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 ! <[EMAIL PROTECTED]> 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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