[Rd] Why is there no c.factor?

[Hadley Wickham]

> I'd propose the following: If the sets of levels of all arguments are the
> same, then c.factor() would return a factor with the common set of levels;
> if the sets of levels differ, then, as Hadley suggests, the level-set of the
> result would be the union of sets of levels of the arguments, but a warning
> would be issued.

I like this compromise (as long as there was an argument to suppress
the warning)

Hadley


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