Hi, May be this helps: #Creating some dummy data.
set.seed(24) lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE))) set.seed(49) lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE))) ??Find_Max_CCF() #No vignettes or demos or help files found with alias or concept or #title matching ‘Find_Max_CCF’ using regular expression matching. Found a function with the same name from http://r.789695.n4.nabble.com/ccf-function-td2288257.html capture.output(do.call(rbind,lapply(seq_along(lst1),function(i) Find_Max_CCF(lst1[[i]],lst2[[i]]))),file="output.txt") #output.txt # cor lag #10 0.4799088 -1 #17 0.2060688 6 #16 0.3716986 5 #6 0.3701101 -5 #8 0.3964724 -3 #4 0.2942228 -7 #15 0.3191763 4 #9 0.3654471 -2 A.K. capture.output(do.call(rbind,lapply(seq_along(lst1),function(i) Find_Max_CCF(lst1[[i]],lst2[[i]])))) #[1] " cor lag" "10 0.4799088 -1" "17 0.2060688 6" "16 0.3716986 5" #[5] "6 0.3701101 -5" "8 0.3964724 -3" "4 0.2942228 -7" "15 0.3191763 4" #[9] "9 0.3654471 -2" Hi, I am using following code to Find lag at which cross correlation is maximum ccf( ), where Find_Max_CCF(x,y) returns the max cross correlation. where x[i], y[i] are two different time series, i=1,2 ,...,8 for( i in 1:8) { {c=Find_Max_CCF ( x [ i ], y [ i ] ) ) } Now, I want to capture the whole 8 results in the same excel sheet. I used capture.output(Find_Max_CCF ( x [ i ], y [ i ] ) ), file="output.txt") But this is giving me only the last result that is cross correlation for last two time series, x[8], y[8]. Please help... Thanks in advance.. Shilpa Rai ______________________________________________ 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.
