Recode categorical factor with N categories into N binary columns
dataframe, factors, r
Solution
Even better with the help of @AnandaMahto's search capabilities,
model.matrix(~ . + 0, data=df, contrasts.arg = lapply(df, contrasts, contrasts=FALSE))
# v1a v1b v1c v2a v2b v2c
# 1 0 1 0 0 0 1
# 2 1 0 0 1 0 0
# 3 0 0 1 0 0 1
# 4 0 1 0 1 0 0
# 5 0 0 1 0 0 1
# 6 0 0 1 0 1 0
# 7 1 0 0 1 0 0
# 8 1 0 0 0 1 0
# 9 1 0 0 0 0 1
# 10 1 0 0 0 1 0
I think this is what you're looking for. I'd be happy to delete if it's not so. Thanks to @G.Grothendieck (once again) for the excellent usage of `model.matrix`!
cbind(with(df, model.matrix(~ v1 + 0)), with(df, model.matrix(~ v2 + 0)))
# v1a v1b v1c v2a v2b v2c
# 1 0 1 0 0 0 1
# 2 1 0 0 1 0 0
# 3 0 0 1 0 0 1
# 4 0 1 0 1 0 0
# 5 0 0 1 0 0 1
# 6 0 0 1 0 1 0
# 7 1 0 0 1 0 0
# 8 1 0 0 0 1 0
# 9 1 0 0 0 0 1
# 10 1 0 0 0 1 0
Note: Your output is just:
with(df, model.matrix(~ v2 + 0))
Note 2: This gives a `matrix`. Fairly obvious, but still, wrap it with `as.data.frame(.)` if you want a `data.frame`.
Problem
Original data frame: ``` v1 = sample(letters[1:3], 10, replace=TRUE) v2 = sample(letters[1:3], 10, replace=TRUE) df = data.frame(v1,v2) df ``` ``` v1 v2 1 b c 2 a a 3 c c 4 b a 5 c c 6 c b 7 a a 8 a b 9 a c 10 a b ``` New data frame: ``` new_df = data.frame(row.names=rownames(df)) for (i in colnames(df)) { for (x in letters[1:3]) { #new_df[x] = as.numeric(df[i] == x) new_df[paste0(i, "_", x)] = as.numeric(df[i] == x) } } ``` ``` v1_a v1_b v1_c v2_a v2_b v2_c 1 0 1 0 0 0 1 2 1 0 0 1 0 0 3 0 0 1 0 0 1 4 0 1 0 1 0 0 5 0 0 1 0 0 1 6 0 0 1 0 1 0 7 1 0 0 1 0 0 8 1 0 0 0 1 0 9 1 0 0 0 0 1 10 1 0 0 0 1 0 ``` For small datasets this is fine, but it becomes slow for much larger datasets. Anyone know of a way to do this without using looping?