similar to: ccf time units

Hi list, I have the following code to compute the acf of a time series acfresid <- acf(residfit), where residfit is the series when I type acfresid at the prompt the follwoing is displayed Autocorrelations of series ?residfit?, by lag 0.0000 0.0833 0.1667 0.2500 0.3333 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333 1.000 -0.015 0.010 0.099 0.048 -0.014 -0.039 -0.019 0.040 0.018

I have data as below.Please let me know how the ACF and Pacf used to determine the order od arima model. Is there any rules need to be followed to determine order.Please advise > turkey.price.ts Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 2001 1.58 1.75 1.63 1.45 1.56 2.07 1.81 1.74 1.54 1.45 0.57 1.15 2002 1.50 1.66 1.34 1.67 1.81 1.60 1.70 1.87 1.47 1.59 0.74 0.82

Hi all, I have a package that is specific to a task I was repetitively using a few years ago. I now needed to run it again with new data. However I am told it was built with an older version or R and will not work. How can I tweak the package so it will run on 11.1? It was a one-off product and has not been maintained. Is there a way to "unpackage" it and repackage it to work? I

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

Dear list, I've recently become interested in comparing the spectral estimates using the different methods ("pgram" and "ar") in the spectrum() function in the stats package. With many thanks to the authors of these complicated functions, I would like to point out what looks to me like a bit of an inconsistency -- but I would not be surprised if there is good reasoning

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

Hi, I am far from experienced in both R and time series hence the question. The code for spec.pgram() seems to involve a circularity of the kernel (see below) yielding new power estimates to all frequencies computed by FFT. " if (!is.null(kernel)) { for (i in 1:ncol(x)) for (j in 1:ncol(x)) pgram[, i, j] <- kernapply(pgram[, i, j], kernel, circular = TRUE)

Greetings R-helpers, I am attempting to fit an lme() while specifying a correlation structure, but I'm getting into trouble long before I get to that point. I am receiving the error: Error in y[revOrder] - Fitted : non-conformable arrays It doesn't seem to matter how simple or complex the model I specify is, it always gives this same error message. This makes me suspect something is

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Hi all i need to put on one graph 2 functions who's x axis is the same and y not. I mean on horizontal the time, and on vertical left: pressure, on vertical right: rpm of a motor, is R able to do that? i've found this that i could adapt maybe (i don't need time series really?) :/ : (http://tolstoy.newcastle.edu.au/R/help/04/03/1456.html) ## ## Description: A simple function which

Hi All, Just hoping some one can give me a hand with a problem... I have a dataframe (DF) with about 5 million entries that looks something like the following: >DF ID Cl Co Brd Ind A AB AB 1 S-3 IND A BR_F BR_F01 1 0 0 2 S-3 IND A BR_F BR_F01 1 0 0 3 S-3 IND A BR_F BR_F01 1 0 0 4 S-3 IND A BR_F BR_F01 1 0 0 5 S-3 IND A BR_F BR_F01 1 0 0 6 S-3 IND A BR_F

can you recommend a good manual for R that starts with a data set and gives demonstrations on what can be done using R? I downloadedR Langauage definition and An introduction to R but haven't found them overly useful. I'd really like to be able to follow some tutorials using a dataset or many datasets. The datasets I have available on R are Data sets in package 'datasets':

Hi there, I am trying to use linear regression to solve the following equation - y <- c(0.2525, 0.3448, 0.2358, 0.3696, 0.2708, 0.1667, 0.2941, 0.2333, 0.1500, 0.3077, 0.3462, 0.1667, 0.2500, 0.3214, 0.1364) x2 <- c(0.368, 0.537, 0.379, 0.472, 0.401, 0.361, 0.644, 0.444, 0.440, 0.676, 0.679, 0.622, 0.450, 0.379, 0.620) x1 <- 1-x2 # equation lmFit <- lm(y ~ x1 + x2) lmFit Call:

Sun, 14 Mar 2004, Jan Verbesselt wrote: > Dear R specialists, > > I have two time series in a data.frame and want to plot them in the same > plot(), with the left axis scaled to time series 1 (-700,0) and the > right axis scaled to time series 2 (-0.2, 0.4). > > plot(timeserie1) > lines(timeserie2, col=c(2)) => this one should be scaled differently > with a new

Hello, I did a cross-validation using cvlm from DAAG package but wasn't sure how to assess the result. Does this result means my model is a good model? I understand that the overall ms is the mean of sum of squares. But is 0.0987 a good number? The response (i.e. gailRel5yr) has min,1st Quantile, median, mean and 3rd Quantile, and max as follows: (0.462, 0.628, 0.806, 0.896, 1.000, 2.400) ?

Hi, I am reading the book "Mixed Effects Models in S and S-Plus" and come across an example with the Rail data. I tried to use lm(travel~Rail,data=Rail) and got the following result: Call: lm(formula = travel ~ Rail, data = Rail) Residuals: Min 1Q Median 3Q Max -6.6667 -1.0000 0.1667 1.0000 6.3333 Coefficients: Estimate Std. Error t value Pr(>|t|)

>2) Under MacOS 10.2.2. Given the following dataset: > > > summary(missing.data) > Hosp..No. Category Offset Side > r060093: 4 F Post op 0.5 year :62 Min. :-5.8333 L: 63 > r023250: 3 E Pre op 0.5 year :54 1st Qu.:-0.4167 R:141 > r026316: 3 H Post op 1.5 years :44 Median : 7.7083 > r032583: 3 G Post op 1 year

Dear all, I have a matrix, called newdata1, > dim(newdata1) [1] 34176 83 It looks like: EntrezID Name S1 S2 S3 S4 S5..... 1 4076 CAPRIN1 0.1 0.2 0.3... 2 139170 WDR40B 0.4 0.5 0.6... 3 5505 PPP1R2P1 0.3 0.3 0.7... 4 4076 CAPRIN1 0.7 0.3 0.2... 5 139170 WDR40B null 0.8

Hi all, I try to understand the output of the summary(model). I make a simple model. > summary(m1anova) Call: glm(formula = peso ~ gen) Deviance Residuals: Min 1Q Median 3Q Max -4.3114 -2.3788 -0.9167 2.1581 5.7856 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4.8660 0.9808 4.961 0.000101 *** gen2g 3.8504

Hi all. Belove is the example from the cluster-help page wtih the output. I simply cannot figure out how to relate the estimate and robust Std. Err to the p-value. I am aware this a marginal model applying the sandwich estimator using (here I guess) an emperical (unstructered/exchangeable?) ICC. Shouldent it be, at least to some extend, comparable to the robust z-test, for rx :