So I have
A = [1,2,3,4]
I want to check if the array is symmetric. So the output would be False
Another example would be
arr = [1,2,3,3,2,1]
out = fun(arr)
out = True
I have been trying by like
def checksymm(a):
flag = 0
for i in range(0,len(a)/2):
if np.abs(a[i]) == np.abs(a[-i]):
return True
else:
return False
What can I be doing wrong? Thanks.
Corresponding of a[i] should be a[-1-i], not a[-i].
Besides, why not just reverse the list and compare each elements:
def isSymmetric(arr):
return (arr == arr[::-1]).all()
Probably the most efficient way would be to restrict the comparison to each half of the array:
n = len(arr) // 2
result = np.all(arr[:n] == arr[-1:-n - 1:-1])
With absolute values:
result = (np.abs(arr[:n]) == np.abs(arr[-1:-n - 1:-1])).all()
Both formulations should work directly with either lists or arrays.
When you compare, pay attention to the index, 0 with -1, 1 with -2 etc (not 1 with -1), so it should be i and -i-1 . Also instead of returning True after finding one equality , you should return after all of them are equal ...
arr = [1,2,3,3,2,1]
def checksymm(a):
for i in range(0,len(a)//2):
if a[i] != a[-i-1]: # or np.abs(a[i]) != np.abs(a[-i-1]) as in the original code
return False
return True
out = checksymm(arr)
print(out)
>>> True
And :
arr = [1,2,3,3,5,1]
out = checksymm(arr)
print(out)
>>> False
Edit : I tried to find the problem with your code and fix it, but I would recommand the answer of @songziming it's better.
np.abshere?