search for: colmean

...ehs/core/microarrays/docs/heinloth.txt",1)) x = t(x) #now rows are subjects, cols are genes x = x[order(rownames(x)),] #order by treatment group oxygen, ultra-violet, gamma radiation y = x[26:45,1:10] x = x[2:25,1:10] p = ncol(x); p nx = nrow(x); nx ny = nrow(y); ny n = nx+ny; n # (t(x)-colMeans(x)) %*% t(t(x)-colMeans(x)) T2 = nx*ny/n * t(colMeans(x)-colMeans(y)) %*% solve( ( (nx-1)*cov(x)+(ny-1)*cov(y) )/( n-2 ) ) %*% (colMeans(x)-colMeans(y)); T2 library(ICSNP) HotellingsT2(y,x) http://en.wikipedia.org/wiki/Hotelling's_T-square_distribution http://finzi.psych.upenn.edu/R/Rh...

Having searched and searched I still haven't found what's the problem with my data (I've attached the relevant file). Every time I tried to use the CANCOR-Function I got error messages. So I turned to check my two sets of variables separately by using the "colMeans"-Function. With one set no problem .. but with the other. My input : > struktur <- read.delim("struktur.csv", header=TRUE, sep = ";") >colMeans(struktur) And the resulting error: Error in colMeans(x, n, prod(dn), na.rm) : `x' must be numeric Can...

Hi all, I would like to know if it is posible by, someway, to get colMeans from a data.frame with numeric as well as character data, dispersed all over the object. Note that I would like to get colMeans neglecting character data. I am really in need of some function proceeding in that way… All the best Diogo André Alagador [[alternative HTML version deleted]]

...ichael Milton wrote: > I just checked out R-devel and noticed that the new "pipe extractor" > capability coming in 4.3 only works for the 4 extractor operators, but no > other standard operators like +, *, %*% etc, meaning that e.g. mtcars |> > as.matrix() |> _ + 1 |> colMeans() is a syntax error. In addition, we are > still subject to the restriction that the functions on the RHS of a pipe > can't have special names, so mtcars |> as.matrix() |> `+`(1) |> colMeans() is > also a syntax error. > > Either option would be great, as I find it muc...

I read from external data source containing several columns. Each column represents value of a metric. The columns are time series data. I want to sort the resulting dataframe such that the column with the largest mean is the leftmost column, descending in colMean values to the right. I see many solutions for sorting rows based on some column characteristic, but haven't found any discussion of sorting columns based on column characteristics. viz. input data looks like this time met-a met-b met-c 00:00 42 18 99 00:05 88...

...2.33044953 1.0423625 2.7687266 0.2904245 0.70155274 2.4158326 1.2315324 1.4287387 1.5701019 $WW 2008 2009 20101 1.4737028 2.314878 2.6726612 1.6700918 2.609722 2.1124213 3.2387775 7.305766 6.9395364 6.7063592 18.745256 13.278218 afterwards when i applied the "colMeans" command i was not able to calculate the mean as i got the following error >'x' must be an array of at least two dimensions i cant spot the mistake in the last 5 hours. you help is neededthanks in advance eliza -------------- next part -------------- An embedded and ch...

...out (and I think it was in S Poetry) the statement was made that vectorisation befuddles some beginners. So learning how to use this command, on a nice easy topic such as summing or averaging does have some merit. I have to admit using colSums a lot, but I don't think I have ever thought to use colMeans. As an aside; does anyone have a list of optimised functions. That is functions like this one whose main benefit is speed (I think) I guess many of us feel that we are constantly using hammers to crack the nuts, but we still don't really know what's in the toolbox. Tom > -----Origin...

I just checked out R-devel and noticed that the new "pipe extractor" capability coming in 4.3 only works for the 4 extractor operators, but no other standard operators like +, *, %*% etc, meaning that e.g. mtcars |> as.matrix() |> _ + 1 |> colMeans() is a syntax error. In addition, we are still subject to the restriction that the functions on the RHS of a pipe can't have special names, so mtcars |> as.matrix() |> `+`(1) |> colMeans() is also a syntax error. Either option would be great, as I find it much cleaner to coordinate a...

dear useRs, i have a list containing 16 matrices. i want to calculate the column mean of each of them. i tried >sr <- lapply(s,function(x) colMeans(x, na.rm=TRUE)) but i am getting the following error >Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric can it be done in any other way? and why i am getting this error?? thanks in advance.. elisa [[alternative HTML version deleted]]

...fficient) and can someone point me to a possible solution? (It would be great if hours could be reduced to seconds :-). # --------------------------------------------- RanEigen=function(items=x,cases=y,sample=z) { X=matrix(rnorm(cases*items),nrow=cases,byrow=F) S=crossprod(X-rep(1,cases) %*% t(colMeans(X))) EV=eigen((1/sqrt(diag(S))*diag(items))%*%S%*%(1/sqrt(diag(S))*diag(items)),only.values=T)$values for (i in 2:sample) { X=matrix(rnorm(cases*items),nrow=cases,byrow=F) S=crossprod(X-rep(1,cases) %*% t(colMeans(X))) EV=rbind(EV,eigen((1/sqrt(diag(S))*diag(items))%*%S%*%(1/sqrt(diag(S...

