similar to: Calculating rowMeans from different columns in each row?

Dear all, I have monthly data in wide format, I am only providing data (at the bottom of the email) for the first 24 columns but I have 2880 columns in total. I would like to take max of every 12 columns. I have taken the mean of every 12 columns with the following code: byapply <- function(x, by, fun, ...) { # Create index list if (length(by) == 1) { nc <- ncol(x)

Suppose I have the following: x1<-as.vector(rnorm(10)) x2<-as.vector(rnorm(10)) x3<-as.vector(rnorm(10)) x4<-as.vector(rnorm(10)) x5<-as.vector(rnorm(10)) x6<-as.vector(rnorm(10)) x7<-as.vector(rnorm(10)) x8<-as.vector(rnorm(10)) x9<-as.vector(rnorm(10)) x10<-as.vector(rnorm(10)) I would like the mean of x1 to x6 for each vector position. I would do something else

Hi Milu, byapply(df, 12, function(x) apply(x, 1, max)) You might also be interested in the matrixStats package. Best, Ista On Tue, Feb 20, 2018 at 9:55 AM, Miluji Sb <milujisb at gmail.com> wrote: > Dear all, > > I have monthly data in wide format, I am only providing data (at the bottom > of the email) for the first 24 columns but I have 2880 columns in total. > > I

This is what I was looking for. Thank you everyone! Sincerely, Milu <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Mail priva di virus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>

Ista, et. al: efficiency? (Note: I needed to correct my previous post: do.call() is required for pmax() over the data frame) > x <- data.frame(matrix(runif(12e6), ncol=12)) > system.time(r1 <- do.call(pmax,x)) user system elapsed 0.049 0.000 0.049 > identical(r1,r2) [1] FALSE > system.time(r2 <- apply(x,1,max)) user system elapsed 2.162 0.045 2.207 ##

The maximum over twelve columns is the maximum of the twelve maxima of each of the columns. single_col_max <- apply(x, 2, max) twelve_col_max <- apply( matrix(single_col_max, nrow = 12), 2, max ) ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team Biometrie

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

Don't do this (sorry Thierry)! max() already does this -- see ?max > x <- data.frame(a =rnorm(10), b = rnorm(10)) > max(x) [1] 1.799644 > max(sapply(x,max)) [1] 1.799644 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic

#### I checked through every 3 factor * 3 loading case. #### While, 4 factor * 3 loading failed. #### the data is 6 factor * 3 loading require(sem); cor18<-read.moments(); 1 .68 1 .60 .58 1 .01 .10 .07 1 .12 .04 .06 .29 1 .06 .06 .01 .35 .24 1 .09 .13 .10 .05 .03 .07 1 .04 .08 .16 .10 .12 .06 .25 1 .06 .09 .02 .02 .09 .16 .29 .36 1 .23 .26 .19 .05 .04 .04 .08 .09 .09 1 .11 .13 .12 .03 .05 .03

Does anyone know how to pass a list of parameters into a function? for example: somefun=function(x1,x2,x3,x4,x5,x6,x7,x8,x9){ ans=x1+x2+x3+x4+x5+x6+x7+x8+x9 return(ans) } somefun(1,2,3,4,5,6,7,8,9) # I would like this to work: temp=c(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9) somefun(x1=1,x2=2,temp) # OR I would like this to work: temp=list(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9)

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Ista, et. al: efficiency? > (Note: I needed to correct my previous post: do.call() is required for > pmax() over the data frame) > > > x <- data.frame(matrix(runif(12e6), ncol=12)) > > > system.time(r1 <- do.call(pmax,x)) > user system elapsed > 0.049 0.000

Dear friends, I'm doing a simulation on logistic regression model, but the programs can't work well,please help me to correct it and give some suggestions. My programs: data<-matrix(rnorm(400),ncol=8) #sample size is 50 data<-data.frame(data) names(data)<-c(paste("x",1:8,sep="")) #8 independent variables,x1-x8; #logistic regression model is

Hi, You might need to check library(boot).? I have never used that before.? So, I can't comment much.? It is better to post on R-help list.? I had seen your postings on Nabble in the past.? Unfortunately those postings were not accepted in R-help.? You have to directly post at ? r-help at r-project.org after registering at: https://stat.ethz.ch/mailman/listinfo/r-help ?

Thank you for your kind replies. Maybe I was not clear with my question (I apologize) or I did not understand... I would like to take the max for X0...X11 and X12...X24 in my dataset. When I use pmax with the function byapply as in byapply(df, 12, pmax) I get back a list which I cannot convert to a dataframe. Am I missing something? Thanks again! Sincerely, Milu

Dear friends, After running the lm() model, we can get summary resluts like the following: Coefficients: Estimate Std. Error t value Pr(>|t|) x1 0.11562 0.10994 1.052 0.2957 x2 -0.13879 0.09674 -1.435 0.1548 x3 0.01051 0.09862 0.107 0.9153 x4 0.14183 0.08471 1.674 0.0975 . x5 0.18995 0.10482 1.812 0.0732 . x6 0.24832 0.10059 2.469 0.0154 * x7

Failed? What was the error message? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka

It looks like OP uses a data.frame, so in order to use matrixStats (I'm the author) one would have to pay the price to coerce to a matrix before using matrixStats::rowMaxs(). However, if it is that the original data could equally well live in a matrix, then matrixStats should be computational efficient for this task. (I've seen cases where an original matrix was turned into a data.frame