cornjuliox cornjuliox - 1 year ago 196
Python Question

Python - Algorithm to determine if a list is symmetric

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" or not, meaning that row 1 of the array == column 1, row 2 == column 2, row 3 == column 3, etc so on and so forth. The goal is to have a function that returns the boolean True if its symmetric, and False if its not.

I've managed to write a function that works, but it only work on lists whose sizes are perfect squares, (e.g 2 x 2, 4 x 4), and a few of my test cases are of "irregular" sizes (e.g 2 x 5, 3 x 2). For those lists I end up getting a list index out of range error Code here:

def symmetric(square):
final_result = []
x = 0
y = 0
while x < len(square):
row_list = []
col_list = []
while y < len(square[x]):
print "(x, y): %d, %d" % (x, y)
print "(y, x): %d, %d" % (y, x)
y = y + 1
if row_list == col_list:
x = x + 1

for x in final_result:
if x == False:
return False
return True

And the test cases that I'm failing on here:

print symmetric([[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6]])
#Expected result: >>> False
#List index out of range

# This one actually returns the correct result, I'm just including it here
# for reference.
#print symmetric([["cat", "dog", "fish"],
# ["dog", "dog", "fish"],
# ["fish", "fish", "cat"]])
#Expected result: >>> True
#Actual result: >>> True

print symmetric([[1,2,3],
#Expected Result: >>> False
#Actual result: list index out of range

Can someone help me modify the code so that it will work on these "irregularly shaped" arrays?

Answer Source

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
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