similar to: ccf (cross correlation function) problems

I can't understand the results from cross-correlation function ccf() even though it should be simple. Here's my short example: ********* a<-rnorm(5);b<-rnorm(5) a;b [1] 1.4429135 0.8470067 1.2263730 -1.8159190 -0.6997260 [1] -0.4227674 0.8602645 -0.6810602 -1.4858726 -0.7008563 cc<-ccf(a,b,lag.max=4,type="correlation") cc Autocorrelations of series 'X',

hello, i have been using ccf() to look at the correlation between lightning and electrogamnetic data. for the most part it has worked exactly as expected. however, i have come across something that puzzles me a bit: > x <- c(1, 0, 1, 0, 1, 0) > y <- c(0, 0, 0, 0, 0, 0) > ccf(x, x, plot = FALSE) Autocorrelations of series 'X', by lag -4 -3 -2 -1 0

Hi all, I'm seeing this failure on my PPC G4 box running TOT with llvm-gcc 4.2. Is anyone else seeing this? I'm sure it's related to the byval stuff that's recently gone into LLVM. I'm attaching the output of this command: $ llvm-gcc -emit-llvm -O3 -S -o - -emit-llvm /Users/wendling/llvm/ llvm.src/test/CFrontend/2008-01-25-ByValReadNone.c As you can see in it, there

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter

I am new to R and new to time series modeling. I have a set of variables (var1, var2, var3, var4, var5) for which I have historical yearly data. I am trying to use this data to produce a prediction of var1, 3 years into the future. I have a few basic questions: 1) I am able to read in my data, and convert it to a time series format using 'ts.' data_ts <- ts(data, start = 1988, end =

Hi, I am using ccf but I could not figure out how to calculate the actual lag in number of periods from the returned results. The documentation for ccf says:"The lag is returned and plotted in units of time". What does "units of time" mean? For example: > x=ldeaths > x1=lag(ldeaths,1) > results=ccf(x,x1) > results Autocorrelations of series 'X', by lag

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

Unless there is good reason not to, always cc the list -- there are lots of smarter folks than I on it who can help. I may or may not have time to look at this. Hopefully someone else will. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip

Hello, I am a very new R user and not a statistician so please excuse any over explanation, I'm just trying to be as clear as possible. I have performed a cross correlation of two time series (my columns) in a single data setusing: ccf(ts(A[rows,columnX]),(A[rows,columnY]), lag=NULL, type="correlation",plot=F) I?am able to get the results (for example): Autocorrelations of

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Hi, I have missing values in my time series. "na.action = na.pass" works for acf and pacf. Why do I get the following error for the ccf? > ts(matrix(c(dev$u[1:10],dev$q[1:10]),ncol=2),start=1,freq=1) Time Series: Start = 1 End = 10 Frequency = 1 Series 1 Series 2 1 68.00000 138.4615 2 70.00000 355.5556 3 68.76000 304.3200 4 68.00000 231.4286 5 69.74194 357.4963 6

Hi. It's my understanding that a cross-correlation function of vectors x and y at lag zero is equivalent to their correlation (or covariance, depending on how the ccf is defined). If this is true, could somebody please explain why I get an inconsistent result between cov() and ccf(type = "covariance"), but a consistent result between cor() and ccf(type = "correlation")? Or

Hi all, I am getting different results from ccf and cor, Here is a simple example: set.seed(100) N <- 100 x1 <- sample(N) x2 <- x1 + rnorm(N,0,5) ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1] cor(x1[-N], x2[-1]) Results: > ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1] [1] -0.128027 > cor(x1[-N], x2[-1]) [1] -0.1301427 Thanks, Tal ----------------Contact

Dear R-users, I have been using R and its core-packages with great satisfaction now for many years, and have recently started using the "ccf" function (part of the "stats" package version 2.16.0), about which I have a question. The "ccf"-algorithm for calculating the cross-correlation between two time series always calculates the mean and standard deviation per time

I am trying to run a cross-correlation using the "ccf()" function. When I select plot = TRUE in the ccf() I get a graph which has ACF on the y-axis, which would suggest that these y-values are the auto-correlation values. How should I adjust the code to produce a plot that provides the cross-correlation values? Here is my code: w002dat <-

Hi, I have been running the ccf() function to find cross-correlations of time series across various lags. When I give the option of plot=TRUE, I get a plot that gives me 95% confidence interval cut-offs (based on sample covariances) for my cross-correlations at each lag. This gives me a sense of whether my cross-correlations are statistically significant or not. However, I am unable to get R to

Dear R gurus, I would like to use the ccf function on two matrices that are each 196000 x 12. Ideally, I want to be able to go row by row for the two matrices using apply for the ccf function and get one 196000 X 1 array output. The apply function though wants only one array, no? Basically, is there a way to use apply when there are two arrays in order to do something like correlation on a row

Hi I'm computing the correlation between two time-series x_t and y_t-1 (time-series lagged using the lag(y,-1) function) using the cor() function and the returned value is different from the value of ccf() function at the same lag. Any ideas why this is so? Thanks in advance for any hints. Mihnea [[alternative HTML version deleted]]