any() and all() on empty list?

Ron Adam rrr at ronadam.com
Wed Mar 29 09:31:19 CEST 2006


Paul McGuire wrote:
> "Paul Rubin" <http://phr.cx@NOSPAM.invalid> wrote in message
> news:7x3bh1x0ym.fsf at ruckus.brouhaha.com...

> To my mind, the *meaning* of all() is that every element in the list asserts
> True.  But this is with an initial assumption that all() is False, unless I
> test every value and find them to be True.  Since I assume False to begin
> with, I get no values in the list to contradict the assumption, and so
> all([]) returns False.

Looking at in a different way...  If we consider also having a function 
  none() (for comparison), would it be consistent with all()?

def none(S):
    for x in S:
	if x: return False
    return True


any([]) -> False

none([])  -> True    (same as 'not any(S)')

all([])  ->  True ? False


I think I agree that all() should have an initial presumption of being 
False.


Looking at it in yet another way...  (yes, not as efficient)

def all(S):
    S_ = [x for x in S if x]
    return S_ == S

def any(S):
    S_ = [x for x in S if x]
    return S_ != []

def none(S):
    S_ = [x for x in S if x]
    return S_ == []


In this view and empty set can be True for all().  Is it posible 
'all([])'  is undefined?  Here, none() and all() return contradicting 
values.  So maybe the correct version may be...

def all(S):
    if S == []: return False
    for x in S:
	if x return True
    return False

I think a few valid actual use case examples could clear it up.  What 
makes the most sense?

Cheers,
    Ron




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