A better solution to this problem is to use character indexing:
x <- c("Tuesday", "Thursday", "Sunday")
c(Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5,
Saturday = 6, Sunday = 7)[x]
http://adv-r.had.co.nz/Subsetting.html#lookup-tables-character-subsetting
Hadley
On Mon, Dec 2, 2013 at 6:33 PM, Bill <william...@gmail.com> wrote:
> 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)))))))
>
>
> In code like the above, day_of_week is a vector and so day_of_week ==
> "Monday" will result in a boolean vector. Suppose day_of_week is Monday,
> Thursday, Friday, Tuesday. So day_of_week == "Monday" will be
> True,False,False,False. I think that ifelse will test the first element and
> it will generate a 1. At this point it will not have run day_of_week ==
> "Tuesday" yet. Then it will test the second element of day_of_week and it
> will be false and this will cause it to evaluate day_of_week == "Tuesday".
> My question would be, does the evaluation of day_of_week == "Tuesday"
> result in the generation of an entire boolean vector (which would be in
> this case False,False,False,True) or does the ifelse "manage the indexing"
> so that it only tests the second element of the original vector (which is
> Thursday) and for that matter does it therefore not even bother to generate
> the first boolean vector I mentioned above (True,False,False,False) but
> rather just checks the first element?
> Not sure if I have explained this well but if you understand I would
> appreciate a reply.
> Thanks.
>
> [[alternative HTML version deleted]]
>
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