Partial "visual" Matching of matrices

[newbieAuggie2019]

I am thinking that this might be slower...but thanks for replying.

Hi again!

My answer is probably quite verbose, and as I am a newbie, I'm sure that there must be better ways to do it than my code, but I made the program give a visual and self-explanatory output:

A = [[1, 0, 1, 1, 1, 0, 0, 1],
[1, 0, 0, 1, 1, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 1, 0, 0, 1]]

B = [[0, 1, 0, 0, 1, 0, 0, 1],
[1, 1, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 0, 0, 0, 0, 1],
[1, 0, 1, 0, 1, 0, 0, 1]]

C = [[1, 0, 1, 0, 1, 0, 0, 1],
[1, 0, 1, 1, 1, 0, 0, 1],
[1, 0, 1, 0, 0, 0, 0, 1],
[1, 1, 0, 0, 1, 0, 0, 1]]

commonElementsAB = []
commonElementsAC = []

for i in range(4):
    for j in range(8):
        if (A[i][j] == B[i][j]) : 
            commonElementsAB.append(A[i][j]) 
        else : 
            pass

print("\nThis is matrix A:\n")
for elements in A:
    print(*elements)
print("\nThis is matrix B:\n")
for elements in B:
    print(*elements)
print(f"\nThe elements in common in the same \n\
position in matrices A and B are {len(commonElementsAB)}.\n\
These {len(commonElementsAB)} elements in common are: \n\
{commonElementsAB}.\n")

for i in range(4) : 
    for j in range(8) : 
        if (A[i][j] == B[i][j]) : 
            print(A[i][j], end = " ") 
        else : 
            print("*", end = " ")  
    print()

for x in range(4):
    for y in range(8):
        if (A[x][y] == C[x][y]) : 
            commonElementsAC.append(A[x][y]) 
        else : 
            pass

print("\nThis is matrix A:\n")
for elements in A:
    print(*elements)
print("\nThis is matrix C:\n")
for elements in C:
    print(*elements)
print(f"\nThe elements in common in the same \n\
position in matrices A and C are {len(commonElementsAC)}.\n\
These {len(commonElementsAC)} elements in common are: \n\
{commonElementsAC}.\n")

for i in range(4) : 
    for j in range(8) : 
        if (A[i][j] == C[i][j]) : 
            print(A[i][j], end = " ") 
        else : 
            print("*", end = " ")  
    print()

if len(commonElementsAB) == len(commonElementsAC):
    print(f"\nBoth matrices B and C have the same number of\n\
common elements with matrix A.")
elif len(commonElementsAB) > len(commonElementsAC):
    print(f"\nMatrices A and B have a bigger number of\n\
common elements in the same position than\n\
matrices A and C.")    
elif len(commonElementsAB) < len(commonElementsAC):
    print(f"\nMatrices A and C have a bigger number of\n\
common elements in the same position than\n\
matrices A and B.")

and the output is:

Output:
This is matrix A:

1 0 1 1 1 0 0 1
1 0 0 1 1 0 0 1
1 0 1 0 0 0 0 0
1 0 0 0 1 0 0 1

This is matrix B:

0 1 0 0 1 0 0 1
1 1 0 0 0 0 0 1
1 1 1 0 0 0 0 1
1 0 1 0 1 0 0 1

The elements in common in the same 
position in matrices A and B are 22.
These 22 elements in common are: 
[1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1].

* * * * 1 0 0 1 
1 * 0 * * 0 0 1 
1 * 1 0 0 0 0 * 
1 0 * 0 1 0 0 1 

This is matrix A:

1 0 1 1 1 0 0 1
1 0 0 1 1 0 0 1
1 0 1 0 0 0 0 0
1 0 0 0 1 0 0 1

This is matrix C:

1 0 1 0 1 0 0 1
1 0 1 1 1 0 0 1
1 0 1 0 0 0 0 1
1 1 0 0 1 0 0 1

The elements in common in the same 
position in matrices A and C are 28.
These 28 elements in common are: 
[1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1].

1 0 1 * 1 0 0 1 
1 0 * 1 1 0 0 1 
1 0 1 0 0 0 0 * 
1 * 0 0 1 0 0 1 

Matrices A and C have a bigger number of
common elements in the same position than 
matrices A and B.
>>> 

All the best,