Checking for lists of empty values

Question

What is the most effective way to check if a list contains only empty values (not if a list is empty, but a list of empty elements)? I am using the famously pythonic implicit booleaness method in a for loop:

def checkEmpty(lst):
 for element in lst:
 if element:
 return False
 break
 else:
 return True

Anything better around?

Answer 1

if not any(lst):
 # ...

Should work. any() returns True if any element of the iterable it is passed evaluates True. Equivalent to:

def my_any(iterable):
 for i in iterable:
 if i:
 return True
 return False

• Dr. Jan-Philip Gehrcke: Seems to be the most elegant approach :-)
• squiguy: I agree, this is nice and clean.

Answer 2

len([i for i in lst if i]) == 0

Answer 3

Using all:

 if all(item is not None for i in list):
 return True
 else:
 return False

• Steven Rumbalski: I think all's brother any feels slighted.
• squiguy: @StevenRumbalski I like the shorter approach using any, but I thought I would throw in my two cents for an alternate idea.
• Steven Rumbalski: Depending on the OP's definition of "empty", not bool(item) may be better than item is not None.
• squiguy: @StevenRumbalski Thanks for the edit, you can change whatever you find to be suitable.

Answer 4

>>> l = ['', '', '', '']
>>> bool([_ for _ in l if _])
False
>>> l = ['', '', '', '', 1]
>>> bool([_ for _ in l if _])
True

• Steven Rumbalski: This is a bit awkward looking and it doesn't short circuit like not any(l).
• Dr. Jan-Philip Gehrcke: I agree that this solution is far from optimal :)