why change days of the week from a factor to an ordered factor?
On 02 Dec 2013, at 16:35 , Bert Gunter <gunter.berton at gene.com> wrote:
Not true, Rich.
z <-factor(letters[1:3],lev=letters[3:1]) sort(z)
[1] c b a Levels: c b a What you say is true only for the **default** sort order. (Although maybe the code author didn't realize this either)
The coding is certainly clunky (the phrase about writing FORTRAN in any language springs to mind, only with SAS instead of FORTRAN):
dodgers$ordered_day_of_week <- with(data=dodgers,
ifelse ((day_of_week == "Monday"),1,
ifelse ((day_of_week == "Tuesday"),2,
ifelse ((day_of_week == "Wednesday"),3,
ifelse ((day_of_week == "Thursday"),4,
ifelse ((day_of_week == "Friday"),5,
ifelse ((day_of_week == "Saturday"),6,7)))))))
dodgers$ordered_day_of_week <- factor(dodgers$ordered_day_of_week,
levels=1:7,
labels=c("Mon", "Tue", "Wed", "Thur", "Fri", "Sat", "Sun"))
This'll do:
dodgers$ordered_day_of_week <- factor(dodgers$ordered_day_of_week,
levels=c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"),
labels=c("Mon", "Tue", "Wed", "Thur", "Fri", "Sat", "Sun"))
And BTW, it doesn't (and didn't) create an ordered factor, just a factor with a different level ordering.
Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com