similar to: make methods work in lapply - remove lapply's environment

(inline) On Tue, 28 Mar 2017, Rolf Turner writes: > On 28/03/17 04:21, Barry Rowlingson wrote: >> On Mon, Mar 27, 2017 at 1:17 AM, Rolf Turner <r.turner at auckland.ac.nz> wrote: >>> >>> Is there any way to trap/detect the use of an optional argument called >>> "X" and thereby issue a more perspicuous error message? >>> >>>

On Mon, Mar 27, 2017 at 1:17 AM, Rolf Turner <r.turner at auckland.ac.nz> wrote: > > Is there any way to trap/detect the use of an optional argument called > "X" and thereby issue a more perspicuous error message? > > This would be helpful to those users who, like myself, are bears of very > little brain. > > Failing that (it does look impossible) You can

Consider the following expression, in which we pass 'i=', with no value given for the 'i' argument, to lapply. lapply("x", function(i, j) c(i=missing(i),j=missing(j), i=) >From R-2.14.0 (2011-10-31) through R-3.4.4 (2018-03-15) this evaluated to c(i=TRUE, j=FALSE). From R-3.5.0 (2018-04-23) through R-4.0.0 (2020-04-24) this evaluated to c(i=FALSE, j=TRUE). Was

Hello, I am facing a problem with lapply which I ''''think''' may be a bug. This is the most basic function in which I can reproduce it: myfun <- function() { foo = data.frame(1:10,10:1) foos = list(foo) fooCollumn=2 cFoo = lapply(foos,subset,select=fooCollumn) return(cFoo) } I am building a list of dataframes, in each of which I want to keep only column

So I have a function that does lapply's for me based on dimension. Currently only works for length(pivotColumns)=2 because I haven't fixed the rbinds. I have two versions. One runs WAYYY faster than the other. And I'm not sure why. Fast Version: fedb.ddplyWrapper2Fast <- function(data, pivotColumns, listNameFunctions, ...){ lapplyFunctionRecurse <- function(cdata, level=1,

I am faced with a situation wherein I have to use multiple lapply's. The pseudo-code could be approximated to something as below: For each X from i=1 to n For each Y based on j=1 to m For each F from 1 to f Do some calculation based on Fij Store Xi,Yj = Fij End For F End for Y End for X Is there anyway to optimize the processing logic further? I *guess* using the multiple lapply

Would introducing the new frame, with the call to local(), cause problems when you use frame counting instead of <<- to modify variables outside the scope of lapply's FUN, I think the frame counts may have to change. E.g., here is code from actuar::simul() that might be affected: x <- unlist(lapply(nodes[[i]], seq)) lapply(nodes[(i + 1):(nlevels - 1)],

From: Daniel Kaschek <daniel.kaschek at physik.uni-freiburg.de> > ... When I evaluate this list of functions by > another lapply/sapply, I get an unexpected result: all values coincide. > However, when I uncomment the print(), it works as expected. Is this a > bug or a feature? > > conditions <- 1:4 > test <- lapply(conditions, function(mycondition){ >

There seems to be a funny interaction between lapply and "$" -- also, "$" doesn't signal an error in some cases where "[[" does. The $ operator accepts a string second argument in functional form: > `$`(list(a=3,b=4),"b") [1] 4 lapply(list(list(a=3,b=4)),function(x) `$`(x,"b")) [[1]] [1] 4 ... but using an lapply "..."

Hi Bert, On Tue, Nov 14, 2017 at 8:11 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Could someone please explain the following? I did check bug reports, but > did not recognize the issue there. I am reluctant to call it a bug, as it > is much more likely my misunderstanding. Ergo my request for clarification: > > ## As expected: > >> lapply(1:3, rnorm, n = 3)

I need to extract identically named columns from several data frames in a list. the column name is a variable (i.e. not known in advance). the whole thing occurs within a function body. I'd like to use lapply with a variable 'select' argument. example: tt <- function (n) { x <- list(data.frame(a=1,b=2), data.frame(a=3,b=4)) for (xx in x) print(subset(xx, select = n))

|Hello, Given a function with several arguments, I would like to perform an lapply (or equivalent) while holding one or more arguments fixed to some common value, and I would like to do it in as elegant a fashion as possible, without resorting to wrapping a separate wrapper for the function if possible. Moreover I would also like it to work in cases where one or more arguments to the original

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

Dear R experts, recently I tried to debug a R function with an internal lapply call. When debugging I seem not to be able to use the "n" command to debug the inner function called by lapply. How could I achieve this? *For example:* test <- function( ) { lapply( 1:3, function( x ) x + 1 ) } debug( test ) *Start debug:* > test() debugging in: test() debug bei #1:{

Hi, If needed, lapply() tries to convert its first argument into a list before it starts doing something with it: > lapply function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(lapply(X, FUN)) } But in practice, things don't always seem to "work" as suggested by this code (at least to the

Dear R helpers: I wonder how to pass more than one argument to the function called by lapply. For example, #R code below --------------------------- indf <- data.frame(id=I(c('a','b')),y=c(1,10)) #I want to add an addition argument cutoff into the function called by lapply. outside.fun <- function(indf, cutoff) { unlist(lapply(split(indf, indf[,'id']),

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l<p) sup=((l-1)*m)+m > else sup=n >

Dear R developers, Reviewing my code, I have realized that about 80% of the time in the lapply I need to access the names of the objects inside the loop. In such cases I iterate over indexes or names: lapply(names(x), ... [i]), lapply(seq_along(x), ... x[[i]] ... names(x)[i] ), or for(i in seq_along(x)) ... which is rather inconvenient. How about an argument to lapply which would specify the

R-help, I have a list whose elements are data frames. I want to change the colnames attribute in each element of this list but an error message comes up: > lapply(LD_strataNew,function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] Error in lapply(LD_strataNew, function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] : could not find function "lapply<-" >

I'm still getting familiar with lapply I have this date sequence x <- seq(as.Date("01-Jan-2010",format="%d-%b-%Y"), Sys.Date(), by=1) #to generate series of dates I want to apply the function for all values of x . so I use lapply (Still a newbie!) I wrote this test function pFun <- function (x) { print(paste("This is: ",x,sep="")) } When I