similar to: summary in functions

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

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

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

> 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

I have two data frames: > str(ysmd) 'data.frame': 8325 obs. of 6 variables: $ X.stock : Factor w/ 8325 levels "A","AA","AA-",..: 2702 6547 4118 7664 7587 6350 3341 5640 5107 7589 ... $ market.cap : num -1.00 2.97e+10 3.54e+08 3.46e+08 -1.00 ... $ X52.week.low : num 40.2 22.5 27.5 12.2 20.7 ... $

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

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/)

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 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

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 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 <-

Hi, I am trying to aggregate some data and I am confused by the results. I load a data frame "all" from a csv file, and then I do: (FOO,BAR,X,Y come from the header line in the csv file, BTW, how do I rename a column?) byFOO <- aggregate(list(all$BAR,all$QUUX,all$X/all$Y), by = list(FOO=all$FOO), FUN = mean); I expect a data frame with 4

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

Hi folks, I was wondering how to run a mixed models approach to analyze a linear regression with a user-defined covariance structure. I have my model y = xa +zb +e and b ~ N (0, C*sigma_square). (and a is a fixed effects) I would like to provide R the C (variance-covariance) matrix I can easily provide an example, but at this point I am first trying to know what is the best package the

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 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

> * 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

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