Support Vector Regression with R

In this article I will show how to use R to perform a Support Vector Regression.
We will first do a simple linear regression, then move to the Support Vector Regression so that you can see how the two behave with the same data.

A simple data set

To begin with we will use this simple data set:

A simple data set in excel

I just put some data in excel. I prefer that over using an existing well-known data-set because the purpose of the article is not about the data, but more about the models we will use.

As you can see there seems to be some kind of relation between our two variables X and Y, and it look like we could fit a line which would pass near each point.

Let's do that in R !

Step 1: Simple linear regression in R

Here is the same data in CSV format, I saved it in a file regression.csv :

A simple data set in CSV

We can now use R to display the data and fit a line:

# Load the data from the csv file
dataDirectory <- "D:/" # put your own folder here
data <- read.csv(paste(dataDirectory, 'regression.csv', sep=""), header = TRUE)

# Plot the data
plot(data, pch=16)

# Create a linear regression model
model <- lm(Y ~ X, data)

# Add the fitted line
abline(model)

The code above displays the following graph:

The linear regression with our simple data set

Step 2: How good is our regression ?

In order to be able to compare the linear regression with the support vector regression we first need a way to measure how good it is.

To do that we will change a little bit our code to visualize each prediction made by our model

dataDirectory <- "D:/"
data <- read.csv(paste(dataDirectory, 'regression.csv', sep=""), header = TRUE)

plot(data, pch=16)
model <- lm(Y ~ X , data)

# make a prediction for each X
predictedY <- predict(model, data)

# display the predictions
points(data$X, predictedY, col = "blue", pch=4)

This produces the following graph:linear model prediction

For each data point X_i the model makes a prediction \hat{Y}_i displayed as a blue cross on the graph. The only difference with the previous graph is that the dots are not connected with each other.

In order to measure how good our model is we will compute how much errors it makes.

We can compare each Y_i value with the associated predicted value \hat{Y}_i and see how far away they are with a simple difference.

Note that the expression \hat{Y}_i - Y_i is the error, if we make a perfect prediction \hat{Y}_i will be equal to Y_i and the error will be zero.

If we do this for each data point and sum the error we will have the sum of the errors, and if we takes the mean we will get the Mean Squared Error (MSE)

MSE = \frac{1}{n}\sum\limits_{i=1}^n (\hat{Y}_i - Y_i)^2

A common way to measure error in machine learning is to use the Root Mean Squared Error (RMSE) so we will use it instead.

To compute the RMSE  we take the square root and we get the RMSE

RMSE = \sqrt{MSE}

Using R we can come with the following code to compute the RMSE

rmse <- function(error)
{
  sqrt(mean(error^2))
}

error <- model$residuals  # same as data$Y - predictedY
predictionRMSE <- rmse(error)   # 5.703778

We know now that the RMSE of our linear regression model is 5.70. Let's try to improve it with SVR !

Step 3: Support Vector Regression

In order to create a SVR model with R you will need the package e1071. So be sure to install it and to add the library(e1071) line at the start of your file.

Below is the code to make predictions with Support Vector Regression:

  model <- svm(Y ~ X , data)

  predictedY <- predict(model, data)

  points(data$X, predictedY, col = "red", pch=4)

As you can see it looks a lot like the linear regression code. Note that we called the svm function (not svr !)  it's because this function can also be used to make classifications with Support Vector Machine. The function will automatically choose SVM if it detects that the data is categorical (if the variable is a factor in R).

The code draws the following graph:

Support Vector Regression Predictions

This time the predictions is closer to the real values ! Let's compute the RMSE of our support vector regression model.

  # /!\ this time  svrModel$residuals  is not the same as data$Y - predictedY
  # so we compute the error like this
  error <- data$Y - predictedY
  svrPredictionRMSE <- rmse(error)  # 3.157061

As expected the RMSE is better, it is now 3.15  compared to 5.70 before.

But can we do better ?

Step 4: Tuning your support vector regression model

In order to improve the performance of the support vector regression we will need to select the best parameters for the model.

In our previous example, we performed an epsilon-regression, we did not set any value for epsilon ( \epsilon ), but it took a default value of 0.1.  There is also a cost parameter which we can change to avoid overfitting.

