You could use tapply() and cut() in base R, as in
> tapply(d$z, INDEX=list(y=cut(d$y,breaks=seq(900,1300,by=100)),
> x=cut(d$x,breaks=seq(35,60,by=5))), FUN=mean)
x
y (35,40] (40,45] (45,50] (50,55] (55,60]
(900,1e+03] 1644.779 1643.957 NA NA NA
(1e+03,1.1e+03] NA NA NA 1597.028 NA
(1.1e+03,1.2e+03] NA NA NA 1281.988 1846.123
(1.2e+03,1.3e+03] NA NA NA 1990.707 NA
cut() does the binning and tapply() applies the function 'FUN' to each group.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Jul 19, 2014 at 7:57 AM, Mark Miller <mamille...@gmail.com> wrote:
> This is probably a basic question, but I haven't been able to Google
> anything helpful after trying for days.
>
> I have an R dataframe with x,y,z tuples, where z is a response to x and y
> and can be modeled as a surface.
>
> > head(temp)
> x y z
> 1 36.55411 965.7779 1644.779
> 2 42.36912 978.9721 1643.957
> 3 58.34699 1183.7426 1846.123
> 4 53.55439 1232.2696 1990.707
> 5 50.76167 1115.2049 1281.988
> 6 51.37299 1059.9088 1597.028
>
> I would like to create a matrix of mean z values, with rows representing
> binned y values and columns representing binned x values, like
>
> 0<x<40 40<x<60 60<x<80 x>80
> 0<y<800 1000.0 1100.00 1100.00 1000.0
> 800<y<1200 1000.0 1200.00 1200.00 1000.0
> 1200<y<1400 1000.0 1200.00 1200.00 1000.0
> y<1400 1000.0 1100.00 1100.00 1000.0
> thanks
> Mark
>
> [[alternative HTML version deleted]]
>
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