Connecting points to the regression line in 3d

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3

This is the 3d version of this other question.

Consider the following data: 

Income2<-structure(list(X = 1:30, Education = c(21.5862068965517, 18.2758620689655, 
12.0689655172414, 17.0344827586207, 19.9310344827586, 18.2758620689655, 
19.9310344827586, 21.1724137931034, 20.3448275862069, 10, 13.7241379310345, 
18.6896551724138, 11.6551724137931, 16.6206896551724, 10, 20.3448275862069, 
14.1379310344828, 16.6206896551724, 16.6206896551724, 20.3448275862069, 
18.2758620689655, 14.551724137931, 17.448275862069, 10.4137931034483, 
21.5862068965517, 11.2413793103448, 19.9310344827586, 11.6551724137931, 
12.0689655172414, 17.0344827586207), Seniority = c(113.103448275862, 
119.310344827586, 100.689655172414, 187.586206896552, 20, 26.2068965517241, 
150.344827586207, 82.0689655172414, 88.2758620689655, 113.103448275862, 
51.0344827586207, 144.137931034483, 20, 94.4827586206897, 187.586206896552, 
94.4827586206897, 20, 44.8275862068966, 175.172413793103, 187.586206896552, 
100.689655172414, 137.931034482759, 94.4827586206897, 32.4137931034483, 
20, 44.8275862068966, 168.965517241379, 57.2413793103448, 32.4137931034483, 
106.896551724138), Income = c(99.9171726114381, 92.579134855529, 
34.6787271520874, 78.7028062353695, 68.0099216471551, 71.5044853814318, 
87.9704669939115, 79.8110298331255, 90.00632710858, 45.6555294997364, 
31.9138079371295, 96.2829968022869, 27.9825049000603, 66.601792415137, 
41.5319924201478, 89.00070081522, 28.8163007592387, 57.6816942573605, 
70.1050960424457, 98.8340115435447, 74.7046991976891, 53.5321056283034, 
72.0789236655191, 18.5706650327685, 78.8057842852386, 21.388561306174, 
90.8140351180409, 22.6361626208955, 17.613593041445, 74.6109601985289
)), .Names = c("X", "Education", "Seniority", "Income"), class = "data.frame", row.names = c(NA, 
-30L))

How to make a 3d graph of a regression model, including the points, the model's hyperplane, and a line connecting the points to the hyperplane?

I would like to see different ways of doing this type of chart, especially with base graphs, lattice and rgl .

in my SOEN question.

    
asked by anonymous 06.04.2014 / 05:19

2 answers

3

Another solution, somewhat simpler and perhaps not so pretty graphically, is the plotPlane function of the rockchalk package. The advantage is that you can make a 3D chart with just one command line. Using the above data: 

require(rockchalk)
model <- loess(Income ~ Education + Seniority, data = Income2)
plotPlane(model, plotx1 = "Education", plotx2 = "Seniority",
                 drawArrow = TRUE, pch = 20, pcol = 2,
                 npp = 100, theta = 20, phi = 30)

In which%% is the variable on each axis, plotx traces lines, drawArrow is the number of points for which predicted values are calculated, and npp and theta serve to rotate the graph horizontally and vertically. The result is this:

    
17.04.2014 / 01:30
1

Doing with the base graphics:

First, run the template: 

model <- loess(Income ~Education + Seniority, data=Income2)

Create sequences of x and y : 

x <-range(Income2$Education)
x <- seq(x[1], x[2], length.out=50)    
y <- range(Income2$Seniority)
y <- seq(y[1], y[2], length.out=50)

Create the values of z with all combinations of x and y : 

z <- outer(x,y, 
           function(Education,Seniority)
                     predict(model, data.frame(Education,Seniority)))

Graph the model with persp : 

p <- persp(x,y,z, theta=30, phi=30, 
           col="lightblue",expand = 0.5,shade = 0.2,
           xlab="Education", ylab="Seniority", zlab="Income")

Plot the points in 3d for 2d with trans3d and add the points (with points ) and segments (with segment ): 

obs <- trans3d(Income2$Education, Income2$Seniority,Income2$Income,p)
pred <- trans3d(Income2$Education, Income2$Seniority,fitted(model),p)
points(obs, col="red",pch=16)
segments(obs$x, obs$y, pred$x, pred$y)

    
06.04.2014 / 05:22
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