Regression models (both linear and non-linear) are used for predicting a real value, like salary for example. If your independent variable is time, then you are forecasting future values otherwise, your model is predicting present but unknown values. Regression technique vary from Linear Regression to SVR and Random Forests Regression.

In this part, you will understand and learn how to implement the following Machine Learning Regression models:

  1. Simple Linear Regression
  2. Multiple Linear Regression
  3. Polynomial Regression
  4. Support Vector for Regression (SVR)
  5. Decision Tree Classification
  6. Random Forest Classification

Simple Linear Regression

It is assumed that the two variables are linearly related. Hence, we try to find a linear function that predicts the response value(y) as accurately as possible as a function of the feature or independent variable(x).

For generality, we define:

x as feature vector, i.e x = [x_1, x_2, …., x_n],

y as response vector, i.e y = [y_1, y_2, …., y_n]

for n observations (in above example, n=10).

A scatter plot of above dataset looks like:-

Now, the task is to find a line which fits best in above scatter plot so that we can predict the response for any new feature values. (i.e a value of x not present in a dataset)

This line is called the regression line.

The equation of the regression line is represented as:

Here,

  • h(x_i) represents the predicted response value for ith observation.
  • b_0 and b_1 are regression coefficients and represent y-intercept and slope of regression line respectively.

To create our model, we must “learn” or estimate the values of regression coefficients b_0 and b_1. And once we’ve estimated these coefficients, we can use the model to predict responses!

In this article, we are going to use the Least Squares technique.

Now consider:

Here, e_i is a residual error in ith observation.
So, our aim is to minimize the total residual error.

We define the squared error or cost function, J as:

and our task is to find the value of b_0 and b_1 for which J(b_0,b_1) is minimum!

Without going into the mathematical details, we present the result here:

where SS_xy is the sum of cross-deviations of y and x:

and SS_xx is the sum of squared deviations of x:

Note: The complete derivation for finding least squares estimates in simple linear regression can be found here.

Given below is the python implementation of the above technique on our small dataset:

import numpy as np 
import matplotlib.pyplot as plt
def estimate_coef(x, y):
# number of observations/points
n = np.size(x)
# mean of x and y vector
m_x, m_y = np.mean(x), np.mean(y)
# calculating cross-deviation and deviation about x
SS_xy = np.sum(y*x) - n*m_y*m_x
SS_xx = np.sum(x*x) - n*m_x*m_x
# calculating regression coefficients
b_1 = SS_xy / SS_xx
b_0 = m_y - b_1*m_x
return(b_0, b_1)def plot_regression_line(x, y, b):
# plotting the actual points as scatter plot
plt.scatter(x, y, color = "m",
marker = "o", s = 30)
# predicted response vector
y_pred = b[0] + b[1]*x
# plotting the regression line
plt.plot(x, y_pred, color = "g")
# putting labels
plt.xlabel('x')
plt.ylabel('y')
# function to show plot
plt.show()
def main():
# observations
x = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
y = np.array([1, 3, 2, 5, 7, 8, 8, 9, 10, 12])
# estimating coefficients
b = estimate_coef(x, y)
print("Estimated coefficients:\nb_0 = {} \
\nb_1 = {}".format(b[0], b[1]))
# plotting regression line
plot_regression_line(x, y, b)
if __name__ == "__main__":
main()

The output of the above piece of code is:

Estimated coefficients:
b_0 = -0.0586206896552
b_1 = 1.45747126437