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);
}
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