similar to: Possible bug with glm.nb and starting values (PR#1695)

Hello, I have a question on unit root test with urca toolbox. First, to run a unit root test with lags selected by BIC, I type: > CPILD4UR<-ur.df(x1$CPILD4[5:nr1], type ="drift", lags=12, selectlags ="BIC") > summary(CPILD4UR) The results indicate that the optimal lags selected by BIC is 4. Then I run the same unit root test with drift and 4 lags:

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates &lt;- as.Date(data.df[,1]) selection&lt;-which(dates&gt;=as.Date("1986-1-1") &amp; dates&lt;=as.Date("2007-12-31")) dep &lt;- ts(data.df[selection,c("dep")]) indep.ret1

Hello guys .. I would like to have some help about as.table . I made a table with the autocorrelations of the returns whit 10 lags and i get this : autocorrelazione2 <- as.table(c((cor(r2[-1151,],lag(r2))),(cor(r2[- c(1151,1150),],lag(r2, k=2))),(cor(r2[- c(1151,1150,1149),],lag(r2, k=3))),(cor(r2[- c(1151,1150,1149,1148),],lag(r2, k=4))),(cor(r2[- c(1151,1150,1149,1148,1147),],lag(r2,

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Hello all, I need to figure out a way to lag a variable in by a number of days without using the zoo package. I need to use a remote R connection that doesn't have the zoo package installed and is unwilling to do so. So that is, I want a function where I can specify the number of days to lag a variable against a Date formatted column. That is relatively easy to do. The problem arises when I

Dear Members, Greetings! I would like to know how to create the lag variable for my data. # Load data and create time series object ---- oil <- read_xlsx("crudefinal.xlsx") pricet=ts(oil$price, start = c(2020, 22), frequency = 365) roilt=ts(diff(log(oil$price))*100,start=c(2020,22),freq=365) # Fit MSW model ---- roilt.lag0 =

Hi, is there any function in R that shifts elements of a vector to the opposite direction of what Lag() of the Hmisc package does? (something like, Lag(x, shift = -1) ) Thanks Zava -------------------------------------------------------- This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}

Hello I'm only user of R and have many little knowledge in programming but I permit to send you some whishes/suggestions for R. which.min like which(), which.min() should also include an argument arr.ind. Note that one can have it with which(a==min(a), arr.ind=TRUE) but if there is a reason to build a special function which.min, why not add also this nice argument? lag() If one wants to

I am using R 63.0. Now let's try this simple image plot. Here is the data file: ============================ lag1 lag2 cif2d 1 1 11 1 2 12 1 3 13 2 1 21 2 2 22 2 3 23 3 1 31 3 2 32 3 3 33 ==================== data<-read.table("~/r/rt/data/unif/junk.out",header=TRUE) x<-unique(data$lag1) y<-unique(data$lag2) z<-matrix(data$cif2d,length(y),length(x)) At this point, see

Hi all, I have run a ridge regression on a data set 'final' as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it

I am using R on unix version 63.0 I am doing an image plot of the following data file: ================================ lag1 lag2 cif2d 0.000 0.000 NaN 0.000 1.000 0.500000 0.000 2.000 0.489831 0.000 3.000 0.492986 0.000 4.000 0.493409 0.000 5.000 0.492727 0.000 6.000 0.494485 1.000 0.000 0.500000 1.000 1.000 NaN 1.000 2.000 0.495098 1.000 3.000 0.489831 1.000 4.000 0.492986 1.000 5.000

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

I'm currently working with some time series data with the xts package, and would like to generate a forecast 12 periods into the future. There are limited observations, so I am unable to use an ARIMA model for the forecast. Here's the regression setup, after converting everything from zoo objects to vectors. hire.total.lag1 <- lag(hire.total, lag=-1, na.pad=TRUE) lm.model <-

Hi, I was wondering if there is a more efficient way of handling the following method of creating a lagged value in a data frame without using the recursive 'for(i in 1:n)' loop and without using as.ts #Steps to creating a lag variable in a data frame 'my.dat.fr' # with 275 columns, 2400 rows of numbers and factors . The #variable x is a factor of #with five different levels the

Suppose we observe N individuals, for each of which we have a time-series. How do we correctly create a lagged value of the time-series variable? As an example, suppose I create: A <- data.frame(year=rep(c(1980:1984),3), person= factor(sort(rep(1:3,5))), wage=c(rnorm(15))) > A year person wage 1 1980 1 0.17923212 2 1981

I think my problems are coused by a fundamental R incompatibility in how matrices are stored and the usual way of specifying Cartesian coordinates. When I do data<-read.table("~/r/rt/data/unif/6cbcif2d.out",header=TRUE) x<-unique(data$lag1) y<-unique(data$lag2) z<-matrix(data$cif2d,length(y),length(x)) This z matrix is printed apparently correctly from a Cartesian point of

Dear all, I have weekly data by city (variable citycode). I would like to take the average of the previous two, three, four weeks (without the current week) of the variable called value. This is what I have tried to compute the average of the two previous weeks; df = df %>% mutate(value.lag1 = lag(value, n = 1)) %>% mutate(value .2.previous = rollapply(data = value.lag1,

I am sure that this sort of thing has been asked and answered before, so in case my suggestions don't work for you, just search the archives a bit more. I am also sure that it can be handled directly by numerous functions in numerous packages, e.g. via time series methods or by calculating running means of suitably shifted series. However, as it seems to be a straightforward task, I'll

Hi R, I have doubt. >x= c(4,5,6,3,2,4,5) >acf(x,plot=F,lag.max=1) Autocorrelations of series 'x', by lag 0 1 1.000 0.182 But if I actually calculate the autocorrelation at lag1 I get, >cor(x[-1],x[-length(x)]) [1] 0.1921538 Even in excel I get 0.1921538 value. So, I want to know what the 'acf' function is calculating here....