[R] Basic question on concatenating factors

[jim holtman]

You are right.  union used 'unique(c(x,y))' and I am not sure if
'unique' preserves the order, but the help page seems to indicate that
 "an element is omitted if it is identical to any previous element ";
this might mean that the order is preserved.

On Sat, Nov 22, 2008 at 11:43 PM, Stavros Macrakis
<macrakis at alum.mit.edu> wrote:
> On Sat, Nov 22, 2008 at 10:20 AM, jim holtman <jholtman at gmail.com> wrote:
>>  c.Factor <-
>> function (x, y)
>> {
>>    newlevels = union(levels(x), levels(y))
>>    m = match(levels(y), newlevels)
>>    ans = c(unclass(x), m[unclass(y)])
>>    levels(ans) = newlevels
>>    class(ans) = "factor"
>>    ans
>> }
>
> This algorithm depends crucially on union preserving the order of the
> elements of its arguments. As far as I can tell, the spec of union
> does not require this.  If union were to (for example) sort its
> arguments then merge them (generally a more efficient algorithm), this
> function would no longer work.
>
> Fortunately, the fix is simple.  Instead of union, use:
>
>     newlevels <- c(levels(x),setdiff(levels(y),levels(x))
>
> which is guaranteed to preserve the order of levels(x).
>
>             -s
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?