similar to: create a data frame with the given column names

> 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

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 <-

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 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")) >

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:

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")

Hi, I have data with multiple sub-second entries: 2011/03/15 09:32:15.035619,-0.403103,1.09664,48.6,126.92,117.32 2011/03/15 09:32:15.069331,-0.39851,1.09874,48.6,126.92,117.32 2011/03/15 09:32:15.289135,-0.402463,1.10084,48.59,126.92,117.32 2011/03/15 09:32:15.296110,-0.450244,1.10063,48.59,126.92,117.32 2011/03/15 09:32:15.451358,-0.438813,1.10273,48.59,126.93,117.32 2011/03/15

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

Tinn-R (http://www.sciviews.org/Tinn-R/) is one of the topmost suggestions when googling an R-(text-)editor for Windows. However, to me it appears dissappointing that Tinn-R does not handle utf-8 (mac-roman, or any other) encoded R-scripts or, in general, text files. Besides Emacs and the R built-in editor, could you recommend a good editor for Windows, even some commmercial for a small

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

Hello, I am looking for an elegant one-liner for the following operation: x <- rnorm(10) y <- runif(10) c(mean(x)-mean(y), mean(x)+mean(y)) I thought about apply(data.frame(x, y), 2, mean) but I don't know how to apply the +- operation on the result of apply. Thanks, *S* -- Sascha Vieweg, saschaview at gmail.com

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. --

Hello, Vector y is an alphabetically sorted version of vector x. Will both samples, X and Y, be "absolutely" random or will they have systematic differences? And: Should I sort or shuffle a vector before sampling? Thank you, *S* x <- as.factor(LETTERS[sequence(10:1)]) y <- sort(x) X <- sample(x, 5) Y <- sample(y, 5) -- Sascha Vieweg, saschaview at gmail.com

> 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

Hi, When qqnorm on a vector of length 10M+ I get a huge pdf file which cannot be loaded by acroread or evince. Any suggestions? (apart from sampling the data). Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://mideasttruth.com http://honestreporting.com http://camera.org http://openvotingconsortium.org http://pmw.org.il

Hi there, I have a numeric vector let say: Vect <- c(12.234, 234.5675, 1.5) Now I want a string vector like: changedVec <- c("012.234", "234.568", "001.500") Would be grateful if somebody help me how can I do that. Thanks and regards, [[alternative HTML version deleted]]

Hello How can I apply functions along "layers" of a data matrix? Example: daf <- data.frame( 'id' = rep(1:5, 3), matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4, sep='') )), rep = rep(1:3, each=5) ) The data frame "daf" contains 3 repetitions/layers (rep) of 4 variables of 5 persons (id). For some reason, I want to calculate

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 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