Linear Regression

crates/nu-command/src/math/nushell-liniar-regression/Cargo.toml

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+[package]
+name = "nushell-liniar-regression"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]

crates/nu-command/src/math/nushell-liniar-regression/README.md

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+# Matrix operation
+The structures the store variables in lines and columns:
+```rust
+pub struct MatrixMN {
+    pub values: Vec<Vec<f64>>,
+}
+```
+
+## Creating a matrix
+```rust
+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
+```
+
+## Accessing an element
+```rust
+mat.values[i][j]
+```
+- element on the line `i - 1` and `j - 1` column
+- indexing starts from `0`
+
+
+# Linear Regression
+
+```rust
+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);
+    }
+}
+
+```
+
+
+# Explaining The Algorithm
+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`.
+
+## How to approximate the matrix X?
+```
+               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