Sundar Dorai-Raj sundar.dorai-raj at pdf.com
Thu Feb 5 21:58:39 CET 2004

```How about simply

F1 <- factor(c("b", "a"))
F2 <- factor(c("c", "b"))
F3 <- factor(c(levels(F1)[F1], levels(F2)[F2]))

-sundar

Spencer Graves wrote:

>  > F1 <- factor(c("b", "a"))
>  > F2 <- factor(c("c", "b"))
>  > k1 <- length(F1)
>  > k2 <- length(F2)
>  > F12.lvls <- unique(c(levels(F1), levels(F2)))
>  > F. <- factor(rep(F12.lvls, k1+k1), levels=F12.lvls)
>  > F.[1:k1] <- F1
>  > F.[-(1:k1)] <- F2
>  > F.
>  b a c b
> Levels: a b c
>
>      This saves converting the factors to characters, which might save
> computer time at the expense of your time.      hope this helps.
> spencer graves
>
> Svetlana Eden wrote:
>
>> I have two factors l1, l2, and I'd like to merge them.
>>
>> (Remark:       The factors can not be converted to charaters)
>>
>> Function c() does not give me the result I want:
>>
>>
>>
>>
>>> l1 = factor(c('aaaa', 'bbbb'))
>>> l2 = factor(c('ccc', 'dd'))
>>> lMerge = factor(c(l1, l2))
>>> lMerge
>>>
>>
>>  1 2 1 2
>> Levels: 1 2
>>
>>
>> I'd like to merge l1 and l2 and to get lMerge
>> ----------------------------------------------
>>
>>  aaaa bbbb ccc dd
>> Levels: aaaa bbbb ccc dd
>>
>>
>>  1 2 1 2
>> Levels: 1 2
>>
>>
>>
>> How should I do that without converting the factors back to strings.
>> -------------------------------------------------------------------
>>
>>
>>
>
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