如何根据另一个矩阵的行和列的乘积分配一个 python 矩阵？

Question

I have a matrix M that I intend to assign based on one condition checked on another matrix Q. The python code should assign 1 if the product of i and j of Q is greater than 0.5 else that index should be assigned 0. The code for both the first matrix and the second one is below

x=3
y=3
matx=[[0 for i in range(x)]for j in range(y)]
matx[0][0]=0.3
matx[0][1]=0.1
matx[0][2]=0.6
matx[1][0]=0.1
matx[1][1]=0.6
matx[1][2]=0.3
matx[2][0]=0.3
matx[2][1]=0.4
matx[2][2]=0.3
Q=matx
##check the product of the column and row of i and assign, am stuck below 
M=[[1 if ... else 0]]

Answer 1

Assuming I understood your question correctly, you can do it with loops by first calculating the product of each row and column, then looping through each element of the new matrix calculate the product of the row and column:

x=3
y=3
matx=[[0 for i in range(x)]for j in range(y)]
matx[0][0]=0.3
matx[0][1]=0.1
matx[0][2]=0.6
matx[1][0]=0.1
matx[1][1]=0.6
matx[1][2]=0.3
matx[2][0]=0.3
matx[2][1]=0.4
matx[2][2]=0.3
Q=matx

# Calculate row products
row_products = []
for i in range(x):
    product = 1
    for j in range(y):
        product *= Q[i][j]
    row_products.append(product)

# Calculate column products
column_products = []
for j in range(y):
    product = 1
    for i in range(x):
        product *= Q[i][j]
    column_products.append(product)

# Calculate matrix values
M=[[0 for i in range(x)]for j in range(y)]
for i in range(x):
    for j in range(y):
        if row_products[i] * column_products[j] > 0.5:
            M[i][j] = 1

Answer 2

Using the prod() function from the math module you can get the row and column products of the matrix and combine (cross product) them in a list comprehension:

Q = [[0.3, 0.1, 0.6],
     [0.1, 0.6, 0.3],
     [0.3, 0.4, 0.3]]

from math import prod

rProd = [*map(prod,Q)]         # row products
cProd = [*map(prod,zip(*Q))]   # column products
M = [int(r*c/Qij>0.5) for row,r in zip(Q,rProd) for Qij,c in zip(row,cProd)] 

print(M)
[0, 0, 0, 0, 0, 0, 0, 0, 0]

Note that even with the division by Qij (which avoids having the Qij value twice in the product), the results will all be well below 0.5

[r*c/Qij for row,r in zip(Q,rProd) for Qij,c in zip(row,cProd)]

[0.00054, 0.00432, 0.00162, 
 0.00162, 0.00072, 0.00324, 
 0.00108, 0.00216, 0.00648]