similar to: Take the maximum of every 12 columns

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

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

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

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

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

I hit this while testing HMM with nouveau on linux-5.5-rc5. I'm not a lockdep expert but my understanding of this is that an invalidation callback could potentially call kzalloc(GFP_KERNEL) which could cause another invalidation and recursively deadlock. Looking at the drivers/gpu/drm/nouveau/nvkm/ layer, I do see a number of places where GFP_KERNEL is used for allocations and I don't see

Dear R Staff You can see my data.csv file in the annex. I try to count non-zero values in dataset but I need to exclude NA in this calculation My code is very long (following), How can I write this code more efficiently and shortly? ## [NA_Count] - Find NA values data.na =sapply(data[,3:ncol(data)], function(c) sum(length(which(is.na (c))))) ## [Zero] - Find zero values

What are the limitations on size of matrix for MDS functions? steve -- Steve Antos Priva-Technologies 847-640-9020 x3114 cell (540)409-1231 [[alternative HTML version deleted]]

Hola Eric, A la que me refiero es a la que se utiliza para probar la independencia de dos variables entre sí, mediante la presentación de los datos en tablas de contingencia. Pero no se si se puede realiza esa prueba Chi-cuadrado ajustando por otra variable. Y no me refiero a si quiero probar la independencia de X1 y X2 y tengo además la variable *sexo*, realizar la prueba para entre X1 y X2

Hello all, I'm sorry if my question seems basic. Im studying a responses (Yes,No) in a survey and, thanks to GLM I obtain the following relation with my variables : (Yes,No)~ ?0 + Age We note this this certain type of (Yes,No) response is linked to age (p<0.05 in glm) . After that we calculated : model1=glm(cbind(Yes,No) ~ Age + Times + Type, family=binomial) summary(model1)