similar to: list to matrix?

suppose I have two sets of vectors: x1,x2,...,xN and y1,y2,...,yN. I want N correlations: cor(x1,y1), cor(x2,y2), ..., cor(xN,yN). my sets of vectors are arranged as data frames x & y (vector=column): x <- data.frame(a=rnorm(10),b=rnorm(10),c=rnorm(10)) y <- data.frame(d=rnorm(10),e=rnorm(10),f=rnorm(10)) cor(x,y) returns a _matrix_ of all pairwise correlations: cor(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

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

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

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 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

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

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, what does this mean? --8<---------------cut here---------------start------------->8--- > graph <- graph.data.frame(merged[!v,], vertices=ve, directed=FALSE) cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't data.table aware cedta decided 'igraph' wasn't

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 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

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 ...

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 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 large data.frame Z (2,424,185,944 bytes, 10,256,441 rows, 17 columns). I want to get the result of table(aggregate(Z$V1, FUN = length, by = list(id=Z$V2))$x) alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is 24.3G, and no end in sight. both V1 and V2 are characters (not factors). Is there anything I could do to speed this up? Thanks. -- Sam Steingold

Hi, It appears that deal does not support missing values (NA), so I need to remove them (NAs) from my data frame. how do I do this? (I am very new to R, so a detailed step-by-step explanation with code samples would be nice). Some columns (variables) have quite a few NAs, so I would rather drop the whole column than sacrifice all the rows (observations) which have NA in that column. How do I

why? > rle Run Length Encoding lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ... values : chr [1:1650061] "4bbf9e94cbceb70c BG bg" "4fbbf2c67e0fb867 SK sk" ... > as.data.frame(rle) Error in as.data.frame.default(vertices.rle) : cannot coerce class '"rle"' into a data.frame it seems that rle.df <-

summary() for a factor prints: ColName SNDK : 72 VXX : 36 MWW : 30 ACI : 28 FRO : 28 (Other):1801 it would have been much more useful if it additionally printed frequency stats as if by summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x) -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://jihadwatch.org

Hi, batch is a vector of lines returned by readLines from a NL-line-terminated file, here is the relevant section: ========================================================= AA BB CC DD EE FF GG H H JJ KK LL MM ========================================================= as you can see, a line is corrupt; two CRLF's are inserted. This is okay, I drop the bad lines, at least I hope I do:

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.