Python不使用库进行矩阵运算详解

class Solution:
    def multiplyofmatrix(self, A, f):
        """
        Compute matrix A * matrix f.
        For example:
        Input:
        A: [[1, 2], [3, 4]]    Row of A: 2 Col of A: 2
        f: [[1, 2], [3, 4]]    Row of f: 2 Col of f: 2
        Output:
        ans: [[7, 10], [15, 22]]
        s.t. A.col == f.col
        """
        Arow, Acol = len(A), len(A[0])
        frow, fcol = len(f), len(f[0])
        if Acol != frow:
            return 'False!'
        ans = [[0] * fcol for _ in range(Arow)]
        for i in range(fcol):
            for j in range(Arow):
                for k in range(frow):
                    ans[j][i] += A[j][k]*f[k][i]
        return ans
if __name__=='__main__':
    a = Solution().multiplyofmatrix([[1, 2], [3, 4]], [[1, 2], [3, 4]])
    print(a)

    def transposeofmatrix(A):
        """
        Matrix transpose.
        For example:
        Input:
        A = [[1, 2],
             [3, 4]]
        Output:
        A_transpose = [[1, 3],
                       [2, 4]]
        """
        return [list(raw) for raw in zip(*A)]

    def submatrix(A, i, j):
        # 矩阵A第i行第j列元素的余矩阵
        m, n = len(A), len(A[0])
        C = [[A[x][y] for y in range(n) if y != j] for x in range(m) if x != i]  # 列表推导式
        return C

    def detofmatrix(A):
        m = len(A)  # 矩阵的行数
        n = len(A[0])  # 矩阵的列数
        if (m == 1 and n == 1):
            return A[0][0]
        else:
            value = 0
            for j in range(n):
                value += ((-1) ** (j + 2)) * A[0][j] * detofmatrix(submatrix(A, 0, j))
            return value

    def inverseofmatrix(A):
        m = len(A)  # 矩阵的行数
        n = len(A[0])  # 矩阵的列数
        C = [[0] * n for _ in range(m)]
        d = self.detofmatrix(A)
        for i in range(m):
            for j in range(n):
                C[i][j] = ((-1) ** (i + j + 2)) * detofmatrix(submatrix(A, j, i))
                C[i][j] = C[i][j] / d
        return C