R help - May 2008 - how to not sort factors when plotting

Hello,

When making a graph, plot and boxplot automatically sort my factor
levels (y axis) into alphabetical order.  Is there a way to make it
NOT do this?  I've tried:

sort.by="none", sorted=FALSE, sort=FALSE, reorder=FALSE.

Thank you.

The short answer is 'they don't'.  They use the order of the levels
of the
factor that you supply, and the answer is for you to create the factor 
with the levels in the order you want -- otherwise factor creation uses 
alphabetical order.  See ?factor.

On Thu, 1 May 2008, Lydia N. Slobodian wrote:
> Hello,
>
> When making a graph, plot and boxplot automatically sort my factor
> levels (y axis) into alphabetical order.  Is there a way to make it
> NOT do this?  I've tried:
>
> sort.by="none", sorted=FALSE, sort=FALSE, reorder=FALSE.
>
> Thank you.
>
> ______________________________________________
> R-help at 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.
>
-- 
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

Hello.  I'm trying find the ratios between each of the integers in a
vector.  I have:

for (n in x) {
   ratio <- (x[n]/x[n-1])
   ratio.all <- c(ratio.all, ratio)
}

Of course this doesn't work, nor does diff(n)/diff(n-1).  Is there a
way to specify a pair of integers in a vector?

Thank you,

Lydia

Hi Lydia,

I compared my ratio function with Dimitris and Phil's suggestions. Please do
NOT use my approach because it's painfully slow for a large vector (as Phil
told me). Here is why (using Win XP SP2, Intel Core- 2 Duo 2.4 GHz, R 2.7.0
Patched):


# Vector
x=rnorm(100000,0,1)

# Suggestion
new.ratio=function(x) x[2:length(x)]/x[1:(length(x)-1)]

# My horrible function
my.ratio=function(x){
 temp=NULL
 for (n in 1:length(x)) temp=c(temp,x[n]/x[n-1])
 temp
 }

 # System time
t=system.time(my.ratio(x))
tnr=system.time(new.ratio(x))
 t
   user  system elapsed
  38.79    0.06   39.31
 tnr
   user  system elapsed
      0       0       0


Thanks to all,

Jorge



On Mon, May 12, 2008 at 11:15 AM, Phil Spector <spector@stat.berkeley.edu>
wrote:
> Another alternative would be to take advantage of R's vectorization:
>
>  x=c(1,2,3,2,1,2,3)
> > x[2:length(x)]/x[1:(length(x)-1)]
> >
> [1] 2.0000000 1.5000000 0.6666667 0.5000000 2.0000000 1.5000000
>
> The solution using your ratio function will be painfully slow
> for a large vector.
>
>                                       - Phil Spector
>                                         Statistical Computing Facility
>                                         Department of Statistics
>                                         UC Berkeley
>                                         spector@stat.berkeley.edu
>
>
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