You need to span the range of your data. That is if rx<-range(x) ## and your set of breaks is mybreaks then you get the error if rx[1] < min(mybreaks) || rx[2] > max(mybreaks) returns T. So you migh
Tristan, I think you mean to find all rows of X whose 4th column contains a value from test, try using %in%, Y<-X[X[,4]%in%test,] FYI: get("%in%") function(x, table) !is.na(match(x, table, nomatch =
I use a university wide license for version 6.2 and 7.0, but even in the academic environment license management is difficult. While I could use the features of large data objects, I don't want to de
Yes by name is needed, try the following: clean.all <-function(id) { obj.name<-paste("station",id,sep="") value<-clean.dates(get(where=1,obj.name)$weather.date) eval(substitute(X$weather.date<-VALUE
Hi, maybe I am missing something here. Columns in a data frame all have the same length, so what are you replacing the say three values of 1.5 with? If you mean that you have a list and order is not
Good morning. I have written many functions that perform similarly. If you can use a single function on the whole matrix, using matrix mathematics then that is the fastest. For example the unbiased c
Hello, in the function predict.glm, while it seems the calculation of link and response confidence limits are correct those for the prediction intervals are not since the variance component due to th
Sonia, I don?t have any fast methods, but here is a start. There are two ways to do it. The first way is just to take a small incremental say delta h, and try simple derivatives. For example: dIG.da<
You can use the missing data library. If you have the imputations, you can do the following from the missing library: 1. Run you linear model for each set of imputations and put into a list: fit<-lis
Use eval(parse(paste(text=...))) form. lapply(xxx, function(x) try(eval(parse(text=paste("library(help=",x,")") )))) sections without help files. This generates a lot of readme.txt notepad windows (