The base solution could be the following:
dat <- data.frame(time = c(20000616, 20000616, 20000616, 20000616, 20000616, 20000616),
hour = c(1, 2, 3, 4, 5, 6),
Money = c(9.35, 6.22, 10.65, 11.42, 10.12, 7.32),
day = c(5, 5, 5, 5, 5, 5) )
dat$dummy_day <- factor(dat$day, levels = 1:7)
model.matrix(~time + hour + Money + day + dummy_day, dat,
contrasts = list(dummy_day = "contr.SAS"))
It returns a matrix:
(Intercept) time hour Money day dummy_day1 dummy_day2 dummy_day3 dummy_day4 dummy_day5 dummy_day6
1 1 20000616 1 9.35 5 0 0 0 0 1 0
2 1 20000616 2 6.22 5 0 0 0 0 1 0
3 1 20000616 3 10.65 5 0 0 0 0 1 0
4 1 20000616 4 11.42 5 0 0 0 0 1 0
5 1 20000616 5 10.12 5 0 0 0 0 1 0
6 1 20000616 6 7.32 5 0 0 0 0 1 0
attr(,"assign")
[1] 0 1 2 3 4 5 5 5 5 5 5
attr(,"contrasts")
attr(,"contrasts")$dummy_day
[1] "contr.SAS"
