This one seems to have slipped through the cracks, but I'll take a stab at
it:

I don't think there's an (easy) general solution but there are two direct
ways to get at the idea.

If you are generally only seeking to do things like: x[order(x)] would
sort(x) work for you?

More generally, this can be done in not so hard a fashion with a function
wrapper:

e.g.,

# Not tested...
RecFunc <- function(df, FUN) {
    df[FUN(df),]
}

so your code would look something like

RecFunc(aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean)),order)

Does this do what you were looking for?

Michael Weylandt

On Tue, Sep 13, 2011 at 12:36 PM, Benjamin Polidore <polidore@gmail.com>wrote:

> If I create an aggregation like this:
>
> aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean)
>
> I'll get a new data frame, which I can order if I assign it like this:
>
> newFrame <- aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean)
> newFrame[order(newFrame$TEMP),]
>
> But.. if I just want to do this in one statement, what can I do?  If this
> was perl, I'd be using something like $_:
>
>
> aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean)[order($_$TEMP),]
>
> So basically, I want to call order within the square braces, and I want to
> reference the dataframe that preceeds the square brace.  Is there a way to
> do this in R?
>
> thanks,
> bp
>
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