...same result weather you use 1 or 2, i.e. same result for columns than for means..:( Thanks a lot for the help, m = matrix( c(1,4,7,4,5,8,7,8,9), nrow = 3 ) divideByMean = function( v ){ return( v/mean(v)) } rowMean = c( mean( m[1,]),mean( m[2,]),mean( m[3,]) ) rowMean colMean = c( mean( m[,1]),mean( m[,2]),mean( m[,3]) ) colMean m print("ByRow") apply( m, 1, divideByMean) print("ByCol") apply( m, 2, divideByMean) [[alternative HTML version deleted]]

I have a fair bit of experience with S-Plus and have been asked to port some of my S-Plus bootstrapping functions to R, to which I am relatively new, Needless to say, I've run into some problems. In particular, I need to perform bootstrap resampling of the colMeans function using a moving blocks bootstrap, where the blocks are row indices. It's easy to build a function that allows me to perform a bootstrap on colMeans, but I'm flummoxed at the moving blocks bootstrap step. I could simply hack together a function that, when given block bootstrap...

...head, but not for mean. Obviously we importFrom(utils, head), but don't need to do this for mean, which is imported automatically from base. If we export(mean) rather than exportMethods(mean), it works as intended. A similar problem arises for a new generic function and an associated method, colmean, where the use of exportMethods(colmean) fails, but export(colmean) succeeds. We can build and use the package by using export instead of exportMethods, but we suspect that we're missing something that might be important to understand. Any hints or links to something we missed would be apprec...

...> > I just checked out R-devel and noticed that the new "pipe extractor" > > capability coming in 4.3 only works for the 4 extractor operators, but no > > other standard operators like +, *, %*% etc, meaning that e.g. mtcars |> > > as.matrix() |> _ + 1 |> colMeans() is a syntax error. In addition, we are > > still subject to the restriction that the functions on the RHS of a pipe > > can't have special names, so mtcars |> as.matrix() |> `+`(1) |> > colMeans() is > > also a syntax error. > > > > Either option wou...

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678

Is it possible to compute the variance of every column in a matrix without a loop? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch

...int main (int argc, char** argv) { SEXP e,val; int errorOccurred; Rf_initEmbeddedR(argc, argv); // library("fPortfolio") PROTECT(e = lang2(install("library"), mkString("fPortfolio"))); R_tryEval(e, R_GlobalEnv, NULL); UNPROTECT(1); // colMeans(SWX.RET) PROTECT(e = lang2(install("colMeans"), install("SWX.RET"))); val = (R_tryEval(e, NULL, &errorOccurred)); Rf_PrintValue(val); return 0; } When I tried : >mean(SWX.RET) in the R prompt I got the following output: SBI SPI...

...I also tried following code for plotting the line point exactly at the center of two bars: x = c("a","b","c","d") y= cbind(c(50,40,30,20),c(40,30,20,10)) y2 = c(0.80,0.65,0.75,0.50) barplot(t(y),beside = TRUE) a = barplot(t(y),beside = TRUE) par(new=T) plot(colMeans(a),y2,type="l",col="black",lwd= 1,lty=1,xaxt="n",yaxt="n",xlab="",ylab="") points(colMeans(a),y2,pch=20) Please help me in plotting this graph -- View this message in context: http://r.789695.n4.nabble.com/Barplot-with-Secondary-axis-...

...ows the text message "Error using packet n replacement has length zero." where n is the panel number. > data(iris) > cl <- cclust(iris[,-5], k=3) > barplot(cl, mcol=NULL) #works fine > barchart(cl, mcol=NULL) # oops FWIW, as the data of the clusters is scaled showing the colMeans for the centers is misleading. The global mean of each column is 0, not the colMean of the centers.

...nconsistency recently came to my attention: > df <- data.frame(A = 1:10, B = rnorm(10)) > min(df) [1] -1.768958 > max(df) [1] 10 > mean(df) [1] NA Warning message: In mean.default(df) : argument is not numeric or logical: returning NA I recall the times where `mean(df)` would give `colMeans(df)` and this behaviour was deemed inconsistent. It seems though that the change has removed one inconsistency and replaced it with another. Am I missing good reasons why there couldn't be a `mean.data.frame()` method which worked like `max()` etc when given a data frame? Namely that they ret...