[R] How to bin x,y,z vectors into matrix?

[Sarah Goslee]

Hi Mark,

First you need to use cut() to bin the values. Here's an example:

# generate some fake data
temp <- data.frame(x=runif(20), y=runif(20), z=runif(20))
# bin x and y
temp$x <- cut(temp$x, seq(0, 1, by=.2))
temp$y <- cut(temp$y, seq(0, 1, by=.2))

>From there, there are all sorts of ways to get your desired table. Here are two:

xtabs(z ~ x + y, data=aggregate(z ~ x + y, data=temp, FUN="mean"))

or

library(ecodist)
with(temp, crosstab(x, y, z, "mean"))

On Sat, Jul 19, 2014 at 10:57 AM, Mark Miller <mamillerpa at gmail.com> wrote:
> This is probably a basic question, but I haven't been able to Google
> anything helpful after trying for days.

Using the targeted www.rseek.org search engine is a whole lot more
effective than using base Google for R searches.

Sarah


> 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


-- 
Sarah Goslee
http://www.functionaldiversity.org