R split factor to m-columns with original length

Question

Suppose I have a factor (in a data.frame) which represents years:

  year
1 2012
2 2012
3 2012
4 2013
5 2013
6 2013
7 2014
8 2014
9 2014

I would to to create (in this case) three new columns in the data.frame and end up with:

  y2012 y2013 y2014
1     1     0     0
2     1     0     0
3     1     0     0
4     0     1     0
5     0     1     0
6     0     1     0
7     0     0     1
8     0     0     1
9     0     0     1

I can of course write a bunch of ifelse-statements, but that seems very unhandy.

Answer 1

We can use mtabulate from qdapTools

library(qdapTools)
mtabulate(df1$year)
#  2012 2013 2014
#1    1    0    0
#2    1    0    0
#3    1    0    0
#4    0    1    0
#5    0    1    0
#6    0    1    0
#7    0    0    1
#8    0    0    1
#9    0    0    1

---

Or using some options in base R.

1. model.matrix. We convert the 'year' column to factor class and use that in the model.matrix to get the binary columns.

   model.matrix(~0+factor(year), df1)
   

2. table. We can get the expected output using table of the sequence of rows of df1 and the column 'year'.

   table(1:nrow(df1), df1$year)
   

Answer 2

Also maybe

library(dplyr)
library(tidyr)
df %>%
  mutate(id = 1L) %>%
  spread(year, id, fill = 0L)

#   2012 2013 2014
# 1    1    0    0
# 2    1    0    0
# 3    1    0    0
# 4    0    1    0
# 5    0    1    0
# 6    0    1    0
# 7    0    0    1
# 8    0    0    1
# 9    0    0    1

---

Maybe this too (as can't think of a better way)

library(data.table)
dcast(setDT(df)[, `:=`(indx = .I, indx2 = 1L)], indx ~ year, fill = 0L)
#    indx 2012 2013 2014
# 1:    1    1    0    0
# 2:    2    1    0    0
# 3:    3    1    0    0
# 4:    4    0    1    0
# 5:    5    0    1    0
# 6:    6    0    1    0
# 7:    7    0    0    1
# 8:    8    0    0    1
# 9:    9    0    0    1