如何使用 for 循环乘以矩阵 - Python

Question

I have no idea how to even begin doing this It needs to be a for loop to multiply mtrixes

for example

[[1,2],[3,4]] * [[3,4],[5,6]]

[1 , 2] , [3 , 4]

[3 , 4] *[5 , 6]

Need help much appreciated I know 90% of dont want to code for me so that's ok

It only needs to be two square matrixes

i'm pretty sure the pattern is looking at it in the list thing

a[1][1]*b[1][1]+a[1][2]*b[2][1]       a[1][1]b[1][2]+a[1][2]b[2][2]

a[2][1]b[1][1]+a[2][2]b[2][1]         a[2][1]b[1][2]+a[2][2]b[2][2]

Answer 1

result = [] # final result
for i in range(len(A)):

    row = [] # the new row in new matrix
    for j in range(len(B[0])):
        
        product = 0 # the new element in the new row
        for v in range(len(A[i])):
            product += A[i][v] * B[v][j]
        row.append(product) # append sum of product into the new row
        
    result.append(row) # append the new row into the final result


print(result)

Answer 2

If you look at how matrix multiplication works:

[ 1 2 ] x [ 5 6 ] = [ 1*5+2*7 1*6+2*8 ]
[ 3 4 ]   [ 7 8 ]   [ 3*5+4*7 3*6+4*8 ]

then you can determine a method to calculate this, eg if you are multiplying for element i , j of the output matrix, then you need to multiply everything in row i of the LHS matrix by everything in the column j of the RHS matrix, so that is a single for loop (as the number of elements in the row i is equal to column j ).

You also need to cover every combination of i and j for the dimensions of the output matrix, which is a for loop for the columns nested inside a for loop for the rows.

The actual code is, of course, an exercise for you to implement.

Answer 3

Break it down. Before you try to write a function that multiplies matrices, write one that multiplies vectors. If you can do that, multiplying two matrices is just a matter of multiplying row i and column j for every element i,j of the resultant matrix.

Answer 4

def matmul(matrix1_,matrix2_):
    result = [] # final result

    for i in range(len(matrix1_)):

        row = [] # the new row in new matrix
        for j in range(len(matrix2_[0])):

            product = 0 # the new element in the new row

            for v in range(len(matrix1_[i])):
                 product += matrix1_[i][v] * matrix2_[v][j]
            row.append(product) # append sum of product into the new row

        result.append(row) # append the new row into the final result
    return result

Answer 5

from numpy import *
m1 = array([[1, 2, 3],[4, 5, 6] ])
m2 = array([[7, 8],[9, 10],[11, 12]])
r = array([[0, 0],[0, 0]])
s = 0
for i in range(2):
    for j in range(2):
        for k in range(3):
            s = s + m1[i][k]*m2[k][j]

        r[i][j] = s
        s = 0
print(r)

I think append function is not working in a two-dimensional array when we are using numpy module, so this is the way I have solved it.

Answer 6

>>> A=[[1,2],[3,4]]
>>> B=[[3,4],[5,6]]
>>> n=2
>>> ans=[[0]*n for i in range(n)]
>>> ans
[[0, 0], [0, 0]]
>>> for i in range(n):
...     for j in range(n):
...             ans[i][j]=sum((A[i][v]*B[v][j] for v in range(n)))
... 
>>> ans
[[13, 16], [29, 36]]

I think you just need to simplify the formula of matrix multiplication.

We have A*B=C then: Cij= the value in the ith row and jth column of the answer. For example above we have C12=16 and C11=13.. (note that this is the 0th position in the array so often we start from 0 instead of 1)

Cij= dot_product(row_i_of_A,column_j_of_B)=sum(row_i_of_A(v)*column_j_of_B(v) for v in range(n))

Because we want the whole answer (all of C), we need to work out all possible Cij. This means we need to try all possible pairs ij, so we loop through i in range(n), j in range(n) and do this for each possible pair.

Answer 7

u and v are constructed for visualization purpose.

from typing import List
A = [[1,0,0],[-1,0,3]]
B = [[7,0,0],[0,0,0],[0,0,1]]

def mult_mat(A:List[List[int]], B:List[List[int]]) -> List[List[int]]:
    n = len(A)      # Number of rows in matrix A
    m = len(B[0])   # Number of columns in matrix B

    ret = [[0 for i in range(m)] for j in range(n)]

    for row in range(n):
        u = A[row]

        for col in range(m):
            v = [B[i][col] for i in range(len(B))]
            # Here you can calculate ret[row][col] directly without v
            # But v is constructed for visualization purpose
            ret[row][col] = sum([x*y for x,y in zip(u,v)])
    return ret

if __name__ == '__main__':
    print(mult_mat(A,B))