nly 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:[EMAIL PROTECTED]
Gesendet: Mittwoch, 9. Januar 2008 17:17
An: Ma
There probably is an easier way but you put the
value.labels and corresponding numbers into a
data.frame, stick d[,'Y6] into a data frame and merge
the two. Then just add or replace d[,Y6] in the
original data.frame.
Example
# Set up basic data
aa <- 1:15
bb <- letters[1:15]
# Apply att
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 <[EMAIL PROTECTED]> wrote:
> I couldn't find out how the following to s
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
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