I am still thinking about this problem. The solution could look something like this (it's net yet working):
k<-lapply(h, function (x) x*0) # I keep the same format as h, but set all values to 0 years<-c(1997:1999) # I define the years for (t in 1:length(years)) { year = as.character(years[t]) ids = rownames(h[year][[1]]) } for (m in 1:length(relevant_firms)) { k[year][[m]]<-lapply(k[year], function (col) k[year][[1]][,m] = h[year][[1]][,m] & k[year][[1]][m,] = h[year][[1]][m,]) } # I am creating new list objects that should look like this k$'1999'$'8029' and I replace the values in the 8029 column and row by the original ones in h Any takes on this problem? Thank you very much! Best -- View this message in context: http://r.789695.n4.nabble.com/Multiply-list-objects-tp3595719p3603871.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.