R help - Oct 2002 - merge: How to preserve the original order?

I tried:
    x.vals <- data.frame(
          id = c( 'A1', 'C2', 'B3' )
        , ref = c( 'Ref1', 'Ref2' ,'Ref1' )
        , val = c( 1.11, 2.22, 3.33 )
        )
    x.labels <- data.frame(
          ref = c( 'Ref1', 'Ref2' )
        , label = c( 'Label01', 'Label02' )
        )

    merge( x.vals, x.labels, by='ref', all.x = T, sort=F )

I received:
         ref  id  val   label
    1   Ref1  A1 1.11 Label01
    2   Ref1  B3 3.33 Label01
    3   Ref2  C2 2.22 Label02

Alltough 'sort=F' is set, the original order: id = A1, C2, B3 is
not preserved. - Is there a possibility to preserve the original
order (when there is no key field which can be ordered after merging).
(Perhaps 'merge' is not the right solution for this problem?)

Wolfram
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You did get the original order as described in help(merge).  You can't
have the original order for x *and* the original order for y, and it is
clearly documented that you get that for y.

Try

merge( x.labels, x.vals, by='ref', sort=FALSE )

after reading the help page.  (Why did you specify all.x=T for
this example?)


On Fri, 25 Oct 2002, Wolfram Fischer - Z/I/M wrote:
> I tried:
>     x.vals <- data.frame(
>           id = c( 'A1', 'C2', 'B3' )
>         , ref = c( 'Ref1', 'Ref2' ,'Ref1' )
>         , val = c( 1.11, 2.22, 3.33 )
>         )
>     x.labels <- data.frame(
>           ref = c( 'Ref1', 'Ref2' )
>         , label = c( 'Label01', 'Label02' )
>         )
>
>     merge( x.vals, x.labels, by='ref', all.x = T, sort=F )
>
> I received:
>          ref  id  val   label
>     1   Ref1  A1 1.11 Label01
>     2   Ref1  B3 3.33 Label01
>     3   Ref2  C2 2.22 Label02
>
> Alltough 'sort=F' is set, the original order: id = A1, C2, B3 is
> not preserved. - Is there a possibility to preserve the original
> order (when there is no key field which can be ordered after merging).
> (Perhaps 'merge' is not the right solution for this problem?)
If you have an ordered id field (who would have guessed in this example
that those were ordered?) you can always sort on it. Here's another
solution.

x.vals$order <- seq(len=nrow(x.vals))
m <- merge( x.vals, x.labels, by='ref', all.x = T, sort=F )
m[sort.list(m$order), -4]

-- 
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 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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