How to generate a smoothed surface for a three dimensional dataset?

On Dec 4, 2013, at 8:56 AM, Duncan Murdoch wrote:

On 04/12/2013 11:36 AM, Jun Shen wrote:

Hi,

I have a dataset with two independent variables (x, y) and a response
variable (z). I was hoping to generate a response surface by plotting x, y,
z on a three dimensional plot. I can plot the data with rgl.points(x, y,
z). I understand I may not have enough data to generate a surface. Is there
a way to smooth out the data points to generate a surface? Thanks a lot.

There are many ways to do that.  You need to fit a model that predicts z from (x, y), and then plot the predictions from that model.
An example below follows yours.

Jun

===========================

An example:

x<-runif(20)
y<-runif(20)
z<-runif(20)

library(rgl)
rgl.points(x,y,z)

Don't use rgl.points, use points3d() or plot3d().  Here's the full script:


x<-runif(20)
y<-runif(20)
z<-runif(20)

library(rgl)
plot3d(x,y,z)

fit <- lm(z ~ x + y + x*y + x^2 + y^2)

 Newcomers to R may think they would be getting a quadratic in x and y. But R's formula interpretation will collapse x^2 to just x and then it becomes superfluous and is discarded. The same result is obtained with z ~ (x + y)^2).   I would have thought that this would have been the code:

fit <- lm(z ~ poly(x,2) +poly(y,2) + x:y )

xnew <- seq(min(x), max(x), len=20)
ynew <- seq(min(y), max(y), len=20)
df <- expand.grid(x = xnew,
                 y = ynew)

df$z <- predict(fit, newdata=df)

surface3d(xnew, ynew, df$z, col="red")

With the modified fitting formula one sees a nice saddle (for that particular random draw) using rgl.snapshot().


The result with the earlier formula is a more restrained:




Continued thanks to you Duncan for making this great tool available.

--
David.

Duncan Murdoch

David Winsemius
Alameda, CA, USA