Reproducible example: x <- list(list(list(), list())) unlist(x) *> Error in as.character.factor(x) : malformed factor* What should happen: unlist(x)> NULLR.version platform x86_64-apple-darwin15.6.0 arch x86_64 os darwin15.6.0 system x86_64, darwin15.6.0 status major 3 minor 5.0 year 2018 month 04 day 23 svn rev 74626 language R version.string R version 3.5.0 (2018-04-23) nickname Joy in Playing -- Steven Nydick PhD, Quantitative Psychology M.A., Psychology M.S., Statistics -- "Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before, Bokonon tells us. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way." -Kurt Vonnegut [[alternative HTML version deleted]]
On 08/05/2018 1:48 PM, Steven Nydick wrote:> Reproducible example: > > x <- list(list(list(), list())) > unlist(x) > > *> Error in as.character.factor(x) : malformed factor*The error comes from the line structure(res, levels = lv, names = nm, class = "factor") which is called because unlist() thinks that some entry is a factor, with NULL levels and NULL names. It's not legal for a factor to have NULL levels. Probably it should never get here; the earlier test if (.Internal(islistfactor(x, recursive))) { should have been false, and then the result would have been .Internal(unlist(x, recursive, use.names)) (with both recursive and use.names being TRUE), which returns NULL. Duncan Murdoch> > What should happen: > > unlist(x) >> NULL > > R.version > platform x86_64-apple-darwin15.6.0 > arch x86_64 > os darwin15.6.0 > system x86_64, darwin15.6.0 > status > major 3 > minor 5.0 > year 2018 > month 04 > day 23 > svn rev 74626 > language R > version.string R version 3.5.0 (2018-04-23) > nickname Joy in Playing >
On 08/05/2018 2:58 PM, Duncan Murdoch wrote:> On 08/05/2018 1:48 PM, Steven Nydick wrote: >> Reproducible example: >> >> x <- list(list(list(), list())) >> unlist(x) >> >> *> Error in as.character.factor(x) : malformed factor* > > The error comes from the line > > structure(res, levels = lv, names = nm, class = "factor") > > which is called because unlist() thinks that some entry is a factor, > with NULL levels and NULL names. It's not legal for a factor to have > NULL levels. Probably it should never get here; the earlier test > > if (.Internal(islistfactor(x, recursive))) { > > should have been false, and then the result would have been > > .Internal(unlist(x, recursive, use.names)) > > (with both recursive and use.names being TRUE), which returns NULL.And the problem is in the islistfactor function in src/main/apply.c, which looks like this: static Rboolean islistfactor(SEXP X) { int i, n = length(X); switch(TYPEOF(X)) { case VECSXP: case EXPRSXP: if(n == 0) return NA_LOGICAL; for(i = 0; i < LENGTH(X); i++) if(!islistfactor(VECTOR_ELT(X, i))) return FALSE; return TRUE; break; } return isFactor(X); } One of those deeply nested lists is length 0, so at the lowest level it returns NA_LOGICAL. But then it does C-style logical testing on the results. I think to C NA_LOGICAL counts as true, so at the next level up we get the wrong answer. A fix would be to rewrite it like this: static Rboolean islistfactor(SEXP X) { int i, n = length(X); Rboolean result = NA_LOGICAL, childresult; switch(TYPEOF(X)) { case VECSXP: case EXPRSXP: for(i = 0; i < LENGTH(X); i++) { childresult = islistfactor(VECTOR_ELT(X, i)); if(childresult == FALSE) return FALSE; else if(childresult == TRUE) result = TRUE; } return result; break; } return isFactor(X); }