similar to: lapply sequence

Hi Everyone, I would like to do sequential subtractions within a group so that I know the time between separate observations for a group of individuals. My data: data <- structure(list(group = c("IND1", "IND1", "IND2", "IND2", "IND2", "IND3", "IND4", "IND5", "IND6", "IND6"), date_obs =

Hi llvm-dev, Does anybody know why we're not using the `add_test` feature in CMake [0] for unit tests? In particular, compiler-rt (sanitizers and xray) has a number of unit tests which could really just be built as normal binaries and invoked appropriately. If we're avoiding ctest [1], then for the unit tests in compiler-rt I'd like to see whether just using normal binaries can make

Hello R users I recently learned how to use this command: lapply(datum,"[[","ColumnName") Unfortunately, I don't know how exactly it works, what it's called (in particular the "[[" part], and what other things you can do with it (retrieve multiple columns?). Given datum is a list of dataframes with the same column, but different number of rows,

Dear all, I am trying to learn lapply. I would like, as a test case, to try the lapply alternative for the Shadowlist<-array(data=NA,dim=c(dimx,dimy,dimmaps)) for (i in c(1:dimx)){ Shadowlist[,,i]<-i } ---so I wrote the following--- returni <-function(i,ShadowMatrix) {ShadowMatrix<-i} lapply(seq(1:dimx),Shadowlist[,,seq(1:dimx)],returni) So far I do not get same results

Of course it's *possible*, in a fundamental sense. It's even pretty easy to get right in a compiler back end (in a conceptual sense). You have to touch a LOT of code, but all the changes are trivial. We did this at Tartan Labs back in the 90s. Done with only a bit of care, it makes debugging possible at any optimization level. The idea is to make the debug information reflect what the

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 "..."

I've defined my own version of summary.default, that gives a better summary for highly skewed vectors. If I call summary(x) the method is used. If I call summary(data.frame(x)) the method is not used. I've traced this to lapply; this uses the new method: lapply(list(x), function(x) summary(x)) and this does not: lapply(list(x), summary) If I make a copy of lapply, WITHOUT the

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

Hello R-users, I have the following problem, that I want to solve efficiently: I have a named list, for example: > l <- list(a = 1, b = 3, c = 'asd') > l $a [1] 1 $b [1] 3 $c [1] "asd" I know that I can iterate through it using lapply() function, but I would also like to able to get the list names or some attributes of l in the lapply(). For example if I use

Hi, I would like to get lapply() to work in the natural way on a class I've defined. As far as I can tell, lapply() needs the class to be coercible to a list. Even after I define as.list() and as.vector(x, mode="list") methods, though, I still get an "Error in as.vector(x, "list") : cannot coerce to vector". What am I doing wrong? # dummy class