Checking for alternating parity

Question

I am struggling to find a way to check if every other number in my list is of an alternating parity (i.e. even, odd, even, odd, etc.)

[1,2,3,4] # odd, even, odd, even (True, because they alternate)
[1,3,2,4] # odd, odd, even, even (False, because they don't alternate)

Anyone have any idea how I can check for this?

It's not an easy question...been trying my best. Can't seem to get past: `for i in range(len(lst)-1): if lst[i] % 2 == 0 and lst[i+1] % 2 != 0: return True` I seem to only be checking the first and second element and going nowhere. — djennacs, Oct 02 '18 at 23:43
Iterate through your array and check if somewhere the parity of the current element is same as that of the next element. If so you don't have alternating parity. Only if you can go through the full array (except last element), parity is alternating. — Michael Butscher, Oct 02 '18 at 23:55

score 3 · Answer 1 · answered Oct 03 '18 at 00:07

You can use the Python modulus operator % on adjacent values within an input list and check for equality while iterating.

If you choose this route and efficiency is a concern, I advise you use 3rd party libraries such as NumPy / numba so the iteration does not take place at Python level:

from numba import jit

@jit(nopython=True)
def check_alt_parity(L):
    for i in range(len(L)-1):
        if L[i] % 2 == L[i+1] % 2:
            return False
    return True

def check_alt_parity_list(L):
    for i in range(len(L)-1):
        if L[i] % 2 == L[i+1] % 2:
            return False
    return True

A = [1, 2, 3, 4] * 10000
B = [1, 3, 2, 4] * 10000

%timeit check_alt_parity(A)       # 780 µs
%timeit check_alt_parity_list(A)  # 9.09 ms

U13-Forward · Answer 2 · 2018-10-03T00:06:26.647

1

Try this function, l for list comprehension and getting bool for odd or even, then checking if all are true (the every second even index element is all True or all False, and same with every odd index element:

def oddeven_alter(l):
    l=[i%2 for i in l]
    return all([any([all(l[::2]),all(not i for i in l[::2])]),any([all(l[1::2]),all(not i for i in l[1::2])])])
)
print(oddeven_alter([1,2,3,4]))
print(oddeven_alter([1,3,2,4]))

Output:

True
False

Kind of complex looking tho

edited Oct 03 '18 at 00:06

answered Oct 02 '18 at 23:54

U13-Forward

69,221
14
89
114

3

Can you make it a little be more cryptic please? I can still understand how that works... – Julien Oct 02 '18 at 23:55
1

Hi, I was wondering if you could explain how you decide to slice the array in the second part? – djennacs Oct 02 '18 at 23:57
@Julien I explained now – U13-Forward Oct 02 '18 at 23:58

blueheart · Accepted Answer · 2018-10-03T00:00:56.310

1

Here

def alter(ls):
  for i in range(len(ls)-1):
      if ls[i]%2 == ls[i+1]%2:
         return False
  return True

edited Oct 03 '18 at 00:00

answered Oct 02 '18 at 23:56

blueheart

194
1
7

2

`else: continue` is useless – Julien Oct 02 '18 at 23:59

score 1 · Answer 4 · answered Oct 02 '18 at 23:58

You can iterate over all indices of your sequence and then compare it with the next one:

def is_alternating_parity(seq):
    for i in range(len(seq) - 1):
        if seq[i] % 2 == seq[i + 1] % 2:
            return False
    return True

print(is_alternating_parity([1, 2, 3, 4]))  # True
print(is_alternating_parity([1, 3, 2, 4]))  # False

Checking for alternating parity

4 Answers4