R – How to plot SVM classification hyperplane

libsvmmachine learningrsvm

Here is my sample code for SVM classification.

    train <- read.csv("traindata.csv")
    test <- read.csv("testdata.csv")

    svm.fit=svm(as.factor(value)~ ., data=train, kernel="linear", method="class")

    svm.pred = predict(svm.fit,test,type="class")

The feature value in my example is a factor which gives two levels (either true or false). I wanted to
plot a graph of my svm classifier and group them into two groups. One group
those with a "true" and another group as false. How do we produce a 3D or 2D SVM plot? I tried with plot(svm.fit, train) but it doesn't seem to work out for me.
There is this answer i found on SO but I am not clear with what t, x, y, z, w, and cl are in the answer.

Plotting data from an svm fit – hyperplane

i have about 50 features in my dataset which the last column is a factor. Any simple way of doing it or if any one could help me explain his answer.

Best Answer

The short answer is: you cannot. Your data is 50 dimensional. You cannot plot 50 dimensions. The only thing you can do are some rough approximations, reductions and projections, but none of these can actually represent what is happening inside. In order to plot 2D/3D decision boundary your data has to be 2D/3D (2 or 3 features, which is exactly what is happening in the link provided - they only have 3 features so they can plot all of them). With 50 features you are left with statistical analysis, no actual visual inspection.

You can obviously take a look at some slices (select 3 features, or main components of PCA projections). If you are not familiar with underlying linear algebra you can simply use gmum.r package which does this for you. Simply train svm and plot it forcing "pca" visualization, like here: http://r.gmum.net/samples/svm.basic.html.

library(gmum.r)

# We will perform basic classification on breast cancer dataset 
# using LIBSVM with linear kernel
data(svm_breast_cancer_dataset)

# We can pass either formula or explicitly X and Y
svm <- SVM(X1 ~ ., svm.breastcancer.dataset, core="libsvm", kernel="linear", C=10)
## optimization finished, #iter = 8980
pred <- predict(svm, svm.breastcancer.dataset[,-1])

plot(svm, mode="pca")

which gives

for more examples you can refer to project website http://r.gmum.net/

However this only shows points projetions and their classification - you cannot see the hyperplane, because it is highly dimensional object (in your case 49 dimensional) and in such projection this hyperplane would be ... whole screen. Exactly no pixel would be left "outside" (think about it in this terms - if you have 3D space and hyperplane inside, this will be 2D plane.. now if you try to plot it in 1D you will end up with the whole line "filled" with your hyperplane, because no matter where you place a line in 3D, projection of the 2D plane on this line will fill it up! The only other possibility is that the line is perpendicular and then projection is a single point; the same applies here - if you try to project 49 dimensional hyperplane onto 3D you will end up with the whole screen "black").

Providing a minimal dataset

Usually, sharing huge data sets is not necessary and may rather discourage others from reading your question. Therefore, it is better to use built-in datasets or create a small "toy" example that resembles your original data, which is actually what is meant by minimal. If for some reason you really need to share your original data, you should use a method, such as dput(), that allows others to get an exact copy of your data.

Built-in datasets

You can use one of the built-in datasets. A comprehensive list of built-in datasets can be seen with data(). There is a short description of every data set, and more information can be obtained, e.g. with ?iris, for the 'iris' data set that comes with R. Installed packages might contain additional datasets.

Creating example data sets

Preliminary note: Sometimes you may need special formats (i.e. classes), such as factors, dates, or time series. For these, make use of functions like: as.factor, as.Date, as.xts, ... Example:

d <- as.Date("2020-12-30")

where

class(d)
# [1] "Date"

Vectors

x <- rnorm(10)  ## random vector normal distributed
x <- runif(10)  ## random vector uniformly distributed    
x <- sample(1:100, 10)  ## 10 random draws out of 1, 2, ..., 100    
x <- sample(LETTERS, 10)  ## 10 random draws out of built-in latin alphabet

Matrices

m <- matrix(1:12, 3, 4, dimnames=list(LETTERS[1:3], LETTERS[1:4]))
m
#   A B C  D
# A 1 4 7 10
# B 2 5 8 11
# C 3 6 9 12

