Will something like this work for you:

> d <- read.table(textConnection("ZIP         DATA
+ 94111     12135.545
+ 93105     321354.65654
+ 94111     545.555
+ 94706     558858.66"), header=TRUE)
> closeAllConnections()
> aggregate(d$DATA, list(Zip = d$ZIP), FUN=median, na.rm=T)
    Zip         x
1 93105 321354.66
2 94111   6340.55
3 94706 558858.66
> do.call(rbind, by(d$DATA, list(Zip=d$ZIP), function(x){
+     c(Mean=mean(x), Median=median(x))
+ }))
           Mean    Median
93105 321354.66 321354.66
94111   6340.55   6340.55
94706 558858.66 558858.66
>
> t(sapply(split(d$DATA, d$ZIP), function(x) c(Mean=mean(x),
Median=median(x))))
           Mean    Median
93105 321354.66 321354.66
94111   6340.55   6340.55
94706 558858.66 558858.66


On Fri, May 16, 2008 at 10:18 AM, Mike ! <mc510@hotmail.com> wrote:

>
>
> I've got a data frame having numerical data by zip code:
>
> ZIP         DATA
> 94111     12135.545
> 93105     321354.65654
> 94111     545.555
> 94706     558858.66
> ...           ...
>
> I'm using this function to group records by ZIP and calculate the median of
> DATA:
>
> aggregate(d$DATA, list(Zip = d$ZIP), FUN=median, na.rm=T)
>
> but what I really want to do is to calculate several statistics (median,
> mean, etc.) for each group of records, so that I'll get a result organized
> like:
>
> Zip       median(DATA)    mean(DATA)
> 94706   565                    555
> 94111   59585                 66666
> 93105   595685               5555666
> ...
>
> I tried using FUN=funstofun(median,mean), but found that doesn't work in
> aggregate.  Is there a straightforward way to do what I'm after (I'm pretty
> new to R)?  thanks
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>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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