similar to: which duplicated rows to delete

Hi; It seems to me that has probably been asked in the past. But I cannot find the track. I usually need to extract elements from a list and contruct vector from them; e.g., to create a table. Perhaps there is a way to directly extract them without looping? Simple example: > S.lst $sublist.1 $sublist.1$age [1] 24.58719 $sublist.1$weight [1] 60.82861 $sublist.2 $sublist.2$age [1] 32.39551

Hello R-ologists, Imagine you have a list "list" like so: >list [[1]] [1] "IPI00776145.1" "IPI00776187.1" [[2]] [1] "Something" "IPI00807764.1" "IPI00807887.1" [[3]] [1] "IPI00807764.1" [[4]] [1] "Somethingelse" What I need to achieve is a filtered list "list2" like so: >list2 [[1]] [1]

Hello everyone, I need some help with lists inside lists (a good way to emulate a struct)\ Assume that there is a small list called fred: fred <- list(happy = 1:10, name = "squash") and a big list called bigfred that includes fred list 5 times bigfred <- rep(fred,5) Is it possible somehow to index all these sublists(fred) inside bigfred with a more direct way like this:

my problem is more complex than below but I think below can suffice. i have a list and the name of it at the top level is GGG. so, if i do an lapply and operate on lower components in the sublist, then I can do as shown in EXAMPLE 1 and what will come back will be named GGG at the top level. but, suppose that , the function inside the lapply function was more complex and i wanted to

Hi everyone, Suppose I have a list named "lst", see below: > lst $sub1 ... $sub1$x ... $sub1$y .... $sub2 ... $sub2$x ... $sub2$y ? $sub3 ... ... ... Now, I want to extract the sub-sublist $y from every sublist(sub1, sub2...) and then storage them to a new list. I know how to extract them by subscript or list name one by one, but I wonder if there exist some tricks to finish this

Hello We would like to perform a logistic regression analysis weighting the independent variable in a temporal fashion, i.e. events occuring most recent get highest weight. Does anyone know how to do this in R?? Regards S. Merser and S. Lophaven

hi there tried to use subset with drop=TRUE, but all the 'old' levels are preserved, i.e. when calling e.g. ftable a lot of zeros are displayed >x<-subset(LREG, (kir=='AA' | kir=='BB') & (type=='t1' | otype=='t2'), drop=TRUE, select=c(event, kir, type)) > ftable(x) i explicit have to call factor like

hi there is there a way that i can coerce a vector of integers to ONE string with the numbers comma separated like: 1:500 -> "1,2,3, ..., 500" i've tried deparse, but it divides the result into a vector of string (depending on the setting of width with a max of 500) regards soren

Hi, i was looking through the docs and couldn''t find a good clean solution for ordering a partial with an :order clause. say i was iterating over categories and thier corresponding sub-categories (regular belongs to) and wanted to order the subcategories according to a certain column; <% for category in @categories %> <% render: partial => ''sublist'',

[if the format of my email is changed or is difficult to understand, a text file is attached for easy understanding]Dear useRs, i was wondering that if its possible in R to automatically generate plots and get it saved at the desired location? i have data of cancer patients, from about 1000 cities around the world. i have converted that data into a list (called tcp) and that list has 1000

I'm using win XP R 1.51 (Rterm) emacs-20.7 ess-5.1.20 and everything except from command completion works Is it possible to have this functionality using a windows enviroment? thanks soren -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

I have a list of dataframes i.e. each list element is a dataframe with three columns and differing number of rows. The third column takes on only two values. I wish to split the list into two sublists based on the value of the third column of the list element.  Second issue with lists as well. I would like to reduce each of the sublist based on the range of the second column, i.e. if the range of

hi is there a faster way to do this? i <- 1 for(x in vector1) for(y in vector2) { m[[i]] <- (x - y) i <- i + 1 } regards soren -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in

Using R 2.6.2, say I have the following list of lists, "comb": data1 <- list(a = 1, b = 2, c = 3) data2 <- list(a = 4, b = 5, c = 6) data3 <- list(a = 3, b = 6, c = 9) comb <- list(data1 = data1, data2 = data2, data3 = data3) So that all names for the lowest level list are common. How can I most efficiently access all of the sublist items "a" indexed by the outer

Dear all, The below has been tested quite thoroughly by now, including performance-testing by the way of using a modified compiler that triggers the below while compiling at least an old part of LLVM ["Function.cpp"] and sorting a symbol table with >7000 global variables. Unfortunately, the optimization I have been working on for which I _thought_ I needed the ability to sort a

hi I need 'rw1062.zip' since i can't get excel, R1070 or R1071 to work with the R (D)COM Server everything worked fine in the god rw1062 days the lapack routines can't be loaded 'unable to load shared library c:\r/moduleslapack.dll, the specified library could not be found' don't know if it's the forward slashes that anoys windows btw, the libraries are in the

hi there is there a way to convert objects of class ftable into LaTex code preserving the 'look' with row and column infomation? xtable() {xtable} can't handle such objects and latex() {Hmisc} just texify the number matrix, without row/column information regards soren

Could anyone please point me to the library readline41-4.1-10.i386.rpm which is requiered for installing the R-base package on Redhat 7.2 regards -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

hi there say that i have this table >x<-table(adoc, oarb) >x oarb 0 1 adoc ab 1 0 am 5 1 ba 14 1 cc 271 3 ch 87 2 dz 362 6 fl 7 0 fs 84 2 is there an easy way to get the row names or row

[text file is also attached in case you find the format of email difficult to understand] Dear useRs,You must all the planning for the christmas, but i am stucked in my office on the following issue i had a file containg information about station name, year, month, day, and discharge information. i opened it by using following command > dat1<-read.table("EL.csv",header=TRUE,