similar to: Zoo rolling window with increasing window size

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

Replace "sum" with your custom function's name. I don't see any reason why that wouldn't work, and the problem with my solution is not clear in your response. r <- rollapplyr(x, seq_along(x), yourCustomFunctionGoesHere) On Thu, Aug 10, 2017 at 1:39 PM, Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > Hi Joshua, thanks for your prompt reply. However

Use a `width` of integer index locations. And you likely want = "right" (or rollapplyr(), as I used). R> set.seed(21) R> x <- rnorm(10) R> rs <- rollapplyr(x, seq_along(x), sum) R> cs <- cumsum(x) R> identical(rs, cs) [1] TRUE On Thu, Aug 10, 2017 at 1:28 PM, Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > Hi again, > > I am

Hi, I have a 'zoo' time series as below : Zoo_TS = zoo(5:1, as.Date(Sys.time())+0:4) Now I want to calculate First order difference of order 1, rolling window basis i.e. (Zoo_TS[2] - Zoo_TS[1] ) / Zoo_TS[1] (Zoo_TS[3] - Zoo_TS[2] ) / Zoo_TS[2] ..... Is there any direct function available to achieve this? Thanks,

Zoo_TS/lag(Zoo_TS) - 1 On Tue, Apr 24, 2018 at 9:29 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > Hi, > > I have a 'zoo' time series as below : > > Zoo_TS = zoo(5:1, as.Date(Sys.time())+0:4) > > Now I want to calculate First order difference of order 1, rolling > window basis i.e. > > (Zoo_TS[2] - Zoo_TS[1] ) / Zoo_TS[1] >

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().

I have a zoo object on daily data for 10 years. Now I want to create a list, wherein each member of that list is the monthly observations. For example, 1st member of list contains daily observation of 1st month, 2nd member contains daily observation of 2nd month etc. Then for a particular month, I want to divide all observations into 3 parts (arbitrary) and then want to calculate some statistics

  Hi,   The code below does exactly what I want in sequential mode. But, it is slow and I want to run it in parallel mode. I examined some windows version packages (parallel, snow, snowfall,..) but could not solve my specific problem. As far as I understood, either I have to write a new function like sfRollapplyr or I have to change my code in a way that it utilizes lapply, or sapply instead of

Dear all, please consider my following workbook: library(zoo) lis1 <- vector('list', length = 2) lis2 <- vector('list', length = 2) lis1[[1]] <- zooreg(rnorm(20), start = as.Date("2010-01-01"), frequency = 1) lis1[[2]] <- zooreg(rnorm(20), start = as.yearmon("2010-01-01"), frequency = 12) lis2[[1]] <- matrix(1:40, 20) lis2[[2]] <-

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

Hello: I want to get a rolling estimation of the stdev of my data. Searching the document, I found the function "rollapply" in the zoo package. For example, my series is "c", and i want get a period of 10 days, so i write the command below: roll.sd = rollapply( c, 10, sd, na.pad = TRUE, align = 'right' ) but there is an error in it ,and the computing cannot be

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 =