similar to: problem with urca package

Dear R users, I have the joined txt file in the following direct directory C:// I have written the following lines: library(urca); PIBTUN<-read.table("C:/AF.txt", header=F); ur.df(PIBTUN,type='none',lags=1) but I have obtained the following message: Error in embed(z, lags) : 'x' is not a vector or matrix I don't What's the problem, can you please help me Thank

Bonjour à tous depuis maintenant un certain temps je tire mes cheveux et ne comprends pas la source de mon problème. après une migration de samba 3 pdc vers samba 4.8.5 AD, au démarrage d’un client Windows, l’ordinateur gpo n’est pas appliqué au démarrage. Dans les journaux Windows, il y a 1058 erreurs d'objet de stratégie de groupe et samba côté serveur. Voici les journaux: Mise à jour du

Bonjour, je voulais savoir si wxruby2 est intégré à l''interpréteur ruby? Que veut dire $0 ? merci Sebastien _______________________________________________ wxruby-users mailing list wxruby-users@rubyforge.org http://rubyforge.org/mailman/listinfo/wxruby-users

Thank you Matthieu for your helpful suggestions. Unfortunately I still have problems. I have tried to compile it via your suggestion. " this is strange... you should have the usual summary... I have on my machine library(urca) test2<-ur.df(nottem, type="none", lags=1) summary(test2) " When I type in "library(urca)" I receive the following. "Error in

I can't figure out how to get the "summary" method in the URCA package to work. E.g. when I use the following code fragment in the help for the "ca.jo" function, it always tries to use the "summary" method from the "base" package, not the "urca" package. How do I force it use the "summary" method of the "urca" package?

Hi I am using the package urca and I am interested about the KPSS test. That works fine except the method "summary" did not work in the script, only when it is typed direct in the console the results are shown( not a source file). Is there any problem with these method ?

I'm tring the functions to check the cointegration of a matrix. I'm using **Phillips & Ouliaris Cointegration Test** The function in *tseries* package is **po.test** and **ca.po** in *urca* The results with **URCA** are: > ca.po(prices, demean='none') ######################################## # Phillips and Ouliaris Unit Root Test #

Dear R-list member, an update of package 'urca' has been uploaded to CRAN (Mirror: Austria). In the updated release unit root and cointegration tests encountered in applied econometric analysis are implemented. The package is written in 'pure' R and utilises S4 classes. In particular, the Johansen procedure with likelihood ratio tests for the inclusion of a linear trend,

Dear R Core Team, I am using package {urca} to do cointegration and estimate ECM model, but I have the following two problems: (1) I use ca.jo() to do cointegration first and can get the cointegration rank, alpha and beta. The next step is to test some restrictions on beta with blrtest(),bh5lrtest(), and bh6lrtest(). But none of them can add restrictions on all the cointegration

A question on p-value of Unit Root Tests using “urca” toolbox There’s a function “punitroot” on unit root probability value punitroot(q, N = Inf, trend = c("c", "nc", "ct", "ctt"), statistic = c("t", "n"), na.rm = FALSE) To my knowledge, “c” means “const” or intercept, “nc” means no const or intercept, and “ct” means time trend

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Hi, I can plot the ca.jo package to view the cointegrated relationship for each eigen value. Or I can use the normalized eigen vector to reconstruct the cointegrated relationship series. However, since the package can plot that for me, I wonder is there any specific slot/method in the class from where I can invoke to get this relationship instead of doing a duplicated work? Thank you. - adschai

Hello!   I estimate vector error correction model (vecm) model. I have only one cointegratio relationship. I write :   joh.vecm.rls <- cajorls(joh.vecm, r=1) The output estimation is : Call: lm(formula = substitute(form1), data = data.mat) Coefficients:                up.d            expl.d        upd.d           r.d      ect1      -1.34e-01   4.55e+02   6.91e+00   2.43e+03 constant 

Hi R users! I've come across using kpss tests for time series analysis and i have a question that troubles me since i don't have much experience with time series and the mathematical part underlining it. x<-c(253, 252, 275, 275, 272, 254, 272, 252, 249, 300, 244, 258, 255, 285, 301, 278, 279, 304, 275, 276, 313, 292, 302, 322, 281, 298, 305, 295, 286, 327, 286, 270, 289, 293, 287,

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:

Dear R users, I would like to simulate, for 20000 replications, an autoregressive process: y(t)=0.8*y(t-1)+e(t) where e(t) is i.i.d.(0,sigma*sigma), Thank you in advance ____________________________________________________ Écoutez gratuitement le nouveau single de Noir Désir et découvrez d'autres titres en affinité avec vos goûts musicaux

Hi everybody, I'am very interested with the basic structural model of time series. So I used the command "StructTS" but I failed to obtain a desirable output, in fact when I write in R Console the following lines: > x=(1,2,3,4,5,2,25,14,12,13,11,6,9,24,12,13,14,12,12,14,11,12,14,15,20,21,22,23,21,25,28) >(fit <- StructTS(x,type = "BSM")) I obtained the following

Dear R people, what functions generate respectively vectors with each element is respectively zero and one. sorry for my credulous questions and many thanks in advance. ____________________________________________________ Écoutez gratuitement le nouveau single de Noir Désir et découvrez d'autres titres en affinité avec vos goûts musicaux

Dear R users, I would like to construct a matrix of seasonal dummy variables, such matrix can be written as follows(i.e format(T,4)) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 . . . . . . . . etc I have written the following small program:

Hi everybody. The question: I get two vectors 'iFalseFalse' and 'i2'. I think they should be the same but they are not. Is it because R does not handle complicated logical expressions in such cases or I do something wrong? > z1 = c(NA, "", 3, NA, "", 3) > z2 = c("", "", 3, NA, 3, NA) > cV = (as.character(z1)==as.character(z2))