similar to: replace a for loop

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} (

Hi everybody! I would like to export the results of a test statistic in a *.csv file, but get an error. The code is below. The data are attached as .txt with tab as separator. I tried to get a sample dataset, but for some reason it didn't work, so please excuse the length of the file. I also attached a txt with the output (copy/paste) in case that helps. ---- > library(WRS) >

Dear users, My problem concerns save() within a for loop. Here is my code: for (i in 1:4) { temp <- data.frame(a=(i+1):(i+10), b=LETTERS[(i+1):(i+10)]) filename <- paste("file", i, sep="") assign(filename, temp) save(filename, file=paste(filename, ".rda", sep="")) } As you can see, save() doesn't work as I would like: (1) the object

Hey all, I am writing a function in which I will have a matrix of 4 columns and a variable amount of rows. The first to columns will always contain be of the Character type, the third and fourth columns can be a variation of data types, usually characters and integers, but sometimes lists or matrices. At one point the code makes, for each row, a copy of that row, then it makes some adjustments

Dear users, I would like to process all the lists from all *.rda files that I have in one folder. Up to now, I can load all the *.rda files without any problem. The problem is when I want to access the list saved within each *.rda file (only one list per rda file). Here is my code: fpath <- "D:/R" listnames <- list.files(path=fpath, pattern=glob2rx("*.rda"),

Given: > filename<-paste(tempdir(),"\\","myfile.txt",sep=""); > sub("\\\\","/",filename); [1] "C:/DOCUME~1\\AJAYAS~1\\LOCALS~1\\Temp\\Rtmp2f270T\\myfile.txt" Note: Only the first occurence of \\ is replaced. What is the syntax for replacing every occurence? Thanks. [[alternative HTML version deleted]]

Simple one here ... but can't get it to work ... for (i in 1:4){ paste("stuff",[i]),sep="") <- 3 + i } ls() rm(list=ls()) I just want it to create 4 new variables called stuff1, stuff2, stuff3, stuff4 with the corresponding assignments. I realise that there are more elegant functions but this is just a model of a bigger situation. Thanks Jim

Dear All, I have a data frame "data" with Jan , Feb as column. I using like this data$Jan to take Jan column. I have the Jan in another variable var1 <- "Jan" data$var1 I need to replace the variable value there . Any help will be greatly appreciated. Thanks & Rg Mohan L [[alternative HTML version deleted]]

Dear users, I have a weird questions. A friend of mine, some years ago, supposedly trying to access help files without Internet connection, got something like this error message: "The Pythia is not available, please go to Delphi" When I've heard about it, I found it very funny and I'd like to be able to get it. Does anyone of you know how to get this error message again?

Dear All, Sorry for asking a newbie question. I want to ask how to import 1000 datasets whose file names are labelled from data1.dat to data1000.dat into R so that they are named M[1, , ] to M[1000, , ] accordingly. Thank you very much. Best Regards, Ray [[alternative HTML version deleted]]

The loop is correct, you just need to make sure that your result is computed and stored as the n-th element that is returned by the loop. Pick up any manual of R, and looping will be explained there. Also, I would recommend that you draw a random number for every iteration of the loop. Defining the random vectors outside the loop make sense to me only if they are the same length as n.

Dear all, I know this topic has already been covered in other posts (at least the for loop Vs apply family of function), but I am looking for fresh / up-to-date opinion and feedback on those 3 methods to run unavoidable loops in R. I realise that it may be too general question for many, so any feedback appreciated. 1. apply Vs for loop >> Seems apply is (was?) supposed to be faster than

Hello, I would like to run a script in which a loop is included. Since I'm new to R, I cannot manage the following problem. I really hope someone could help me out. Data in the variable Y should be removed from the simulated data set with probability 0.50 if the variable X has a value below zero, and with probability 0.10 if X has a value above zero (see script). However, the total number of

Dear users, I would like all the ticks on a boxplot (x and y) to be labeled I have checked all the par() arguments but couldn't find what I'm looking for Here is an example to show it: df <- structure(list(SPECSHOR = structure(c(1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 4L, 4L), .Label = c("cotau", "dibic", "eqgre", "gicam"), class =

Hello everyone, I have versions 2.7.2 and 2.10.1 installed on a machine that has no access to internet. In 2.7.2 I can use ? to get help on functions, which in 2.10.1 that does not work, all I see is "starting httpd help server...done" and then nothing. Have I downloaded 2.10.1 incorrectly (=forgot to tick some box for local help file repository) or is the internet help now the

Dear users, I have looked on different sources and found different functions to prompt the user to provide input. However, I couldn't find one that does exactly what I'm looking for. select.list() and menu() are nice because a graphic window appears to prompt the user. However, the user can only choose from a predefined list of choices. readline() and scan() are more free in the

sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. > Hello > I have a data array with soil variables (caperf), in which the variable "clay" is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered > >

Hi all. I was having some trouble with a for loop and I found the problem is the following. Does anyone have some idea why I got the following R result? Since mone is equal to 3, why mu1 only have 2 components? library(MASS) > p0 <- seq(0.1, 0.9,by=0.1) > m <- 10 > > > p0 <- p0[7] > > ## data generation > > mzero <- p0*m > mone <- m-mzero >

Hi, I'm a novice R user, much more used to SAS. My problem is pretty simple - basically, in a data frame, I have variables named x1,....,x10 and y1,...,y10; and I would like to create r1 = x1 / y1 etc Apologies if this is way too rudimentary - but I couldn't find any posts online which solve this exact issue. Cheers, Andre

Hi, I have tried looking at the archives but havent found any answer that works till now (Sorry if i have missed anything) I am a newbie to R and i am trying to carry out hierarchical clustering using hclust -> as.dendrogram and then plotting the results as a dendrogram using the plot function plot(object). My question is : In the function "plot", can one decrease the leaf label