similar to: strsplit with a vector split argument

> x <- rnorm(1000) > h <- hist(x,plot=FALSE) > sum(h$density) [1] 2 ----------------------------- shouldn't it be 1?! > h <- hist(x,plot=FALSE, breaks=(-4:4)) > sum(h$density) [1] 1 ----------------------------- now it's 1. why?! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://www.memritv.org

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

How can I create lists with element names created on the fly? --8<---------------cut here---------------start------------->8--- > list (foo = 10) $foo [1] 10 > list ("foo" = 10) $foo [1] 10 > list (paste("f","oo",sep="") = 10) Error: unexpected '=' in "list (paste("f","oo",sep="") ="

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")

At the end of a for loop its variables are still present: for (i in 1:10) { x <- vector(length=100000000) } ls() will print "i" and "x". this means that at the end of the for loop body I have to write rm(x) gc() is there a more elegant way to handle this? Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

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:

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/

(summary.default): show the vector length in addition to quantiles diff -u -i -p -F '^(def' -b -w -B /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old /home/sds/src/R-3.0.1/src/library/base/R/summary.R --- /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old 2013-03-05 18:02:33.000000000 -0500 +++ /home/sds/src/R-3.0.1/src/library/base/R/summary.R 2013-09-10 10:19:02.682946339

> 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

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

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 want to count attributes of IDs: --8<---------------cut here---------------start------------->8--- z <- data.frame(id=c(10,20,10,30,10,20), a1=c("a","b","a","c","b","b"), a2=c("x","y","x","z","z","y"),

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 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

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. --

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

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 =

How do I convert a list to a matrix? --8<---------------cut here---------------start------------->8--- list(c(50000, 101), c(1e+05, 46), c(150000, 31), c(2e+05, 17), c(250000, 19), c(3e+05, 11), c(350000, 12), c(4e+05, 25), c(450000, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,]

this function is supposed to canonicalize the language: --8<---------------cut here---------------start------------->8--- canonicalize.language <- function (s) { s <- tolower(s) long <- nchar(s) == 5 s[long] <- sub("^([a-z]{2})[-_][a-z]{2}$","\\1",s[long]) s[nchar(s) != 2 & s != "c"] <- "unknown" s }