Most concise way to check whether a list is empty or contains only None?

Question

Most concise way to check whether a list is empty or contains only None?

I understand that I can test:

if MyList:
    pass

and:

if not MyList:
    pass

but what if the list has an item (or multiple items), but those item/s are None:

MyList = [None, None, None]
if ???:
    pass

Answer 1

One way is to use all and a list comprehension:

if all(e is None for e in myList):
    print('all empty or None')

This works for empty lists as well. More generally, to test whether the list only contains things that evaluate to False , you can use any :

if not any(myList):
    print('all empty or evaluating to False')

Answer 2

You can use the all() function to test is all elements are None:

a = []
b = [None, None, None]
all(e is None for e in a) # True
all(e is None for e in b) # True

Answer 3

You can directly compare lists with == :

if x == [None,None,None]:

if x == [1,2,3]

Answer 4

If you are concerned with elements in the list which evaluate as true:

if mylist and filter(None, mylist):
    print "List is not empty and contains some true values"
else:
    print "Either list is empty, or it contains no true values"

If you want to strictly check for None , use filter(lambda x: x is not None, mylist) instead of filter(None, mylist) in the if statement above.