没有硬编码的双重传播

Question

I am stuck here. I tried using spread twice from tidyr , I tried joining. But none of these methods give the right solution without some hard coding.

Is there any way to tranform this data:

    cat1   cat2 title
1      A      G    AB
2      B      G    BC
3      C      B    CD
4      D      G    DE
5      E      H    EF
6      F      A    FG

into this:

   A B C D E F G H
AB 1 0 0 0 0 0 1 0
BC 0 1 0 0 0 0 1 0
CD 0 1 1 0 0 0 0 0
DE 0 0 0 1 0 0 1 0
EF 0 0 0 0 1 0 0 1
FG 1 0 0 0 0 1 0 0

Sample data:

df<-data.frame(cat1=LETTERS[1:6],
               cat2=c('G','G','B','G','H','A'),
               title=paste0(LETTERS[1:6],LETTERS[2:7]))

Since I usually get dplyr answers faster: Base R or tidyr only solutions are also very welcome

Answer 1

I don't know if this qualifies as not hard coding for the op

df %>% 
  tidyr::gather(key = vars, value = values, cat1, cat2) %>% 
  dplyr::mutate(vars = 1) %>% 
  tidyr::spread(key = values, value = vars, fill = 0)
#   title A B C D E F G H
# 1    AB 1 0 0 0 0 0 1 0
# 2    BC 0 1 0 0 0 0 1 0
# 3    CD 0 1 1 0 0 0 0 0
# 4    DE 0 0 0 1 0 0 1 0
# 5    EF 0 0 0 0 1 0 0 1
# 6    FG 1 0 0 0 0 1 0 0

Answer 2

Just melt first, then cast:

require(reshape2)

melt(df, id="title") %>% dcast(title ~ value, length)

  title A B C D E F G H
1    AB 1 0 0 0 0 0 1 0
2    BC 0 1 0 0 0 0 1 0
3    CD 0 1 1 0 0 0 0 0
4    DE 0 0 0 1 0 0 1 0
5    EF 0 0 0 0 1 0 0 1
6    FG 1 0 0 0 0 1 0 0

melt puts all the values in a single column to cast.