The process of choosing these parameters is called hyperparameter optimization, or model selection.

The standard way of doing it is by doing a grid search. It means we will train a lot of models for the different couples of \epsilon and cost, and choose the best one.

  # perform a grid search
  tuneResult <- tune(svm, Y ~ X,  data = data,
                ranges = list(epsilon = seq(0,1,0.1), cost = 2^(2:9))
  )
  print(tuneResult)
  # Draw the tuning graph
  plot(tuneResult)

There is two important points in the code above:

  •  we use the tune method to train models with \epsilon = 0, 0.1, 0.2, ... ,1  and cost = 2^2, 2^3, 2^4, ... ,2^9 which means it will train  88 models (it can take a long time)
  • the tuneResult returns the MSE, don't forget to convert it to RMSE before comparing the value to our previous model.

The last line plot the result of the grid search:

support-vector regression performance 1

 

On this graph we can see that the darker the region is the better our model is (because the RMSE is closer to zero in darker regions).

This means we can try another grid search in a narrower range we will try with \epsilon values between 0 and 0.2. It does not look like the cost value is having an effect for the moment so we will keep it as it is to see if it changes.

  tuneResult <- tune(svm, Y ~ X,  data = data,
                     ranges = list(epsilon = seq(0,0.2,0.01), cost = 2^(2:9))
  ) 

  print(tuneResult)
  plot(tuneResult)

We trained different 168 models with this small piece of code.

As we zoomed-in inside the dark region we can see that there is several darker patch.
From the graph you can see that models with C between 200 and 300 and \epsilon between 0.08 and 0.09 have less error.

support vector regression performance 2
Hopefully for us, we don't have to select the best model with our eyes and R allows us to get it very easily and use it to make predictions.

  tunedModel <- tuneResult$best.model
  tunedModelY <- predict(tunedModel, data) 

  error <- data$Y - tunedModelY  

  # this value can be different on your computer
  # because the tune method  randomly shuffles the data
  tunedModelRMSE <- rmse(error)  # 2.219642  

We improved again the RMSE of our support vector regression model !

If we want we can visualize both our models. The first SVR model is in red, and the tuned SVR model is in blue on the graph below :

support vector regression comparaison
I hope you enjoyed this introduction on Support Vector Regression with R.
You can get the source code of this tutorial. Each step has its own file.

If you want to learn more about Support Vector Machines, you can now read this article:
An overview of Support Vector Machines

188 thoughts on “Support Vector Regression with R”

  1. Hí,

    I have been studying and using the SVM. I work in a financial institution, and one of my responsabilities is to forecast the quarter budget for some variables, in this case non-performance-loans stock reduction. It happens that I trained a SVM model that fits my historic data very well.
    My question is : Ok you have this model, Then What ?! How could i use this trained model to predict future values ? ( I this case the next Quarter !)

    Reply
    • Your problem is not with SVM but with any machine learning model you could use. If your model fit your data and you make the assumption that it correctly represent and underlying unknown relation, then you input new data and use their result as prediction.

      Reply
  2. Hi Alexandre,
    very nice explanation, thank you very much!
    Just a couple of questions (I am using a different dataset than yours):

    1. when i predict the new data (my test set in my case) using the command:
    tunedModelY <- predict(tunedModel, data) --in my case data = test_set
    I get this error:
    "Error in scale.default(newdata[, object$scaled, drop = FALSE], center = object$x.scale$"scaled:center", :
    length of 'center' must equal the number of columns of 'x'"
    However, if I use the normal svm formula including the new tuned parameters manually, the command works (see code below):
    tuned_regressor = svm(formula = Y ~ .,
    data = training_set,
    type = "eps-regression",
    kernel = "radial",
    cost = 256,
    epsilon = 0)
    y_pred_best = predict(tuned_regressor, newdata = test_set)
    Do you have any idea why?

    2. Is it ever possible for the not tuned model to return a smaller RMSE than the tuned model? That's what happened in my case.

    Thank you very much for taking the time to answer this.

    Enrico

    Reply
    • Hello Enrico,
      For 1. I can't give you help it really depends on your data.
      For 2. Yes it is possible, that means that tuning did not improve the model. You can try to increase the range of the trained hyper parameters and see if it find a better model.

      Reply

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