[R] labels to values
Matthias Wendel
office at matthiaswendel.de
Thu Jan 10 12:48:38 CET 2008
Hello, Henrique,
thank you for the tip, but it was not quite what has been desired:
> d[, 'Y6']
[1] 6 3 8 11 8 9 6 8 3 5 10 15 NA 9 8 3 8 16 6 6 NA 10 5 2 7 7 6 16 7 15 7 10 12 8 7 12 12 16 7 6 8 8
15 6 NA 8 99 7 12 8 9 16 7 16 8 7 7 1 15
[60] 12 8 7 10 7 8 7 8 9 8 6 6 8 6 16 11 5 11 11 1 11 3 7 7 10 10 10 6 11 16 NA 1 3 2 10 99 10 3 3 9 7 16
99 16 1 10 2 13 13 13 13 13 13 13 13 13 13 13 13
[119] 13 13 13 13 13 NA 10 16 16 NA 6 10 5 11 11 1 1 1 1 16 1 16 1 1 1 1 6 6 6 16 8 16 16 16 16 5 6 10 99 11 11 10
6 6 1 1 6 1 11 11 16 9 11 16 6 8 8 16 16
[178] 8 6 16 16 12 12 12 12 12 12 12 16 9 16 15 12 12 15 10 16 15 4 1 2 14 4 4 2 5 NA 1 5 5 7 9 5 12 12 NA 16 12 12
12 12 12 12 12 12 12 99 NA 12 12 NA 1 16 1 7 11
[237] 5 6 7 1 13 6 8 16 2 1 5 16 16 9 8 8 8 7 16 8 8 2 8 5 4 6 14 5 14 8 8 14 4 4 8 14 8 14 6 2 3 14
3 16 5 15 15 15 15 15 15 15 15 15 15 15 13 13 13
[296] 13 13 13 13 13 13 13 13 13 13 15 6 NA 12 3 9 9 NA 10 16
attr(,"value.labels")
n.a. Verwaltung Servicegesellschaft Waldfriede (SKW)
Kurzzeitpflege Waldfriede
99 16 15
14
Sozialstation Krankenpflegeschule Med. Technischer Dienst
Pflege OP
13 12 11
10
Funktionsdienst Pflege Gynäkologie Pflege Chirurgie
Pflege Innere
9 8 7
6
Ärzte Anästhesie, Röntgen Ärzte Gynäkologie Ärzte Chirurgie
Ärzte Innere
5 4 3
2
Patientenberatung/-betreuung
1
> x1 <- as.factor(d[,'Y6'])
> levels(x1)[levels(x1) %in% attributes(x1)[1]] <- c(names(unlist(lapply(levels(x1),function(z)which(z==attributes(x1)[[1]])))))
Fehler in levels(x1)[levels(x1) %in% attributes(x1)[1]] <- c(names(unlist(lapply(levels(x1), :
inkompatible Typen (von NULL nach character) in subassignment Typ fix
>
Meanwhile, I solved the problem by takin another route: the data.frame d was imported my read.spss; read.spss imports the factors as
factors only if number of distinct labels is equal to number of distinct values. By assuring this (it was not the case for Y6), I've
got what I Want.
Regards,
Matthias
-----Ursprüngliche Nachricht-----
Von: Henrique Dallazuanna [mailto:wwwhsd at gmail.com]
Gesendet: Mittwoch, 9. Januar 2008 17:17
An: Matthias Wendel
Cc: r-help at stat.math.ethz.ch
Betreff: Re: [R] labels to values
If I understand your question, you can do:
x1 <- as.factor(d[,'Y6'])
levels(x1)[levels(x1) %in% attributes(x)[1]] <- c(names(unlist(lapply(levels(x1),function(z)which(z==attributes(x)[[1]])))))
On 09/01/2008, Matthias Wendel <office at matthiaswendel.de> wrote:
> I couldn't find out how the following to solve:
> There is a column in a data.frame:
>
> > d[, 'Y6']
> [1] 6 3 8 11 8 9 6 8 3 5 10 15 NA 9 8 3 8 16 6 6 NA
> 10 5 2 7 7 6 16 7 15 7 10 12 8 7 12 12 16 7 6 8 8
> 15 6 [45] NA 8 99 7 12 8 9 16 7 16 8 7 7 1 15 12 8 7 10
> 7 8 7 8 9 8 6 6 8 6 16 11 5 11 11 1 11 3 7 7 10 10 10 6
> 11 [89] 16 NA 1 3 2 10 99 10 3 3 9 7 16 99 16 1 10 2 13 13
> 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 NA 10 16
> 16 NA 6 10 5 11 [133] 11 1 1 1 1 16 1 16 1 1 1 1 6 6 6
> 16 8 16 16 16 16 5 6 10 99 11 11 10 6 6 1 1 6 1 11 11
> 16 9 11 16 6 8 8 16 [177] 16 8 6 16 16 12 12 12 12 12 12 12 16
> 9 16 15 12 12 15 10 16 15 4 1 2 14 4 4 2 5 NA 1 5 5
> 7 9 5 12 12 NA 16 12 12 12 [221] 12 12 12 12 12 12 99 NA 12 12 NA 1
> 16 1 7 11 5 6 7 1 13 6 8 16 2 1 5 16 16 9 8 8
> 8 7 16 8 8 2 8 5 4 6 14 5 [265] 14 8 8 14 4 4 8 14 8 14
> 6 2 3 14 3 16 5 15 15 15 15 15 15 15 15 15 15 15 13 13
> 13 13 13 13 13 13 13 13 13 13 13 15 6 NA [309] 12 3 9 9 NA 10 16
> attr(,"value.labels")
> Verwaltung Servicegesellschaft Waldfriede (SKW) Kurzzeitpflege Waldfriede
> 16 15 14
> Sozialstation Krankenpflegeschule Med. Technischer Dienst
> 13 12 11
> Pflege OP Funktionsdienst Pflege Gynäkologie
> 10 9 8
> Pflege Chirurgie Pflege Innere Ärzte Anästhesie, Röntgen
> 7 6 5
> Ärzte Gynäkologie Ärzte Chirurgie Ärzte Innere
> 4 3 2
> Patientenberatung/-betreuung
> 1
>
> I'd like to have the column with the value.labels as values. How do I do that?
> Thanks,
> Matthias
>
> ______________________________________________
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>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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