options("digits.secs"=6); > all$X.Time <- strptime(all$X.Time, format = "%H:%M:%OS"); and it, apparently, works:

I am sure a common need is to plot a scatterplot with some fitted line(s) and maybe save to a file. I have this: plot.glm <- function (x, y, file = NULL, xlab = deparse(substitute(x)), ylab = deparse(substitute(y)), main = NULL) { m <- glm(y ~ x) if (!is.null(file)) pdf(file = file) plot(x, y, xlab = xlab, ylab = ylab, main = main) lines(x, y =

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