similar to: how to rename rows/columns of a matrix?

Hi All I am having 2 very long character strings (550chars) and I want to put them as expressions together with c(). The problem is that I also get these double-quotes, as seen below in 'fct'. How can I remove these double-quotes? I tried as.name() but it did not work (because of size?). These are creating trouble with subsequent programs, which I tested with strings that for some

Hi cor(x,apply(x,1,sum)) gives me the correlations of each column with the sums of each row (correct me if I'm wrong, please). What I need are the correlations of each column with the sums of each row except the entry in the given column. It seems that for any one column i I get it by doing: cor(x[,i],apply(x[,-i],1,sum)) But I struggle to get it for all the columns. I was trying things

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 I am trying to replicate a vector in n rows for comparison purposes with another matrix. foo <- c(1,2,3) bar <- rbind(foo,foo) # does the trick for 2 rows bar <- rbind(rep(foo,2)) # does something else How do I generate a matrix with all rows=foo without writing 'foo' n times as arg? Thanks, David

Hi all I am very astonished that R generates a "syntax error" when I want to split up a line with a backslash, which usually works in any shell script. R itself generates the "+" symbols at the beginning of following lines in a splitted line so I've tried with them as well, but also without success. Searching in the h-help archive did not reveal any answer either.

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>

How about: x <- data.frame(matrix(rnorm(1550),c(50,31))) model <- step(lm(x[,1] ~ as.matrix(x[,2:31]))) --Matt -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of rongguiwong Sent: Monday, September 20, 2004 20:52 PM To: r-help at stat.math.ethz.ch Subject: [R] how to take this experiment with R? This message uses

I have a dataset that is basically structureless. Its dimension varies from row to row and sep(s) are a mixture of tab and semi colon (;) and example is HEADER1 HEADER2 HEADER3 HEADER3 A1 B1 C1 X11;X12;X13 A2 B2 C2 X21;X22;X23;X24;X25 A3 B3 C3 A4 B4 C4 X41;X42;X43 A5 B5 C5 X51 etc., say. Note that a blank

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 ##

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

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

Full_Name: Derek Elmerick Version: 2.4.0 OS: Windows XP Submission from: (NULL) (38.117.162.243) hello - i have some code that i run regularly using R version 2.3.x . the final step of the code is to build a multinomial logit model. the dataset is large; however, i have not had issues in the past. i just installed the 2.4.0 version of R and now have memory allocation issues. to verify, i ran

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

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

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

Hello All, I have a question about glm in R. I would like to fit a model with glm function, I have a vector y (size n) which is my response variable and I have matrix X which is by size (n*f) where f is the number of features or columns. I have about 80 features, and when I fit a model using the following formula,? glmfit = glm(y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13

Hi Thierry, Thanks for the patch. I applied on top of linux-next-2017-12-14. Different output this time. [ 11.862495] WARNING: CPU: 1 PID: 254 at drivers/gpu/drm/nouveau/nvkm/subdev/mmu/vmmgf100.c:391 gf100_vmm_new_+0x60/0x128 [nouveau] [ 11.863458] tegra-dpaux 155c0000.dpaux: 155c0000.dpaux supply vdd not found, using dummy regulator [ 11.866197] tegra-sor 15580000.sor: failed to probe

Dear list I am comparing the results of several different experimental setups. With kruskal.test() I can test if there is any difference at all in any of them, if I understand it correctly. But now, when there is a difference, how do I generate a (half-) table of pairwise comparisons, using e.g. wilcox.test(), to find the ones where the difference actually occurs. I guess I don't have to

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

Hello everyone, I want to transform a data.frame into an array (lets call it mydata), where: mydata[[1]] is the first imputed dataset...and for each mydata[[d]], the first p columns are covariates X, and the last one is the outcome Y. Lets assume a simple data.frame: Imputed = data.frame( X1 = c(1,2,1,2,1,2,1,2, 1,2,1,2,1,2,1,2), X2 =