r - Create 3D Plot Colored According to the Z-axis

Question

 library(Sleuth2)

 mlr<-lm(ex1222$Buchanan2000~ex1222$Perot96*ex1222$Gore2000)


for (i in 0:3) {
           assign(paste("betaHat", i, sep=""), 
           summary(mlr)$coeff[i+1,1])
               }

x<-sort(ex1222$Perot96)
y<-sort(ex1222$Gore2000)


z1 <- outer(x, y, function(a,b) betaHat0+betaHat1*a+betaHat2*b+betaHat3*a*b)
nrz <- nrow(z)
ncz <- ncol(z)

# Create a function interpolating colors in the range of specified colors
jet.colors <- colorRampPalette( c("blue", "red") ) 

# Generate the desired number of colors from this palette
nbcol <- 100
color <- jet.colors(nbcol)

# Compute the z-value at the facet centres
zfacet <- z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz]

# Recode facet z-values into color indices
facetcol <- cut(zfacet, nbcol)

persp(x, y, z1, col=color[facetcol],theta=-30, lwd=.3,xlab="Perot 96", ylab="Gore 2000", zlab="Predicted Votes for Buchanan")

Hello,

I am trying to color the above plot. I was thinking I want to have higher values of 'z' colored darker shades of red (or any color really).

Any help on how to make that happen would be greatly appreciated.

Also, feel free to suggest a different function to make this happen as well.

Thank you!

edit....I put my new code after looking at the example on ?persp. I'd like to change the color and I'm not super happy with the readability of the new plot

Answer 1

I modified your code a bit.

library(Sleuth2)

It's generally better practice to use the data argument than to use predictor variables extracted from a data frame via $:

mlr<-lm(Buchanan2000~Perot96*Gore2000,data=ex1222)

We can use expand.grid() and predict() to get the regression results in a clean way:

perot <- seq(1000,40000,by=1000)
gore <-  seq(1000,400000,by=2000)

If you want the facets evaluated at the locations of the observations, you can use perot <- sort(unique(ex1222$Perot96)); gore <- sort(unique(ex1222$Gore2000)) instead.

pframe <- with(ex1222,expand.grid(Perot96=perot,Gore2000=gore))
mlrpred <- predict(mlr,newdata=pframe)

Now convert the predictions to a matrix:

nrz <- length(perot)
ncz <- length(gore)
z <- matrix(mlrpred,nrow=nrz)

I chose to go from light red (#ffcccc, red with quite a bit of blue/green) to dark red (#cc0000, a bit of red with nothing else).

jet.colors <- colorRampPalette( c("#ffcccc", "#cc0000") ) 

You could also use grep("red",colors(),value=TRUE) to see what reds R has built in.

# Generate the desired number of colors from this palette
nbcol <- 100
color <- jet.colors(nbcol)

# Compute the z-value at the facet centres
zfacet <- z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz]
# Recode facet z-values into color indices
facetcol <- cut(zfacet, nbcol)

persp(perot, gore, z,
      col=color[facetcol],theta=-30, lwd=.3,
      xlab="Perot 96", ylab="Gore 2000", zlab="Predicted Votes for Buchanan")

You say you're "not super happy with the readability" of the plot, but that's not very specific ... I would spend a while with the ?persp page to see what some of your options are ...

Another choice is the rgl package:

library(rgl)
## see ?persp3d for discussion of colour handling
vertcol <- cut(z, nbcol)
persp3d(perot, gore, z,
      col=color[vertcol],smooth=FALSE,lit=FALSE,
      xlab="Perot 96", ylab="Gore 2000", zlab="Predicted Votes for Buchanan")

It might also be worth taking a look at scatter3d from the car package (there are other posts on SO describing how to tweak some of its graphical properties).

library(car)
scatter3d(Buchanan2000~Perot96*Gore2000,data=ex1222)