如何将Python中第二个矩阵中每列的矩阵中的每一行乘以？

Question

I know the steps to multiply two matrices are as follow

Step 1: Make sure that the the number of columns in the 1st one equals the number of rows in the 2nd one.

Step 2: Multiply the elements of each row of the first matrix by the elements of each column in the second matrix.

Step 3: Add the products.

How do you do the second step?

For example

A = [[3,4,5],[5,0,6],[5,7,1]]

B = [[2,1,3],[2,6,4]]

So far I got a function to find each column for the second one

def column(B,j):
    col = []
    for column in B:
        col.append(column[j])
    return col

Next I will have to make a function that finds each row for the first one

def rows(A,i):

But then I don't know how to create a function that will multiply them together like

row(A,0) • col(B,0)

row(A,0) • col(B,1)

row(A,1) • col(B,0)

row(A,1) • col(B,1)

row(A,2) • col(B,0)

row(A,2) • col(B,1)

Answer 1

You should probably use numpy:

import numpy as np
np.dot(row(A,0), col(B,0))

However, assuming you don't want to use that, you could do:

def dot(arr1, arr2):
    return sum([x*y for x,y in zip(arr1, arr2)])

dot(row(A,0), col(B,0))

Answer 2

If you insist on using lists for this....

For C = AB, you need

C_{ij} = sum(A_{ik} * B_{kj})

Here, i, j, and k are subscripts, with the first subscript denoting the row and the second denoting the column. i, j, k run over the rows and columns (ie, list indices) of the matrix, so you can just write for loops over i, j, and k.

Answer 3

A has 3 columns, while B has 2 rows. So your example seems to contradict the requirement stated in Step 1. Nevertheless, this might be close to what you are looking for.

In [1]: A = [[3,4,5],[5,0,6],[5,7,1]]

In [2]: B = [[2,1,3],[2,6,4]]

In [3]: [[sum(r*c for r,c in zip(row, col)) for col in B] for row in A]
Out[3]: [[25, 50], [28, 34], [20, 56]]

---

By the way, here is a useful trick which you might find useful: If you want to transpose a matrix, use zip(*B) :

In [4]: zip(*B)
Out[4]: [(2, 2), (1, 6), (3, 4)]

This may be useful to you because it allows you to easily loop through the columns of B .

Answer 4

Here's a worked out example:

>>> from pprint import pprint
>>> def mmul(A, B):
        nr_a, nc_a = len(A), len(A[0])
        nr_b, nc_b = len(B), len(B[0])
        if nc_a != nr_b:
            raise ValueError('Mismatched rows and columns')
        return [[sum(A[i][k] * B[k][j] for k in range(nc_a))
                 for j in range(nc_b)] for i in range(nr_a)]

>>> A = [[1, 2, 3, 4]]
>>> B = [[1],
         [2],
         [3],
         [4]]

>>> pprint(mmul(A, B))
[[30]]

>>> pprint(mmul(B, A), width=20)
[[1, 2, 3, 4],
 [2, 4, 6, 8],
 [3, 6, 9, 12],
 [4, 8, 12, 16]