search for: unlistable

Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1 A 1 2 B 2 [[2]] Name

Dear list, I hope this is the right place to post a feature request. If there's exists a more formal channel (e.g. as for bug reports), I'd appreciate a pointer. I work a lot with named nested lists with arbitrary degrees of "nestedness". In order to retrieve the names and/or values of "bottom layer/bottom tier", I love the functionality of 'unlist()', or

Here is a description and a proposed solution for a bug in unlist(). I've used version 2.7.2 RC (2008-08-18 r46382) to look at this, under linux. unlist(recursive=TRUE) incorrectly returns a factor with zero levels when passed either a nested list of factors, or a data frame containing only factor columns. You can't print() the result. x <- list(list(v=factor("a")))

Hi all, I wrote a function called relist, which is an inverse to the existing unlist function: http://www.econ.upenn.edu/~clausen/computing/relist.R Some functions need many parameters, which are most easily represented in complex structures. Unfortunately, many mathematical functions in R, including optim, nlm, and grad can only operate on functions whose domain is a vector. R has a

Hi the list, unlist respect the all the atomic type except orderd (it change of ordered into factor) : ### integer class(unlist(list(1:5,1:3))) #[1] "integer" ### numeric class(unlist(list(1.2,3.5))) #[1] "numeric" ### character class(unlist(list("e","e"))) #[1] "character" ### factor

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

Hello I want to unlist the attached element getting only the first element in each element of the list. The last element of the list looks as this: [[5065]] [[5065]]$Pluv3Meses [1] 274.4 [[5065]]$PluvMesesMedio [1] 378.2667 [[5065]]$Pluv2UltimosMeses [1] 23.33333 So I would like to get for each element of the list the element called Pluv3Meses. The whole list has 5065 elements but when I try to

Statement 9 using sapply does not seem to give the correct answer (or at least to me). Yet I do what I think is the same thing with statement 11 and I get the answer I'm looking for. 9 : s <-sapply(unlist(v[c(1:length(v))]), max) 11: for(i in 1 :length(v)) v1[i] <- max(unlist(v[i])) Shouldn't I get the same answer ? library(XML) rm(list=ls()) url <-

Is this a bug? > unlist(list(as.Date("2015-01-01"))) [1] 16436 [[alternative HTML version deleted]]

Hi there, I have a problem about lapply, strsplit, and accessing list elements, which I don't understand or cannot solve: I have e.g. a character vector with three elements: x = c("349/077,349/074,349/100,349/117", "340/384.2,340/513,367/139,455/128,D13/168", "600/437,128/903,128/904") The task I want to perform, is to generate a list,

Hi, I have been using R for a few months and I have this working code. Don't seen any problem but this takes a long time. So if I have about 30000 rows it takes a few minutes. If I have 100000 it does not seem to complete. Original Data: Proto Recv-Q Send-Q Local-Address Foreign-Address State tcp 0 0 172.20.100.2:60255

Hi, I have a very rudimentary kind of question on using unlist(). I am parsing a bunch of JSON text using rjson package. Some data elements in a dictionary happen to be null which rjson parses correctly. > fromJSON(json_str='{"query":{"A":10, "B":null, "C":"hello"}, "query":{"A":20, "B":null,

Dear r-helpers, I have a list whose elements are > str(durCut[[1]]) Ord.factor w/ 5 levels "vLow"<"low"<"med"<..: 3 2 5 2 2 2 4 4 3 5 ... How do I unlist durCut into an ordered factor? > str(unlist(durCut)) int [1:3024] 3 2 5 2 2 2 4 4 3 5 ... _____________________________ Professor Michael Kubovy University of Virginia Department of Psychology

On 02 Dec 2016, at 23:13 , Herv? Pag?s <hpages at fredhutch.org> wrote: > More generally one might reasonably expect 'unlist(x)' to be equivalent > to 'do.call(c, x)' on a list 'x' where all the list elements are atomic > vectors: Well, both are generic, and e.g. there is no "Date" method for unlist(), but there is for c(). It is not clear that

unlist(recursive=FALSE) returns NULL elements when passed a nested pairlist containing non-NULL data: x <- pairlist(pairlist(1:2)) unlist(x, recursive=FALSE) ## [[1]] ## NULL version 2.7.2 RC (2008-08-18 r46382) under linux I'm unaware of any motivation for constructing the above data structure, but if unlist is going to operate on it without error I would be surprised if that is the

Dear fellow R users, I am replacing elements of a list like so: pulse_subset[[1:20]]=unlist(pulse[i])[1:20] where pulse is a list of lists, and pulse [i] has >20 values. This gives the error "Recursive Indexing failed at level 2". But, interestingly this instruction is part of a loop which has gone through about 200,000 iterations before giving this error. Actual code: >

Hi, I had a weird results from using apply(). Here is an simple example: > y<-data.frame(list(a=c(1,NA),b=c('2k','0'))) > y ??? a???? b 1? 1?? 2k 2 NA?? 0 > apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k' & !is.na(x[1]) & x[1]=='1') 1 else 0} ) This should print "1 0" as output, as demonstrated by: >

Reproducible example: x <- list(list(list(), list())) unlist(x) *> Error in as.character.factor(x) : malformed factor* What should happen: unlist(x) > NULL R.version platform x86_64-apple-darwin15.6.0 arch x86_64 os darwin15.6.0 system x86_64, darwin15.6.0 status major 3 minor 5.0 year 2018 month 04 day

Hi, I'm looking for some way to pick up the numbers which are contained and buried in a long character. For example, outtree.new="(((B:1204.25,E:1204.25):7581.11,F:8785.36):8353.85,C:17139.21);" num.char =

Hi there, I am looking for some help with a loop which uses three list()-objects: "sequence" has got numbers, "start" and "end" have got the begin/end time points which should be applied to the numbers in "sequence". length() of all three objects is identical, length() of vectors stored in "sequence" is longer than length() of "start"