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...), perl=T) [[1]] [1] 2 3 4 5 7 8 9 10 attr(,"match.length") [1] 4 3 2 1 4 3 2 1 # gsub gives expected output in Versions 2.2.0 & 2.3.0 gsub("[a]{1,}", "_", as.character(x), perl=T) [1] "x_x_x" Thanks in advance for your help. Thomas -- Thomas Girke, Ph.D. 1008 Noel T. Keen Hall Center for Plant Cell Biology (CEPCEB) University of California Riverside, CA 92521 E-mail: thomas.girke at ucr.edu Ph: 951-827-2469 Fax: 951-827-4437

...thought this could be done like in the following command, but it doesn't: lapply(mylist, function(x) { c(names(x), x) } ) I know how to do this in a for loop, but lapply runs so much faster over large lists. Any help on this simple problem will be highly appreciated. Thomas -- Thomas Girke 1008 Noel T. Keen Hall University of California Riverside, CA 92521 Ph: 951-827-2469 Fax: 951-827-4437

...r is in fact due to changes by the R Development Core Team or not. So, first question: is this change intended or not? (My system has not changed otherwise.) (ii) Since for some applications of mine the first behavior above was exactly what I needed, I now have the same (second) question as Thomas Girke before: is there a way to get the first of the two results now in R 2.4.0 (on a Windows XP machine)? Thanks a lot, STG

I have a two part question Part 1) I am trying to remove characters in a string based on the position of a key character in another string.? I have a solution that works but it requires a for-loop.? A vectorized way of doing this has alluded me.? CleanRead<-function(x,y) { ? if (!is.character(x)) ??? x <- as.character(x) ? if (!is.character(y)) ??? y <- as.character(y) ?