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所以我有这三个完整的函数，它们返回所需的矩阵行、矩阵列以及行和列的点积，我需要在第四个函数中以某种方式将它们网格在一起，以便找到两个的乘积矩阵相乘。有什么建议么？

#Test matrices
A = [[2,4], [7,0], [6,3]]
B = [[3,1], [-1,8], [-3, 3]]
C = [[4,1,9], [6,2,8], [7,3,5]]
D = [[2,9], [5,2], [1,0]]

def row(A,i):
    Z = []
    Z.append(A[i])
    return Z[0]


def col(B,j):
    Z=[]
    for i in range(len(B)):
        Z.append(B[i][j])
    return Z

def dotProduct(x,y):
    prod=0
    for i in range(len(x)):
        prod=prod+x[i]*y[i]
    return prod

def matrixMUL(A, B):
    Z = []
    ....
    return Z
4

1 回答 1

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具有 3 个以上辅助功能的解决方案：

from functools import partial


#Test matrices

A = [[2,4], [7,0], [6,3]]
B = [[3,1], [-1,8], [-3, 3]]
C = [[4,1,9], [6,2,8], [7,3,5]]
D = [[2,9], [5,2], [1,0]]

def row(A,i):
    Z = []
    Z.append(A[i])
    return Z[0]

def col(B,j):
    Z=[]
    for i in range(len(B)):
        Z.append(B[i][j])
    return Z

def dotProduct(x,y):
    prod=0
    for i in range(len(x)):
        prod=prod+x[i]*y[i]
    return prod

def shape(A):
    num_rows = len(A)
    num_cols = len(A[0]) if A else 0
    return num_rows, num_cols

def matrix_product_entry(A, B, i, j):
    return dotProduct(row(A, i), col(B, j))

def matrix_make(rows, cols, entry_fn):
    return [[entry_fn(i, j) for j in range(cols)]
            for i in range(rows)]

def matrixMUL(A, B):
    n1, k1 = shape(A)
    print(n1, k1)
    n2, k2 = shape(B)
    if k1 != n2:
        raise ArithmeticError("matrices shapes are not compatible!")
    return matrix_make(n1, k2, partial(matrix_product_entry, A, B))

print (matrixMUL(C, D))
# returns [[22, 38], [30, 58], [34, 69]]
于 2019-10-17T21:19:07.307 回答