[R] ifelse -does it "manage the indexing"?

[William Dunlap]

> It seems so inefficient.

But ifelse knows nothing about the expressions given
as its second and third arguments -- it only sees their
values after they are evaluated.  Even if it could see the
expressions, it would not be able to assume that f(x[i])
is the same as f(x)[i] or things like
   ifelse(x>0, cumsum(x), cumsum(-x))
would not work.

You can avoid the computing all of f(x) and then extracting
a few elements from it by doing something like
   x <- c("Wednesday", "Monday", "Wednesday")
   z1 <- character(length(x))
   z1[x=="Monday"] <- "Mon"
   z1[x=="Tuesday"] <- "Tue"
   z1[x=="Wednesday"] <- "Wed"
or
   LongDayNames <- c("Monday","Tuesday","Wednesday")
   ShortDayNames <- c("Mon", "Tue", "Wed")
   z2 <- character(length(x))
   for(i in seq_along(LongDayNames)) {
      z2[x==LongDayNames[i]] <- ShortDayNames[i]
   }

To avoid the repeated x==value[i] you can use match(x, values).
   z3 <- ShortDayNames[match(x, LongDayNames)]

z1, z2, and z3 are identical  character vectors.

Or, you can use factors.
   > factor(x, levels=LongDayNames, labels=ShortDayNames)
   [1] Wed Mon Wed
   Levels: Mon Tue Wed

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> Of Bill
> Sent: Monday, December 02, 2013 4:50 PM
> To: Duncan Murdoch
> Cc: r-help at r-project.org
> Subject: Re: [R] ifelse -does it "manage the indexing"?
> 
> It seems so inefficient. I mean the whole first vector will be evaluated.
> Then if the second if is run the whole vector will be evaluated again. Then
> if the next if is run the whole vector will be evaluted again. And so on.
> And this could be only to test the first element (if it is false for each
> if statement). Then this would be repeated again and again. Is that really
> the way it works? Or am I not thinking clearly?
> 
> 
> On Mon, Dec 2, 2013 at 4:48 PM, Duncan Murdoch
> <murdoch.duncan at gmail.com>wrote:
> 
> > On 13-12-02 7:33 PM, Bill 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.
> >>
> >
> > See the help for the function.  If any element of the test is true, the
> > full first vector will be evaluated.  If any element is false, the second
> > one will be evaluated.  There are no shortcuts of the kind you describe.
> >
> > Duncan Murdoch
> >
> >
> 
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