HI Anna, I had a similar problem myself and didn't come up with an easy way to solve it, so my approach was res <- matrix(nrow=length(rownames(x)),ncol=2, byrow=T) for (i in 1:length(rownames(x))){ res[i,1] <- ((x[i,4]+x[i,5]+x[i,6])+x[i,7])/4) res[i,2] <- #calculate the SD here } you could probably set this up as a function allowing you to select different columns each time or if can assign your different columns to groups (maybe assign those you want mean and SD for as 1 and those you don't as 0) you could do something like groups <- c(0,0,0,1,1,1,1) calc.means <-function(x, y){ by(x, y, mean) } apply(eset@exprs,1 calc.means, y=groups) if you have a phenoData variable that distinguishes those columns you want from those you don't then even better, as you can use esApply, this can split your data according to a particular phenoData variable calc.means <-function(x){ by(x, <phenodata variable="">, mean) } esApply(eset@exprs,1 calc.means) hope this helps claire > -----Original Message----- > From: Anna Gustafsson [mailto:annag@biotech.kth.se] > Sent: 20 August 2003 15:39 > To: bioconductor@stat.math.ethz.ch > Subject: [BioC] mean of individual rows for subsets of columns in an > ExprSet > > > Dear all, > > I am sorry to disturb with yet a simple question but I have > tried my best and have now given up on solving it myself... :( > > I wonder if somebody knows how I easily can create mean > values and their SD from subsets of columns (representing > cases) over all rows individually (representing the genes) > from an object - eset - (which is an Expression Set with > 12625 genes, 7 samples). > E.g the mean and SD from all rows individually in columns > 4,5,6 and 7 of my data? > > Grateful for help! > > // Anna :o) > > _______________________________________________ > Bioconductor mailing list > Bioconductor@stat.math.ethz.ch > https://www.stat.math.ethz.ch/mailman/listinfo/bioconductor > -------------------------------------------------------- This email is confidential and intended solely for the use o...{{dropped}}

> HI Anna, > > I had a similar problem myself and didn't come up with an easy way to solve it, so my approach was > > res <- matrix(nrow=length(rownames(x)),ncol=2, byrow=T) > for (i in 1:length(rownames(x))){ > res[i,1] <- ((x[i,4]+x[i,5]+x[i,6])+x[i,7])/4) > res[i,2] <- #calculate the SD here > } i could not tell from the e-mail whether column statistics or row statistics were intended. for row statistics, commands of the form apply(x,1,f) can be used. if f returns a scalar on vector input (as does the function mean) then apply(x,1,f) is the vector with ith element f(x[i,]). you could then use apply(x[,4:7],1,mean) to do the first calculation above (and could easily modify to median or trimmed mean with this approach). if you want to be a little more elegant, you can write a function that returns the vector of statistics of interest msd <- function(x) c(mean(x),sqrt(var(x))) now apply(x,1,msd) returns a 2xn matrix where n is the number of rows of x. msdmat <- t(apply(x,1,msd)) is like your "res" above. lessons: use apply and R functions whenever feasible. > > > you could probably set this up as a function allowing you to select different columns each time or if can assign your different columns to groups (maybe assign those you want mean and SD for as 1 and those you don't as 0) you could do something like > > groups <- c(0,0,0,1,1,1,1) > calc.means <-function(x, y){ > by(x, y, mean) > } > apply(eset@exprs,1 calc.means, y=groups) this can be done in one step using subscripting within the apply msdmat <- t(apply(x[,groups==1],1,msd)) # or specify the groups explicitly in the subscripting NB: please don't use the "@" notation if it can be avoided. we provide "accessor" function exprs() that should be used.