Python - Algorithm to determine if a list is symmetric

Question

So I\'m stuck on this problem where I\'ve been asked to write an function in Python that checks to see if an n-dimensional array (is that what they\'re called?) is \"symmetric\"

Answer 1

You can put this check at the start of your function:

for row in square:
    if len(row) != len(square):
        return False

Or maybe shorter

if not all(len(square) == len(row) for row in square): return False

Answer 2

Add this near the beginning:

for row in square:
    if len(row) != len(square):
        return False

Answer 3

A version for Python3

def symmetric(L)
    return all(i==j for i,*j in zip(L ,*L))

Answer 4

Here's an alternative version for the main test:

for i, line in enumerate(matrix):
    for j in range(len(line)):
        if a[i][j] != a[j][i]:
             return False
return True

Of course that all the other answers that advise you to test if the matrix is square hold true.

Answer 5

Value y = 0 should be inside the first while loop. Like this:

def symmetric(square):
    final_result = []
    x = 0
    while x < len(square):
         y = 0
        row_list = []
        .
        .

Answer 6

This bit of code will do it all for you:

def symmetric(square):
    square = [tuple(row) for row in square]
    return square == zip(*square)

In your solution you're doing too much of the work yourself. Python will compare sequences for you, so an easier method is to transpose the square so its rows become columns and vice versa and then compare it to the original value.

We can transpose the square using the zip function. This takes a number of sequences and returns a tuple containing first of each and then a tuple with the second of each and so on. By passing square as *square we pass each row as a sperate argument; this has the effect of transposing the square.

The only complication is that zip returns tuples not lists so we have to make sure square is a list of tuples so the comparison works.