CodeGuy - 1 year ago 113

R Question

I used svm to find a hyperplane best fit regression dependent on q, where I have 4 dimensions: x, y, z, q.

`fit <- svm(q ~ ., data=data,kernel='linear')`

and here is my fit object:

`Call:`

svm(formula = q ~ ., data = data, kernel = "linear")

Parameters:

SVM-Type: C-classification

SVM-Kernel: linear

cost: 1

gamma: 0.3333333

Number of Support Vectors: 1800

I have a 3d plot of my data, where the 4th dimension is color, using plot3d. How can I overlay the hyperplane that svm found? How can I plot the hyperplane? I'd like to visualize the regress hyperplane.

Answer Source

You wrote:

I used svm to find a hyperplane best fit regression

But according to:

```
Call:
svm(formula = q ~ ., data = data, kernel = "linear")
Parameters:
SVM-Type: C-classification
```

you are doing *classification*.

So, first of all decide what you need: to classify or to fit regression, from `?svm`

, we see:

```
type: ‘svm’ can be used as a classification machine, as a
regression machine, or for novelty detection. Depending of
whether ‘y’ is a factor or not, the default setting for
‘type’ is ‘C-classification’ or ‘eps-regression’,
respectively, but may be overwritten by setting an explicit
value.
```

As I believe you didn't change the parameter `type`

from its default value, you are probably solving `classification`

, so, I will show how to visualize this for classification.

Let's assume there are `2`

classes, generate some data:

```
> require(e1071) # for svm()
> require(rgl) # for 3d graphics.
> set.seed(12345)
> seed <- .Random.seed
> t <- data.frame(x=runif(100), y=runif(100), z=runif(100), cl=NA)
> t$cl <- 2 * t$x + 3 * t$y - 5 * t$z
> t$cl <- as.factor(ifelse(t$cl>0,1,-1))
> t[1:4,]
x y z cl
1 0.7209039 0.2944654 0.5885923 -1
2 0.8757732 0.6172537 0.8925918 -1
3 0.7609823 0.9742741 0.1237949 1
4 0.8861246 0.6182120 0.5133090 1
```

Since you want `kernel='linear'`

the boundary must be `w1*x + w2*y + w3*z - w0`

- hyperplane.
Our task divides to 2 subtasks: 1) to evaluate equation of this boundary plane 2) draw this plane.

1) **Evaluating the equation of boundary plane**

First, let's run `svm()`

:

```
> svm_model <- svm(cl~x+y+z, t, type='C-classification', kernel='linear',scale=FALSE)
```

I wrote here explicitly `type=C-classification`

just for emphasis we want do *classification*.
`scale=FALSE`

means that we want `svm()`

to run directly with provided data without scaling data (as it does by default). I did it for future evaluations that become simpler.

Unfortunately, `svm_model`

doesn't store the equation of boundary plane (or just, normal vector of it), so we must evaluate it. From svm-algorithm we know that we can evaluate such weights with following formula:

```
w <- t(svm_model$coefs) %*% svm_model$SV
```

The negative intercept is stored in `svm_model`

, and accessed via `svm_model$rho`

.

2) **Drawing plane**.

I didn't find any helpful function `plane3d`

, so, again we should do some handy work. We just take grid of pairs `(x,y)`

and evaluate the appropriate value of `z`

of the boundary plane.

```
detalization <- 100
grid <- expand.grid(seq(from=min(t$x),to=max(t$x),length.out=detalization),
seq(from=min(t$y),to=max(t$y),length.out=detalization))
z <- (svm_model$rho- w[1,1]*grid[,1] - w[1,2]*grid[,2]) / w[1,3]
plot3d(grid[,1],grid[,2],z) # this will draw plane.
# adding of points to the graphics.
points3d(t$x[which(t$cl==-1)], t$y[which(t$cl==-1)], t$z[which(t$cl==-1)], col='red')
points3d(t$x[which(t$cl==1)], t$y[which(t$cl==1)], t$z[which(t$cl==1)], col='blue')
```

We did it with `rgl`

package, you can rotate this image and enjoy it :)