R help - Oct 2004 - AW: constructing specially ordered factor

Hi Dimitris,

thank you for your reply, 
> does the following work in your data:
> 
> levs <- unique.default(Names)
> factor(Names, levs[order(unique.default(Weights))])
your solution is really shorter, but has two issues to be 
meant here:

1. "unique.default" is applied twice, what might 
   be a bit expensive for strings.

2. your solution brings an implicit prerequisit on 
  "unique.default". Indeed, "unique.default" should 
  preserve the same order for output both working 
  with strings and with numbers. In other words, 
  correspondence must be kept by "unique.default".
  If "unique.default" implementation uses (or will use) 
  lexicographic strings sorting for acceleration
  then this approach fails.


--
Valery

Hi Petr,

Thank you for your reply.
> Factor <- function(f,n, decreasing=TRUE, ...) {
> ooo<-order(levels(factor(n)), decreasing=decreasing)
> my.order<-levels(factor(f))[ooo]
> factor(f, levels=my.order)
> }
it works incorrectly. Indeed, let's apply with your Factor:

  unames <- c("thousands", "units", "dozens",
"hundreds", "thousands",
"dozens")
  u <- c(1000, 1,10,100, 1000, 10)
  Factor(unames, u)

the above produces the following output:

  [1] thousands units     dozens    hundreds  thousands dozens   
  Levels: units thousands hundreds dozens

where "dozens" > "hundreds"

---
Valery

On Mon, 4 Oct 2004, Khamenia, Valery wrote:
> Hi Dimitris,
> 
> thank you for your reply, 
> 
> > does the following work in your data:
> > 
> > levs <- unique.default(Names)
> > factor(Names, levs[order(unique.default(Weights))])
> 
> your solution is really shorter, but has two issues to be 
> meant here:
> 
> 1. "unique.default" is applied twice, what might 
>    be a bit expensive for strings.
> 
> 2. your solution brings an implicit prerequisit on 
>   "unique.default". Indeed, "unique.default" should 
>   preserve the same order for output both working 
>   with strings and with numbers. In other words, 
>   correspondence must be kept by "unique.default".
>   If "unique.default" implementation uses (or will use) 
>   lexicographic strings sorting for acceleration
>   then this approach fails.
Please follow the posting guide and do your homework before posting, in 
this case the help page for unique.default.

     'unique' returns a vector, data frame or array like 'x' but
with
     duplicate  elements removed.

Value:

     An object of the same type of 'x'. but if an element is equal to
     one with a smaller index, it is removed.

so the order is preserved, by definition.

BTW it uses hashing for `acceleration', not something as slow as sorting.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595