similar to: how to concatenate factor vectors?

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

> 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

> * Jorge Cimentada <pvzragnqnw at tznvy.pbz> [2017-11-09 00:02:53 +0100]: > > I'm happy to announce the release of ess 0.0.1 a package designed to > download data from the European Social Survey Given the existence of ESS (Emacs Speaks Statistics - https://ess.r-project.org/) the package name "ess" seems unfortunate. -- Sam Steingold (http://sds.podval.org/) on

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

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

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 <-

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

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="") ="

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")

The following code: --8<---------------cut here---------------start------------->8--- > myfun <- function (x) list(x=x,y=x*x) > z <- as.data.frame(do.call(rbind,lapply(1:3,function(x) c(a=paste("a",x,sep=""),as.list(unlist(list(b=myfun(x),c=myfun(x*x*x)))))))) > z a b.x b.y c.x c.y 1 a1 1 1 1 1 2 a2 2 4 8 64 3 a3 3 9 27 729

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

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 have the following code: --8<---------------cut here---------------start------------->8--- d <- rep(10,10) for (i in 1:100) { a <- sample.int(length(d), size = 2) if (d[a[1]] >= 1) { d[a[1]] <- d[a[1]] - 1 d[a[2]] <- d[a[2]] + 1 } } --8<---------------cut here---------------end--------------->8--- it does what I want, i.e., modified vector d 100 times.

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

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 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

> 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

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")) >