The structures the store variables in lines and columns:

pub struct MatrixMN {
    pub values: Vec<Vec<f64>>,
}

let values: Vec<f64> = vec![1.0, 5.0];
let mat: MatrixMN = MatrixMN::create_matrix(&values, 1, 2);    // a vector of f64, nr of lines, nr of columns

mat.values[i][j]

• element on the line i - 1 and j - 1 column
• indexing starts from 0

let nm1: String = String::from("X-var");
let nm2: String = String::from("Y-var");
let val1: Vec<f64> = vec![1.0, 2.0, 3.0, 4.0];
let val2: Vec<f64> = vec![14.5, 140.1, 201.3, 220.5];

let dt: DataSet = DataSet::new(nm1.clone(), nm2.clone(), val1.clone(), val2.clone());

match dt.compute_linear_regression() {
    Ok(line) => {
        // d : y = a * x + b ; a = slope; b = intercept
        println!("d : y = {} * x + {}", line.slope, line.intercept);
    },
    Err(xbar) => {
        // d : x = constant
        println!("d : x = {}", xbar.x);
    }
}

What is linear regression? The best line that can go through a set of points P1(x1, y1), P2(x2, y2) ... P3(x3, y3)

d : y = a * x + b (x = ?, y = ?) = the equation of the line

We are to find the slope (x) and the intercept (y) of this line.

Assuming that all points are on the same line, we will obtain the following system of equations:

• a * x1 + b = y1
• a * x2 + b = y2
• a * x3 + b = y3
• .....
• a * xn + b = yn

Therefore, we can define the following matrices

    | 1 x1 |                     | y1 |
    | 1 x2 |                     | y2 |
A = | .... |    X = | b |   B =  | .. |
    | 1 xn |        | a |        | yn |

Now, the matrix X is the answear of the equations, which are incompatible, therefore they are part of an indeterminate system A * X = B.

               A * X = B
At *      |    A * X = B
(At*A)^-1 |    At * A * X = At * B
[(At * A)^(-1) * (At * A)] * X = (At * A)^(-1) * At * B
In * X = (At * A)^(-1) * At * B
X = (At * A)^(-1) * At * B

Where:

• In:

1. identity matrix with n lines and n columns
2. has 1 on the diagonal and 0 in rest

• At:

1. the transposed matrix
2. columns of the initial matrix become rows of the transpose
3. rows of the initial matrix become columns of the transpose