Python: How to check if a nested list is essentially empty?

Use the any() function. This returns True if any list within the list is not empty.

alist = [[],[]]
if not any(alist):
    print("Empty list!")

>> Empty list!

see: https://www.programiz.com/python-programming/methods/built-in/any

I have combined the use of isinstance() by Ants Aasma and all(map()) by Stephan202, to form the following solution. all([]) returns True and the function relies on this behaviour. I think it has the best of both and is better since it does not rely on the TypeError exception.

def isListEmpty(inList):
    if isinstance(inList, list): # Is a list
        return all( map(isListEmpty, inList) )
    return False # Not a list

Simple code, works for any iterable object, not just lists:

>>> def empty(seq):
...     try:
...         return all(map(empty, seq))
...     except TypeError:
...         return False
...
>>> empty([])
True
>>> empty([4])
False
>>> empty([[]])
True
>>> empty([[], []])
True
>>> empty([[], [8]])
False
>>> empty([[], (False for _ in range(0))])
True
>>> empty([[], (False for _ in range(1))])
False
>>> empty([[], (True for _ in range(1))])
False

This code makes the assumption that anything that can be iterated over will contain other elements, and should not be considered a leaf in the "tree". If an attempt to iterate over an object fails, then it is not a sequence, and hence certainly not an empty sequence (thus False is returned). Finally, this code makes use of the fact that all returns True if its argument is an empty sequence.