similar to: Rolling window difference for zoo time series

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 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

Hi again, I am wondering there is any function for 'zoo' time series, where I can apply a user defined function rolling window basis, wherein window size is ever increasing i.e. not fixed. For example, let say I have below user defined function and a zoo time series : > library(zoo) > UDF = function(x) sum(x) > TS = zoo(rnorm(10), seq(as.Date('2017-01-01'),

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

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]] <-

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

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 all, can anyone please guide me how to create a sequence of months? Here I have tried following however couldn't get success > library(zoo) > seq(as.yearmon("2010-01-01"), as.yearmon("2010-03-01"), by="1 month") Error in del/by : non-numeric argument to binary operator What is the correct way to do that? Thanks for your time.

I have a time series dataset, saved in a csv file. However date-formatting is : 7/2/1982 7/6/1982 7/7/1982 7/8/1982 7/13/1982 7/14/1982 However if I use following zoo-code, it is not reading data: read.zoo(file="F:/data.csv", format="%m/%d/%y", header=F) Error is Error in read.zoo(file ="F:/data.csv", format = "%m/%d/%y", : index contains NAs Here

Dear all, I have a date related question. Suppose I have a character string "March-2009", how I can convert it to a valid date object like as.yearmon("2009-01-03") in the zoo package? Is there any possibility there? Ans secondly is there any R function which will give the names of of all months as "LETTERS" does? Thanks for your time. [[alternative HTML version

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 ==

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,

Dear all, I would like to ask one question related to statistics, for specifically on defining dummy variables. As of now, I have come across 3 different kind of dummy variables (assuming I am working with Seasonal dummy, and number of season is 4): > dummy1 <- diag(4) > for(i in 1:3) dummy1 <- rbind(dummy1, diag(4)) > dummy1 <- dummy1[,-4] > > dummy2 <- dummy1 >

@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 again, Let say I have an non-negative integer vector (which may be random): Vec <- c(0, 13, 10, 4) And I have a date: > Date <- as.Date(Sys.time()) > Date [1] "2013-03-09" Using these 2 information, I want to get following date-vector: New_Vec <- c("2013-03-01", "2014-04-01", "2014-01-01", "2013-07-01") Basically the

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,