Dear Ivan, first, it would pay-off in terms of readability to employ line breaks and second to provide a reproducable code snippet and third which package you have used. Now to your questions: 1) What happens if you provide colnames for your objects? 2) What happens if you omit the $ after count?
Best, Bernhard ps: the function seems to have been ported from the package 'vars'. In this package the function causality() is included which returns a named list with elements of class htest. > -----Ursprüngliche Nachricht----- > Von: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] Im Auftrag von ivan > Gesendet: Donnerstag, 14. April 2011 19:37 > An: r-help@r-project.org > Betreff: [R] Automatically extract info from Granger causality output > > Dear Community, > > this is my first programming in R and I am stuck with a > problem. I have the following code which automatically > calculates Granger causalities from a variable, say e.g. "bs" > as below, to all other variables in the data frame: > > log.returns<-as.data.frame( lapply(daten, function(x) > diff(log(ts(x))))) y1<-log.returns$bs > y2<- log.returns[,!(names(log.returns) %in% "bs")] > Granger<- function(y1,y2) {models=lapply(y2, function(x) > VAR(cbind(x,y1),ic="SC") ); results=lapply(models,function(x) > causality(x,cause="y1")); print(results)} > Count<-Granger(y1,y2) > > which produces the following output (I have printed only part > of it (for Granger causality of bs on ml)): > > $ml > $ml$Granger > > Granger causality H0: y1 do not Granger-cause x > > data: VAR object x > F-Test = 0.2772, df1 = 1, df2 = 122, p-value = 0.5995 > > > $ml$Instant > > H0: No instantaneous causality between: y1 and x > > data: VAR object x > Chi-squared = 19.7429, df = 1, p-value = 8.859e-06 > > My questions: > > 1)How can I edit the function above so that the output writes: Granger > causality H0: bs do not Granger-cause ml rather than Granger > causality H0: y1 do not Granger-cause x? > > 2) I want to extract the p-values of the tests into a data > frame for instance. The problem is that the output has a 3 > layer structure. > Thus, for the above p-value I need to write count$ml$Granger$p.value. > I thought of a loop of something like for(i in > 1:length(count)) {z=count$[[i]]$Granger$p.value} but it didn't work. > > Thank you very much for your help. > > Best Regards. > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ***************************************************************** Confidentiality Note: The information contained in this ...{{dropped:10}} ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
