similar to: Help needed regarding acf

help . I have a file which contains a table that has 2 columns of which the first one is a time sereis. I need to extract the first column alone and do an acf plot on it. I dont know how to extract the first column and pass it to the acf function in the ts package. Can anybody pls help me. end -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list

I am trying to understand the function acf stats:::acf shows me the function I am having trouble understanding the usage "$acf" in the following acf <- array(.C(R_acf, as.double(x), as.integer(sampleT), as.integer(nser), as.integer(lag.max), as.integer(type == "correlation"), acf = double((lag.max + 1L) * nser * nser), NAOK =

I am guessing this is a very simple question, but this is only my second day with R so it is all still a bit imposing. I am trying to run an autocorrelation. I imported a CSV file, which has one column labeled "logistic". I ran the command: ACF(data$logistic,maxLag=10) However, I received the error: Error in UseMethod("ACF") : no applicable method for 'ACF'

Full_Name: Ian McLeod Version: 2.3.1 OS: Windows Submission from: (NULL) (129.100.76.136) > There is a simple bug in acf as shown below: > > z <- 1 > acf(z,lag.max=1,plot=FALSE) > Error in acf(z, lag.max = 1, plot = FALSE) : > 'lag.max' must be at least 1 > This is certainly a bug. There are two problems: (i) the error message is wrong since lag.max is

Hi, Im new to R so this question is quite fundamental. Im trying to compare some autocorrelations generated by the acf function to some theoretical correlations. How can I have acces to just the autocorrelations, for computation? This is some of my code: > acf.data<-c(acf(x)) > acf.data This is the R output: $acf , , 1 [,1] [1,] 1.000000000 [2,]

Hi, I don't quite understant how can I access to the acf values from the list produced by the acf function Example: library(ts) t <- acf(ts.union(ts(1:10), ts(11:20))) t$acf > tmp$acf , , 1 [,1] [,2] [1,] 1.00000000 1.00000000 [2,] 0.70000000 0.70000000 [3,] 0.41212121 0.41212121 [4,] 0.14848485 0.14848485 [5,] -0.07878788 -0.07878788 [6,] -0.25757576

Hello all, I found a funny problem with y-axis labels when plotting acf(matrix) - the labels are too close to one of the margins and cut in half. Here's the problem: test<-matrix(rnorm(200),ncol=4) acf(test) This doesn't fix the problem: test<-matrix(rnorm(200),ncol=4) par(mar=c(3,3,2,0.2),oma=c(0,0,0,0)) acf(test) This does fix the margin. I understand why, but not sure why ONLY

Hi everyone, I'm trying to use the acf() function to calculate the autocorrelation of each column in a matrix. The trouble is that I can only seem to get the function to work if I extract the data in the column into a separate matrix and then apply the acf() function to this column. I have something like this: acf(mat,lag.max=10,na.action=na.pass) ...but I would really like to apply the

Shalini Raghavan 3M Pharmaceuticals Research Building 270-03-A-10, 3M Center St. Paul, MN 55144 E-mail: sraghavan at mmm.com Tel: 651-736-2575 Fax: 651-733-5096 ----- Forwarded by Shalini Raghavan/US-Corporate/3M/US on 08/16/2004 11:25 AM ----- Shalini

Hello, I was wondering if anybody could help me with this? I have plotted an acf function for a time series and am very happy with it. Now I am interested in calculating for myself the two values for the confidence intervals that are plotted on the graph of the acf. The confidence intervals do not appear to be returned from the acf function (is this true?). So far I haven't managed to

Hi R users, where can I find the equations used by acf function to calculate autocorrelation? I think I misunderstand acf. Doesn't acf use following equation to calculate autocorrelation? [image: R(\tau) = \frac{\operatorname{E}[(X_t - \mu)(X_{t+\tau} - \mu)]}{\sigma^2}\, ,] If it does, then the autocorrelation of a sine function should give a cosine; however, the following code gives a

Hello, I am attending a course in Computational Statistics at ETH and in one of the assignments I am asked to prove that a time series is not autocorrelated using the R function "acf". I tried out the acf function with the given data, according to what I found here: http://landshape.org/enm/options-for-acf-in-r/ this test data does not look IID but rather shows some trends so how can I

Hi List, I'm trying to calculate the autocorrelation coefficients for a time series using acf at various lags. This is working well, and I can get the coefficients without any trouble. However, I don't seem to be able to obtain the significance of these coefficients from the returned acf object, largely because I don't know where I might find them. It's clear that the acf

(Moved from r-devel to r-bugs) On 3/24/2006 5:03 AM, Antonio, Fabio Di Narzo wrote: > Hi all. > There's a bug in plot.acf, when plotting acf for multivariate time series. > Here a reproducible example: > > X <- rnorm(1000) > Y <- -X + rnorm(1000, sd=0.6) > Z <- cbind(X,Y) > > In > acf(Z) > cross-correlation plot y-axis is limited to 0-1. But: >

Hi, I would like to use acf.plot on a correlogram that is computed externally. In other words, I would like to "fake out" the acf object. Is this possible?-- any help would be appreciated. TIA Martin

Hi I'm getting the following error that do not make sense to me, what am Idoing wrong ? > acf(Recsim[1,], lag.max=1) Error in acf(Recsim[1, ], lag.max = 1) : 'lag.max' must be at least 1 Regards EJ

Dear R list As suggested by Prof Brian Ripley, I have tried to read acf literature. The main problem is I am not the statistician and hence have some problem in understanding the concepts immediately. I came across one literature (http://www.stat.nus.edu.sg/~staxyc/REG32.pdf) on auto-correlation giving the methodology. As per that literature, the auto-correlation is arrived at as per following.

Hi, Can anyone please provide the formula used to compute ACF(nlme). I believe the one that is used in R is of the type mentioned on the website. Please correct me if I am wrong. The normalization of the numerator (Ch) has been done by 'N' where as I would like to do it by 'N-k'. Is there anyway in the present implementation of ACF to normalize it by 'N-k', where

Hi, I noticed that whenever I ran acf or pacf, the plot generated by R always includes two horizontal blue doted lines. Furthermore, these two lines are not documented in the acf documentation. I don't know what they are for, but it seems that they are important. Could someone tell me what they are and how are they calculated? Thanks, -- Tom [[alternative HTML version deleted]]

Dear all, given two numeric vectors x and y, the ACF(x) at lag k is cor(x(t),x(t+k)) while the CCF(x,y) at lag k is cor(x(t),y(t-k)). See below for a simple example. > set.seed(1) > x <- rnorm(10) > y <- rnorm(10) > x [1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078 -0.8204684 0.4874291 0.7383247 0.5757814 -0.3053884 > y [1] 1.51178117 0.38984324