Pythonic way to check if: all elements evaluate to False -OR- all elements evaluate to True

I want the results of the function to be:

• All values evaluate to False (None, 0, empty string) -> True
• All values evaluate to True -> True
• Every other case -> False

This is my try at it:

>>> def consistent(x):
...  x_filtered = filter(None, x)
...  return len(x_filtered) in (0, len(x))
...
>>> consistent((0,1))
Fals开发者_Go百科e
>>> consistent((1,1))
True
>>> consistent((0,0))
True

[Bonus]

What should this function be named?

def unanimous(it):
  it1, it2 = itertools.tee(it)
  return all(it1) or not any(it2)

def all_bools_equal(lst):
    return all(lst) or not any(lst)

See: http://docs.python.org/library/functions.html#all

See: http://docs.python.org/library/functions.html#any

Piggybacking on Ignacio Vasquez-Abram's method, but will stop after first mismatch:

def unanimous(s):
  it1, it2 = itertools.tee(iter(s))
  it1.next()
  return not any(bool(a)^bool(b) for a,b in itertools.izip(it1,it2))

While using not reduce(operators.xor, s) would be simpler, it does no short-circuiting.

def all_equals(xs):
    x0 = next(iter(xs), False)
    return all(bool(x) == bool(x0) for x in xs)

Not so brief, but shortcuts without messing around with 'tee' or anything like that.

def unanimous(s):
   s = iter(s)
   if s.next():
       return all(s)
   else:
       return not any(s)

Just yet another way of doing it, given list l:

sum([int(bool(x)) for x in l]) in (0, len(l))

>>> a=['',None,0,False]
>>> b=[1,True,'ddd']
>>> c=[0,1,False,True,None]
>>> for l in (a,b,c):
...  print sum([int(bool(x)) for x in l]) in (0, len(l))
... 
True
True
False

def AllTheSame(iterable):
    return any(iterable) is all(iterable)