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).…

Accepted Answer

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)
```

Create 3D Plot Colored According to the Z-axis

Solution

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