# Re: [R] Question about creating unique factor labels with the factor function

From: Petr Pikal <petr.pikal_at_precheza.cz>
Date: Wed 20 Jul 2005 - 19:55:07 EST

Hallo

your factor has 3 levels but with only 2 different labels

> hb

[1] 1 1 1 1 1 1 2 2 2
Levels: 1 1 2

but

> str(hb)

Factor w/ 3 levels "1","1","2": 1 1 1 2 2 2 3 3 3

so you gave only one label to level 1 and 2. You can give the same label to any level you choose but I can not see a reason in it.

> hb <- factor(c(1,1,1,2,2,2,3,3,3),

levels=c(1,2,3),labels=c(1,1,1))
> hb

[1] 1 1 1 1 1 1 1 1 1
Levels: 1 1 1

or

> hb <- factor(c(1,1,1,2,2,2,3,3,3),

levels=c(1,2,3),labels=c(3,100,100))
> hb

[1] 3 3 3 100 100 100 100 100 100
Levels: 3 100 100
> str(hb)

Factor w/ 3 levels "3","100","100": 1 1 1 2 2 2 3 3 3

but surprising for me is an ordered behaviour which I find a bit odd.

> hb

[1] 3 3 3 100 100 100 100 100 100
Levels: 3 100 100

> ordered(hb)

[1] 3 3 3 100 100 100 100 100 100
Levels: 3 < 100 < 100

> str(hb)

Factor w/ 3 levels "3","100","100": 1 1 1 2 2 2 3 3 3

> str(ordered(hb))

Ord.factor w/ 3 levels "3"<"100"<"100": 1 1 1 2 2 2 2 2 2

> unique((hb))

[1] 3 100 100 #three levels and three values Levels: 3 100 100

> unique(ordered(hb))
[1] 3 100 # three levels but only 2 values? Levels: 3 < 100 < 100

Can anybody explain why unique(ordered()) results in only 2 displayed levels although this

> str(unique(ordered(hb)))
Ord.factor w/ 3 levels "3"<"100"<"100": 1 2

says that the factor still have 3 levels?

Best regards
Petr

On 19 Jul 2005 at 15:11, Gregory Gentlemen wrote:

> Hi guys,
>
> I ran into a problem of not being able to create unique labels when
> creating a factor. Consider an example below:
>
> hb <- factor(c(1,1,1,2,2,2,3,3,3), levels=c(1,2,3),labels=c(1,1,2)) >
> hb [1] 1 1 1 1 1 1 2 2 2 Levels: 1 1 2
> unique(hb) [1] 1 1 2 Levels:
> 1 1 2
>
> How come there are three unique levels, I thought this would only
> create one unique level?
>
> > unique(as.ordered(hb))
> [1] 1 2
> Levels: 1 < 1 < 2
>
> Is as.ordered the only solution?
>
> Greg
>
>
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