Data frames

set.seed(42)  ## for sake of reproducibility
n <- 6
dat <- data.frame(id=1:n, 
                  date=seq.Date(as.Date("2020-12-26"), as.Date("2020-12-31"), "day"),
                  group=rep(LETTERS[1:2], n/2),
                  age=sample(18:30, n, replace=TRUE),
                  type=factor(paste("type", 1:n)),
                  x=rnorm(n))
dat
#   id       date group age   type         x
# 1  1 2020-12-26     A  27 type 1 0.0356312
# 2  2 2020-12-27     B  19 type 2 1.3149588
# 3  3 2020-12-28     A  20 type 3 0.9781675
# 4  4 2020-12-29     B  26 type 4 0.8817912
# 5  5 2020-12-30     A  26 type 5 0.4822047
# 6  6 2020-12-31     B  28 type 6 0.9657529

^{Note: Although it is widely used, better do not name your data frame df, because df() is an R function for the density (i.e. height of the curve at point x) of the F distribution and you might get a clash with it.}

Copying original data

If you have a specific reason, or data that would be too difficult to construct an example from, you could provide a small subset of your original data, best by using dput.

Why use dput()?

dput throws all information needed to exactly reproduce your data on your console. You may simply copy the output and paste it into your question.

Calling dat (from above) produces output that still lacks information about variable classes and other features if you share it in your question. Furthermore the spaces in the type column make it difficult to do anything with it. Even when we set out to use the data, we won't manage to get important features of your data right.

  id       date group age   type         x
1  1 2020-12-26     A  27 type 1 0.0356312
2  2 2020-12-27     B  19 type 2 1.3149588
3  3 2020-12-28     A  20 type 3 0.9781675

Subset your data

Tho share a subset, use head(), subset() or the indices iris[1:4, ]. Then wrap it into dput() to give others something that can be put in R immediately. Example

dput(iris[1:4, ]) # first four rows of the iris data set

Console output to share in your question:

structure(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6), Sepal.Width = c(3.5, 
3, 3.2, 3.1), Petal.Length = c(1.4, 1.4, 1.3, 1.5), Petal.Width = c(0.2, 
0.2, 0.2, 0.2), Species = structure(c(1L, 1L, 1L, 1L), .Label = c("setosa", 
"versicolor", "virginica"), class = "factor")), row.names = c(NA, 
4L), class = "data.frame")

When using dput, you may also want to include only relevant columns, e.g. dput(mtcars[1:3, c(2, 5, 6)])

_{Note: If your data frame has a factor with many levels, the dput output can be unwieldy because it will still list all the possible factor levels even if they aren't present in the the subset of your data. To solve this issue, you can use the droplevels() function. Notice below how species is a factor with only one level, e.g. dput(droplevels(iris[1:4, ])). One other caveat for dput is that it will not work for keyed data.table objects or for grouped tbl_df (class grouped_df) from the tidyverse. In these cases you can convert back to a regular data frame before sharing, dput(as.data.frame(my_data)).}

Producing minimal code

Combined with the minimal data (see above), your code should exactly reproduce the problem on another machine by simply copying and pasting it.

This should be the easy part but often isn't. What you should not do:

showing all kinds of data conversions; make sure the provided data is already in the correct format (unless that is the problem, of course)
copy-paste a whole script that gives an error somewhere. Try to locate which lines exactly result in the error. More often than not, you'll find out what the problem is yourself.

What you should do:

add which packages you use if you use any (using library())
test run your code in a fresh R session to ensure the code is runnable. People should be able to copy-paste your data and your code in the console and get the same as you have.
if you open connections or create files, add some code to close them or delete the files (using unlink())
if you change options, make sure the code contains a statement to revert them back to the original ones. (eg op <- par(mfrow=c(1,2)) ...some code... par(op) )

Providing necessary information

In most cases, just the R version and the operating system will suffice. When conflicts arise with packages, giving the output of sessionInfo() can really help. When talking about connections to other applications (be it through ODBC or anything else), one should also provide version numbers for those, and if possible, also the necessary information on the setup.

If you are running R in R Studio, using rstudioapi::versionInfo() can help report your RStudio version.

If you have a problem with a specific package, you may want to provide the package version by giving the output of packageVersion("name of the package").

Seed

Using set.seed() you may specify a seed¹, i.e. the specific state, R's random number generator is fixed. This makes it possible for random functions, such as sample(), rnorm(), runif() and lots of others, to always return the same result, Example:

set.seed(42)
rnorm(3)
# [1]  1.3709584 -0.5646982  0.3631284

set.seed(42)
rnorm(3)
# [1]  1.3709584 -0.5646982  0.3631284

¹_{Note: The output of set.seed() differs between R >3.6.0 and previous versions. Specify which R version you used for the random process, and don't be surprised if you get slightly different results when following old questions. To get the same result in such cases, you can use the RNGversion()-function before set.seed() (e.g.: RNGversion("3.5.2")).}