On Dec 3, 2013, at 5:37 AM, Hadley Wickham wrote:
> 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
That does seem more expressive than using match:
x <- c("Tuesday", "Thursday", "Sunday")
match(x, c('Monday', 'Tuesday', 'Wednesday' ,'Thursday' ,
'Friday', 'Saturday', 'Sunday') )
[1] 2 4 7
It would also lend itself well to grouping operations
x <- c("Tuesday", "Thursday", "Sunday")
c(Monday = 'first half', Tuesday = 'first half',
Wednesday = "hump",
Thursday = 'second half', Friday = 'second half',
Saturday = "weekend", Sunday = "weekend")[x]
Tuesday Thursday Sunday
"first half" "second half" "weekend"
The corresponding use case with match would be:
c(rep('first half',2),"hump", rep('second half',2), rep("weekend",2) )[
match(x, c('Monday', 'Tuesday', 'Wednesday' ,'Thursday' ,
'Friday', 'Saturday', 'Sunday') ) ]
[1] "first half" "second half" "weekend"
And your recommended method has the virtue of performing the same partial
matching allowed by charmatch:
x2 <- substr(x, 1,3)
charmatch(x2, c('Monday', 'Tuesday', 'Wednesday' ,'Thursday' ,
'Friday', 'Saturday', 'Sunday') )
#[1] 2 4 7
c(Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5,
Saturday = 6, Sunday = 7)[x]
# Tuesday Thursday Sunday
# 2 4 7
--
David.
>
> 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]]
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> http://had.co.nz/
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
