Another option for doing opertions by grouping variables is to the the
plyr package.
d <- data.frame(x=1:10,
g1=LETTERS[rep(11:12,each=5)],
g2=letters[rep(21:23,c(3,3,4))]
)
library(plyr)
ddply(d, c("g1", "g2"), function(z){
z$x <- z$x / max(z$x)
z
})
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
> -----Oorspronkelijk bericht-----
> Van: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] Namens William Dunlap
> Verzonden: woensdag 22 september 2010 19:51
> Aan: Seth W Bigelow; bill.venab...@csiro.au
> CC: r-help@r-project.org
> Onderwerp: Re: [R] Doing operations by grouping variable
>
> > -----Original Message-----
> > From: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] On Behalf Of Seth W Bigelow
> > Sent: Tuesday, September 21, 2010 4:22 PM
> > To: bill.venab...@csiro.au
> > Cc: r-help@r-project.org
> > Subject: Re: [R] Doing operations by grouping variable
> >
> > Aah, that is the sort of truly elegant solution I have been
> seeking.
> > And it's wrapped up in a nice programming shortcut to boot
> (i.e., the
> > within statement). I retract anything I may have said about tapply
> > being clunky.
> >
> > Many thanks
> >
> > --Seth
> >
> > Dr. Seth W. Bigelow
> > Biologist, USDA-FS Pacific Southwest Research Station
> > 1731 Research Park Drive, Davis California
> >
> > <bill.venab...@csiro.au>
> > 09/21/2010 03:15 PM
> >
> > To <sbige...@fs.fed.us>
> >
> > You left out the subscript. Why not just do
> >
> > d <- within(data.frame(group = rep(1:5, each = 5),
> > variable = rnorm(25)),
> > scaled <- variable/tapply(variable, group, max)[group])
>
> This approach can be tricky when there is more than one
> grouping variable. E.g., suppose we have grouping variables
> g1 and g2:
> > d <- data.frame(x=1:10,
> g1=LETTERS[rep(11:12,each=5)],
> g2=letters[rep(21:23,c(3,3,4))])
> > d
> x g1 g2
> 1 1 K u
> 2 2 K u
> 3 3 K u
> 4 4 K v
> 5 5 K v
> 6 6 L v
> 7 7 L w
> 8 8 L w
> 9 9 L w
> 10 10 L w
> and we want to divide each x value by it max for each
> g1*g2 group (6 possible groups, of which 4 are in the data).
>
> You can extend Bill V.'s approach with
> > with(d, x/tapply(x, list(g1,g2), FUN=max)[cbind(g1,g2)])
> [1] 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000
> [6] 1.0000000 0.7000000 0.8000000 0.9000000 1.0000000 That
> would fail if g1 and g2 were not factors but were integer
> vectors. Try it with
> > di <- data.frame(x=1:10,
> g1=rep(11:12,each=5),
> g2=rep(21:23,c(3,3,4)))
> > with(di, x/tapply(x, list(g1,g2), FUN=max)[cbind(g1,g2)])
> Error in tapply(x, list(g1, g2), FUN = max)[cbind(g1, g2)] :
> subscript out of bounds
>
> To avoid that problem you can call tapply with no FUN to get
> the indices to subscript by
> > with(d, x/tapply(x, list(g1,g2), FUN=max)[tapply(x,
> list(g1, g2))])
> [1] 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000
> [6] 1.0000000 0.7000000 0.8000000 0.9000000 1.0000000
>
> The misleadingly named ave() can avoid the need to do the
> subscripting after tapply but has other problems
> > with(d, x/ave(x, g1, g2, FUN=max))
> [1] 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000
> [6] 1.0000000 0.7000000 0.8000000 0.9000000 1.0000000
> Warning messages:
> 1: In FUN(X[[6L]], ...) : no non-missing arguments to max;
> returning -Inf
> 2: In FUN(X[[6L]], ...) : no non-missing arguments to max;
> returning -Inf It gives the right answer but it is calling
> FUN even for the empty interaction groups. For some FUN's this would
> abort the call, not just give a warning. In any case it
> is a waste of time.
>
> In either case you can also use the interaction() function to
> change the multiple grouping vectors into one:
> > d <- within(d, interaction(g1, g2, drop=TRUE))
> > with(d, x/ave(x, g1g2, FUN=max))
> [1] 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000
> [6] 1.0000000 0.7000000 0.8000000 0.9000000 1.0000000
> > with(d, x/tapply(x, g1g2, FUN=max)[g1g2])
> K.u K.u K.u K.v K.v L.v
> 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000 1.0000000
> L.w L.w L.w L.w
> 0.7000000 0.8000000 0.9000000 1.0000000
> > with(d, x/tapply(x, g1g2, FUN=max)[tapply(x, g1g2)])
> K.u K.u K.u K.v K.v L.v
> 0.3333333 0.6666667 1.0000000 0.8000000 1.0000000 1.0000000
> L.w L.w L.w L.w
> 0.7000000 0.8000000 0.9000000 1.0000000 The names are
> probably unwanted in the tapply cases; use unname to get rid of them.
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
> > and be done with it?
> >
> > (Warning: if you replace the second '<-' above by '=', it will not
> > work.
> > It is NOT true that you can always replace '<-' by '=' for
> assignment.
> > Why?)
> >
> > Bill Venables.
>
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