similar to: extracting row means from a list

Hello! I have a dataset like this: X1 X2 X3 X4 X5 X6 X7 X8 1 2 2 1 2 3 2 6 2 3 2 5 7 9 1 3 1 9 12 6 1 1 3 6 The columns X1-X6 contains ordinary numeric values. X7 contains the number of the first column that the rowMeans should be calculated from and X8 contains the last column

Hi I am trying to create a loop which averages replicates in my data. The original data has many rows. and consists of 40 column zz[,2:41] plus row headings in zz[,1] I am trying to average each set of values (i.e. zz[1,2:3] averaged and placed in average_value[1,2] and so on. below is my script but it seems to be stuck in an endless loop Any suggestions?? for (i in 1:length(average_value[,1])) {

--- On Fri, 15/5/09, Amit Patel <amitrhelp at yahoo.co.uk> wrote: > From: Amit Patel <amitrhelp at yahoo.co.uk> > Subject: Help with loops > To: r-help at r-project.org > Date: Friday, 15 May, 2009, 12:17 PM > Hi > I am trying to create a loop which averages replicates in > my data. > The original data has many rows. and consists of 40 column > zz[,2:41]

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)

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

John, Maybe you can clarify what you want the output to look like. It took me a while to realize what you may want as it is NOT properly described as wanting rowsums. There is a standard function called rowMeans() that probably does what you want if you want the mean of all rows as in: > rowMeans(xxxz) [1] 84.33333 87.00000 89.66667 92.33333 95.00000 97.66667 100.33333 103.66667

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>

Hi, May be this helps: ?set.seed(24) ?mydata4<- as.data.frame(matrix(sample(1:100,10*38,replace=TRUE),ncol=38)) ?dim(mydata4) #[1] 10 38 ?library(matrixStats) res<-do.call(cbind,lapply(1:400, function(i) {permutation<-sample(mydata4); (rowMeans(permutation[,1:27])-rowMeans(permutation[,28:38]))/(rowSds(permutation[,1:27])+rowSds(permutation[,28:38]))} )) ?dim(res) #[1]? 10 400 A.K.

Hello, I am trying to create parcels for a CFA model. I am trying to average 6 sets of 3 variables each into parcels. I don't understand why I am getting an error message as follows: Thanks for your help, Catherine atds1par <- rowMeans(semHW1dat1[, c("atds1", "atds2", "atds3")], na.rm=TRUE) atds2par <- rowMeans(semHW1dat1[, c("atds4",

I have a list of data frames like the following: set.seed(123) a<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10)) b<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10)) L<- list(a,b) All data frames in the list have the same dimensions. I need to calculate the sample means for x and y. The real data are lists of several thousand quite large dataframes, so I need

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

Dear List, I'm not even sure this is an issue or not, but ?rowMeans has: Value: A numeric or complex array of suitable size, or a vector if the result is one-dimensional. The ?dimnames? (or ?names? for a vector result) are taken from the original array. If there are no values in a range to be summed over (after removing missing values with ?na.rm = TRUE?), that

Hi all, I have a matrix called 'data', which looks like: > data[1:4,1:4] Probe_ID Gene_Symbol M1601 M1602 1 A_23_P105862 13CDNA73 -1.6 0.16 2 A_23_P76435 15E1.2 0.18 0.59 3 A_24_P402115 15E1.2 1.63 -0.62 4 A_32_P227764 15E1.2 -0.76 -0.42 > dim(data) [1]

I would like to calculate the mean of tree leader increment growth over 5 years (I1 through I5) where each tree is a row and each row has 5 columns. So far I have achieved this using rowMeans when all columns are numeric type and used in the calculation: Data1 <- data.frame(cbind(I1 = 3, I2 = c(0,3:1, 2:5,NA), I3 =c(1:4,NA,5:2),I4=2,I5=3)) Data1 Data1$mean_5 <- rowMeans(Data1, na.rm =T)

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

Hello, I have some dataset, which i read it from external file using the (data <- read.csv("my file location")) and read as a dataframe > is(data) [1] "data.frame" "list" "oldClass" "vector" but i have also converted this into a matrix and tried to apply my code but didnt work. Anyways, suppose i have the following data.

In R 3.3.3, I observe the following on Ubuntu 16.04 (when building from source as well as for the sudo apt r-base build): > x <- c(NA, NaN) > mean(x) [1] NA > mean(rev(x)) [1] NaN > rowMeans(matrix(x, nrow = 1, ncol = 2)) [1] NA > rowMeans(matrix(rev(x), nrow = 1, ncol = 2)) [1] NaN > .rowMeans(x, m = 1, n = 2) [1] NA > .rowMeans(rev(x), m = 1, n = 2) [1] NaN >

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

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