similar to: cor() on sets of vectors

Displaying 20 results from an estimated 1000 matches similar to: "cor() on sets of vectors"

2012 Feb 13
1
entropy package: how to compute mutual information?
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
2012 Feb 10
2
the value of the last expression
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
2012 Dec 04
3
list to matrix?
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,]
2012 Feb 08
4
"unsparse" a vector
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
2011 Jul 11
1
plot means ?
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
2012 Jul 13
1
LiblineaR: read/write model files?
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/
2012 Feb 24
1
count.fields inconsistent with read.table?
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:
2013 Jan 04
4
non-consing count
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
2012 Feb 10
2
naiveBayes: slow predict, weird results
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
2006 Mar 17
6
removing NA from a data frame
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
2012 Sep 19
2
drop zero slots from table?
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
2012 Nov 09
4
as.data.frame(do.call(rbind,lapply)) produces something weird
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
2012 Sep 14
3
aggregate() runs out of memory
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))$x) alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is 24.3G, and no end in sight. both V1 and V2 are characters (not factors). Is there anything I could do to speed this up? Thanks. -- Sam Steingold
2011 Feb 15
1
summary for factors is not very informative
summary() for a factor prints: ColName SNDK : 72 VXX : 36 MWW : 30 ACI : 28 FRO : 28 (Other):1801 it would have been much more useful if it additionally printed frequency stats as if by summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x) -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://jihadwatch.org
2012 Apr 04
2
recover lost global function
Since R has the same namespace for functions and variables, > c <- 1 kills the global function, which can be restored by > c <- get("c",mode="function") Is there a way to prevent R from overriding globals or at least warning when I do that or at least warning when I replace a functional value with non-functional? thanks. -- Sam Steingold (http://sds.podval.org/)
2011 Feb 15
1
all.equal: subscript out of bounds
When I do > all(all$X.Time == all$Y.Time); [1] TRUE as expected, but > all.equal(all$X.Time,all$Y.Time); Error in target[[i]] : subscript out of bounds why? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://mideasttruth.com http://honestreporting.com http://dhimmi.com http://jihadwatch.org http://pmw.org.il http://ffii.org The dark past once was the
2013 Apr 21
1
cedta decided 'igraph' wasn't data.table aware
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
2012 Apr 04
2
plot with a regression line(s)
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 =
2012 Mar 20
2
igraph: decompose.graph: Error: protect(): protection stack overflow
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. --
2006 May 11
3
cannot turn some columns in a data frame into factors
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")