I have a set of variables coded as binomial.
Pre VALUE_1 VALUE_2 VALUE_3 VALUE_4 VALUE_5 VALUE_6 VALUE_7 VALUE_8
1 1 0 0 0 0 0 1 0 0
2 1 0 0 0 0 1 0 0 0
3 1 0 0 0 0 1 0 0 0
4 1 0 0 0 0 1 0 0 0
I would like to merge the variables (VALUE_1, VALUE_2...VALUE_8) into one single ordered factor, while conserving the column (Pre) as is, duch that the data would look like this:
Pre VALUE
1 1 VALUE_6
2 1 VALUE_5
3 1 VALUE_5
Or even better:
Pre VALUE
1 1 6
2 1 5
3 1 5
I am aware that this exists: Recoding dummy variable to ordered factor
But when I try the code used in that post, I receive the following error:
PA2$Factor = factor(apply(PA2, 1, function(x) which(x == 1)), labels = colnames(PA2))
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?
Any help would be appreciated
A quick solution would be something like
Res <- cbind(df[1], VALUE = factor(max.col(df[-1]), ordered = TRUE))
Res
# Pre VALUE
# 1 1 6
# 2 1 5
# 3 1 5
# 4 1 5
str(Res)
# 'data.frame': 4 obs. of 2 variables:
# $ Pre : int 1 1 1 1
# $ VALUE: Ord.factor w/ 2 levels "5"<"6": 2 1 1 1
OR if you want the actual names of the columns (as Pointed by @BondedDust), you can use the same methodology to extract them
factor(names(df)[1 + max.col(df[-1])], ordered = TRUE)
# [1] VALUE_6 VALUE_5 VALUE_5 VALUE_5
# Levels: VALUE_5 < VALUE_6
OR you can use your own which strategy in the following way (btw, which is vectorized so no need in using apply with a margin of 1 on it)
cbind(df[1], VALUE = factor(which(df[-1] == 1, arr.ind = TRUE)[, 2], ordered = TRUE))
OR you can do matrix multiplication (contributed by @akrun)
cbind(df[1], VALUE = factor(as.matrix(df[-1]) %*% seq_along(df[-1]), ordered = TRUE))