similar to: performance: zoo's rollapply() vs inline

Hi. I'm comparing output from rollapply.zoo, as produced by two versions of R and package zoo. I'm illustrating with an example from a R-help posting 'Zoo - bug ???' dated 2010-07-13. My question is not about the first version, or the questions raised in that posting, because the behaviour is as documented. I'm puzzled as to why na.rm no longer is passed to mean, i.e. why

# Hi there, # I am trying to apply a function over a moving-window for a large number of multivariate time-series that are grouped in a nested set of factors. I have spent a few days searching for solutions with no luck, so any suggestions are much appreciated. # The data I have are for the abundance dynamics of multiple species observed in multiple fixed plots at multiple sites. (I total I

Here is a relatively simple script (with comments as to the logic interspersed): # Some of these libraries are probably not needed here, but leaving them in place harms nothing: library(tseries) library(xts) library(quantmod) library(fGarch) library(fTrading) library(ggplot2) # Set the working directory, where the data file is located, and read the raw data

Hi I would like to understand how to extend the function (FUN) I am using in rollapply below. ###################################### With the following simplified data, test1 yields parameters for a rolling regression data = data.frame(Xvar=c(70.67,70.54,69.87,69.51,70.69,72.66,72.65,73.36), Yvar =c(78.01,77.07,77.35,76.72,77.49,78.70,77.78,79.58)) data.z = zoo(d) test1 =

Hi everyone, I have a rollapply statement that applies a function, in steps, over a data matrix as follows: #Code start testm<-rollapply(mat, 100, by=100, min, na.rm=F) #Code end This moves down matrix 'mat' and calculates the minimum value over a 100 row range, every 100 rows (i.e. no overlaps). NAs are not removed. I want to modify this statement somehow so that the rollapply()

Hi, Sorry ahead of time for not including data with this question. Using rollapply to calculate mean values for 5 day blocks, I'd use this: Roll5mean <- rollapply(data, 5, mean, by=5, align = c("left")) My question is, can someone tell me how to fill in the days between each of these means with the previously calculated mean? If this doesn't make sense, I will clarify and

Dear R users. I have zoo object "size_june" containing market-capital values: > dim(size_june) # market-cap data of 625 firms for 20 years [1] 20 625 > class(size_june) [1] "zoo" > size_june # colnames = "size.firmcode" size.34020 size.4710 size.11050 size.10660 size.9540 size.8060 size.16160 size.8080 size.9280 1988-06-30 NA

Hi all- Thank you for reading my post. Please bear in mind that I'm very much a newbie with R! My question is this: I'm trying to use rollapply() on an irregular time series so I can't simply use the width parameter (I don't think). Rather than last 5 entries, I'd like to rollapply on last 6 months (for example). What would be the proper course of action for this? Thanks!

Hi, I faced with problem when start using function - rollapply(returns, 3 , mean) Error in UseMethod("rollapply") : No suitable Method for "rollaply" How can I fix the problem? Thank you for help. -- Best regards, Andy Fetsun [[alternative HTML version deleted]]

#using "rollapply" to calculate a moving sum or running sum? #I am tryign to use rollapply to calcualte a moving sum? #I tried rollapply and get the error message #"Error in seq.default(start.at, NROW(data), by = by) : # wrong sign in 'by' argument" #example: mymatrix <- ( matrix(data=1:100, nrow=5, ncol=20) ) mymatrix_cumsum <- ( matrix(data=NA, nrow=5,

Dear Colleagues, I have faced with the problem that function rollaply with rolling window for calculation of volatility doesn't give the all results of calculations. I have run the rolling window for calculation in Excel and obtained that the number of outputs for Excel is 36 and for R is 18. The total number of observations is 37. In the attachment you can find pdf of the Excel and Excel

# need zoo to use rollapply() # your data (I called df) df <- structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L), e = c(1L, 5L), f = c(4, 7)), .Names = c("a", "b", "c", "d", "e", "f"), class = "data.frame", row.names = c(NA, -2L)) # transpose and make a zoo object df2 <- zoo(t(df)) #rollapply to get

Hello I have a vector a =(-2,0,0,0,1,0,0,3,0,0,-4) I want to replace all zeros into previous non-zero state. So for instance the above vector should be converted into: a= (-2,-2,-2,-2,1,1,1,3,3,3,-4) I tried many things and finally concluded that probably(?) rollapply may be the best way? I tried f= function(x){ ifelse(x==0,Lag(x),x) } And then, rollappy(a,1,f) and that

Dear List members I want to do the sliding window analysis of some specific values. Here is my code: require(zoo) dat <- read.table("chr1.txt", header = TRUE, sep="\t") dat2 <- cbind(dat[1,3]) #The first column is also important. It represents the position of the site on the chromosome. TS <- zoo(c(dat2)) a <- rollapply(TS, width=1000000, by=200000, FUN=mean,

Hi folks, I am confused whether the following is a bug or it is fine Here is the explanation a <- zoo(c(NA,1:9),1:10) Now If I do rollapply(a,FUN=mean,width=3,align="right") I get > rollapply(a,FUN=mean,width=3,align="right") 3 4 5 6 7 8 9 10 NA NA NA NA NA NA NA NA But I shouldn't be getting NA right ? i.e for index 10 I should get (1/3)*(9+8+7)

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

library(zoo) library(chron) t1 <- chron("1/1/2006", "00:00:00") t2 <- chron("1/31/2006", "23:45:00") deltat <- times("00:15:00") tt <- seq(t1, t2, by = times("00:15:00")) d <- sample(33:700, 2976, replace=TRUE) sin.zoo <- zoo(d,tt) #there are ninety six reading in a day d.max <- rollapply(sin.zoo, width=96, FUN=max)

Good morning, I have discovered what I believe to be a performance regression between Zoo 1.6x and Zoo 1.7-6 in the application of rollapply. On zoo 1.6x, rollapply of my function over my data takes about 20 minutes. Using 1.7-6, the same code takes about 6 hours. R --version R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform:

Hello everyone, I have a dataframe with 1 column and I'd like to replace that column with a moving average. Example: > library('zoo') > mydat <- seq_len(10) > mydat [1] 1 2 3 4 5 6 7 8 9 10 > df <- data.frame("V1" = mydat) > df V1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 > df[df$V1 <- rollapply(df$V1, 3, mean)]

Hi everyone, I'm trying to use apply (with a call to zoo's rollapply within) on the columns of a 1.5Kx165K matrix, and I'd like to make use of the other cores on my machine to speed it up. (And hopefully also leave more memory free: I find that after I create a big object like this, I have to save my workspace and then close and reopen R to be able to recover memory tied up by R, but