python中的矩阵乘法?

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我试图用纯python将两个矩阵相乘。输入(X1为3x3,Xt为3x2):

X1 =  [[1.0016, 0.0, -16.0514], 
       [0.0, 10000.0, -40000.0], 
       [-16.0514, -40000.0, 160513.6437]]
Xt =  [(1.0, 1.0), 
       (0.0, 0.25), 
       (0.0, 0.0625)]

其中XT是另一个矩阵的压缩转置。下面是代码:

def matrixmult (A, B):
    C = [[0 for row in range(len(A))] for col in range(len(B[0]))]
    for i in range(len(A)):
        for j in range(len(B[0])):
            for k in range(len(B)):
                C[i][j] += A[i][k]*B[k][j]
    return C

python给我的错误是:IndexError: list index out of range

提问于
用户回答回答于

这是不正确的初始化。

C = [[0 for row in range(len(A))] for col in range(len(B[0]))]

正确的初始化将是

C = [[0 for col in range(len(B[0]))] for row in range(len(A))]

此外,我建议使用更好的命名约定。在调试方面会有很大的帮助。例如:

def matrixmult (A, B):
    rows_A = len(A)
    cols_A = len(A[0])
    rows_B = len(B)
    cols_B = len(B[0])

    if cols_A != rows_B:
      print "Cannot multiply the two matrices. Incorrect dimensions."
      return

    # Create the result matrix
    # Dimensions would be rows_A x cols_B
    C = [[0 for row in range(cols_B)] for col in range(rows_A)]
    print C

    for i in range(rows_A):
        for j in range(cols_B):
            for k in range(cols_A):
                C[i][j] += A[i][k] * B[k][j]
    return C
用户回答回答于

如果你真的不想用numpy你可以这样做:

def matmult(a,b):
    zip_b = zip(*b)
    # uncomment next line if python 3 : 
    # zip_b = list(zip_b)
    return [[sum(ele_a*ele_b for ele_a, ele_b in zip(row_a, col_b)) 
             for col_b in zip_b] for row_a in a]

x = [[1,2,3],[4,5,6],[7,8,9],[10,11,12]]
y = [[1,2],[1,2],[3,4]]

import numpy as np # I want to check my solution with numpy

mx = np.matrix(x)
my = np.matrix(y)       

结果:

>>> matmult(x,y)
[[12, 18], [27, 42], [42, 66], [57, 90]]
>>> mx * my
matrix([[12, 18],
        [27, 42],
        [42, 66],
        [57, 90]])

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