js之矩阵运算方法

js之矩阵运算方法

一、方法

// 转置矩阵
    function transpose(matrix) {
    let result = new Array(matrix.length).fill(0).map(arr => new Array(matrix[0].length).fill(0));
    for (let i = 0; i < result.length; i++) {
        for (let j = 0; j < result[0].length; j++) {
            result[i][j] = matrix[j][i];
        }
    }
    return result;
}
    // 行列式
    function det(square) {
    // 方阵约束
    if (square.length !== square[0].length) {
        throw new Error();
    }
    // 方阵阶数
    let n = square.length;

    let result = 0;
    if (n > 3) {
        // n 阶
        for (let column = 0; column < n; column++) {
            // 去掉第 0 行第 column 列的矩阵
            let matrix = new Array(n - 1).fill(0).map(arr => new Array(n - 1).fill(0));
            for (let i = 0; i < n - 1; i++) {
                for (let j = 0; j < n - 1; j++) {
                    if (j < column) {
                        matrix[i][j] = square[i + 1][j];
                    } else {
                        matrix[i][j] = square[i + 1][j + 1];
                    }
                }
            }
            result += square[0][column] * Math.pow(-1, 0 + column) * det(matrix);
        }
    } else if (n === 3) {
        // 3 阶
        result = square[0][0] * square[1][1] * square[2][2] +
                 square[0][1] * square[1][2] * square[2][0] +
                 square[0][2] * square[1][0] * square[2][1] -
                 square[0][2] * square[1][1] * square[2][0] -
                 square[0][1] * square[1][0] * square[2][2] -
                 square[0][0] * square[1][2] * square[2][1];
    } else if (n === 2) {
        // 2 阶
        result = square[0][0] * square[1][1] - square[0][1] * square[1][0];
    } else if (n === 1) {
        // 1 阶
        result = square[0][0];
    }
    return result;
}
    // 伴随矩阵
    function adjoint(square) {
    // 方阵约束
    if (square[0].length !== square.length) {
        throw new Error();
    }

    let n = square.length;

    let result = new Array(n).fill(0).map(arr => new Array(n).fill(0));
    for (let row = 0; row < n; row++) {
        for (let column = 0; column < n; column++) {
            // 去掉第 row 行第 column 列的矩阵
            let matrix = [];
            for (let i = 0; i < square.length; i++) {
                if (i !== row) {
                    let arr = [];
                    for (let j = 0; j < square.length; j++) {
                        if (j !== column) {
                            arr.push(square[i][j]);
                        }
                    }
                    matrix.push(arr);
                }
            }
            result[row][column] = Math.pow(-1, row + column) * det(matrix);
        }
    }
    return transpose(result);
}
    // 逆矩阵
    function inv(square) {
        if (square[0].length !== square.length) {
            throw new Error();
        }
        let detValue = det(square);
        let result = adjoint(square);
        
        console.log(JSON.stringify(detValue))
        for (let i = 0; i < result.length; i++) {
            for (let j = 0; j < result.length; j++) {
                result[i][j] /= detValue;
            }
        }
        return result;
    }
    // 矩阵相乘
    function multiply(a, b) {
    // 相乘约束
    if (a[0].length !== b.length) {
        throw new Error();
    }
    let m = a.length;
    let p = a[0].length;
    let n = b[0].length;

    // 初始化 m*n 全 0 二维数组
    let c = new Array(m).fill(0).map(arr => new Array(n).fill(0));

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            for (let k = 0; k < p; k++) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    }

    return c;
}

 二、使用

 let xy = [-1,-1] // -1,-1
    let xy1 = [-3,-2]
    let xy2 = [-2,-2]

    let lonlat = [2,2]
    let lonlat1 = [0,1]
    let lonlat2 = [1,1]

    let A = [xy1, xy2]
    let B = [lonlat1, lonlat2]

    console.log(JSON.stringify(multiply(A,B)))

    console.log(JSON.stringify(inv(A)))

    
    console.log(JSON.stringify((multiply(B,(inv(A))))))
    console.log(JSON.stringify(multiply(xy,(multiply(B,(inv(A)))))))