similar to: Finding solution

Dear all, let say, I have following list object: listObj <- vector("list", length = 3) listObj[[1]] <- rnorm(3) listObj[[2]] <- rnorm(4) listObj[[3]] <- rnorm(5) Now I want to convert above list into a Matrix. Ofcourse I can do it using "Reduce("rbind", listObj)". However as you notice that as elements of that list are arbitrary length vectors,

Dear all, assume I have a matrix with just 1 row. Now suppose I want to fetch 1st few rows from that matrix, however resulting object becomes vector. Here is 1 such example: > matrix(1:5, 1) [,1] [,2] [,3] [,4] [,5] [1,] 1 2 3 4 5 > > matrix(1:5, 1)[,-1] [1] 2 3 4 5 Can somebody point me how to keep resulting object as matrix with same row? Ofcourse I

Hello, I am a new R user. I am trying to use the arima command, but I have a question on intermediate lags. I want to run in R the equivalent Stata command of ARIMA d.yyy, AR(5) MA(5 7). This would tell the program I am interested in AR lag 5, MA lag 5, and MA lag 7, all while skipping the intermediate lags of AR 1-4, and MA 1-4, 6. Is there any way to do this in R? Thank you. -- View this

Dear all, Let say I have following data: RawData <- matrix(1:101, nr = 1); colnames(RawData) <- c("ASD", as.character(as.yearmon(seq(as.Date("2012-03-01"), length.out = 100, by = "1 month")))); rownames(RawData) <- "XYZ" CutOffDate <- as.Date("2012-09-01") NewDateSeries <- as.character(as.yearmon(seq(CutOffDate, to =

Hi again, I was trying to download stock market data from below link : https://www.nseindia.com/products/content/equities/equities/archieve_eq.htm Input choice : Select Report: Bhavcopy Date(DD-MM-YYYY): 03-03-2010 If you put manual input as above, then we will get option for manual download of file : cm03MAR2010bhav.csv.zip However I then tried to use R to have some automatic download :

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =

Hi Joshua, thanks for your prompt reply. However as I said, sum() function I used here just for demonstrating the problem, I have other custom function to implement, not necessarily sum() I am looking for a generic solution for above problem. Any better idea? Thanks, On Fri, Aug 11, 2017 at 12:04 AM, Joshua Ulrich <josh.m.ulrich at gmail.com> wrote: > Use a `width` of integer index

Dear all. Let say I have a group of codes which will be used in many places in my overall R-code files. These group of codes will be used within a for-loop (with a big length, like 10000 times) and also many other places outside of that for loop. As this group of codes are being used in many places, I thought to put them within a user-defined function. Here my question is, is there any speed

Hi friends, there is methods() function to see the all available methods for a particular function, for example: > head(methods("print")) [1] "print.acf" "print.anova" "print.aov" "print.aovlist" "print.ar" "print.Arima" In this list, there are some functions which are asterisked like print.acf().

Bert, Posit (formerly RStudio) has moved from RMarkdown to Quarto. They still support RMarkdown but major new features will be in Quarto. For new users a better choice would be Quarto. See https://quarto.org/docs/faq/rmarkdown.html Secondly, the OP stated he was using the VS-Code IDE, so there is no reason to point him to the Posit/RStudio IDE for this functionality which VS-Code supports very

I happened to see these: > round(.5, 0) [1] 0 > round(1.5, 0) [1] 2 > round(2.5, 0) [1] 2 > round(3.5, 0) [1] 4 > round(4.5, 0) [1] 4 What is the rule here? Should not round(.5, 0) = 1, round(2.5, 0) = 3 etc? Thanks and regards,