similar to: Data Manipulation - make diagonal matrix of each element of a matrix

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

I don't see the issue here. It would be helpful if people would report their sessionInfo() when reporting whether or not they see this issue. Mine is > sessionInfo() R version 3.4.3 (2017-11-30) Platform: x86_64-pc-linux-gnu (64-bit) Running under: Arch Linux Matrix products: default BLAS/LAPACK: /usr/lib/libopenblas_haswellp-r0.2.20.so locale: [1] LC_CTYPE=en_US.UTF-8

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

Hello, I have looked on https://www.r-project.org/bugs.html , but it seems that this is the only way to do it. The issue is that the precision used by write.csv is on consistant for big files. See the following code: First I create a large dataframe filled with random uniform values. Then I write it to .csv and print out the first and last lines. df = data.frame(replicate(100, runif(1000000,

To my surprise, I can confirm on Windows 10 using R 3.4.3 . As tail is not recognized by Windows cmd, I replaced with: system('powershell -nologo "& "Get-Content -Path temp.csv -Tail 1') The last line shows only 7 digits after the decimal, whereas the first have 15 digits after the decimal. I agree with Dirk though, 1.6Gb csv files are not the best way to work with

If you want a character representation of a long formula (or a formula with long names), you can use: as.character(my.formula) However restriction on length of an as.character result returns only the beginning of a long formula, and without comment. In most cases, the following expression provides the complete result: paste(my.formula[[2]], " ~ ",

Dear List, I've around 1000 *.txt files, I've generate with other software. I've now done the following code (below). My question is how can I automate this (with do.call () ?), so it could be done for all the *.txt files. Thanks in advance, Rog??rio names<- list.files() file <- "BLU_Var_%04d.txt" for(i in 1:1000){

Hi - a newbie question, if someone can please help.... I want to change X1, X2,,.....to X.1 X.2 etc in the names below. I am using the Principal Component Regression function (pcr) and it seems to want it this way > datap3.pcr <- pcr(water ~ X, 10, data = datap3, Validation ="cv") Error in model.frame.default(formula = water ~ X, data = datap3) : invalid type (list) for

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)

Hello everyone, I am trying to fit the following model All X. variables are continuous, while the conditions are categoricals. model <- lm(X2

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, I was wondering how it was possible to access the actual cluster exemplars from the APResult class. Currently it only spits it out onto the terminal if you type the object but there is no other way to see which one is the examplar. Would appreciate any help. Thanks, Sachin [[alternative HTML version deleted]]

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

Hallo, All, I have a question about changing the height or scale of the y axis. When I use following two R codes, I can get two plots. Please look at the y axes, the number of indices (x1, x2, ?) on the y axis in the first plot is smaller than that in the second plot, and hence the space between any two indices in the first plot is wider than that in the second plot. As the number of indices

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 ##

I have a list 'ans' from the following code: tt <- rnorm(50) rr <- rnorm(50) ans <- lm(rr~tt) ans[1] is "$coefficients", ans[2] is "$residuals", ans[3] is "$effects", ... and so on up to ans[12]. Is there an easy way to display just these names and not the data they contain? I thought I saw my advisor type "ans$" and they were displayed,

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

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

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