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

Question

Is there a Pythonic way to check if a list (a nested list with elements & lists) is essentially empty? What I mean by

Answer 1

A simple recursive check would be enough, and return as early as possible, we assume it input is not a list or contains non-lists it is not empty

def isEmpty(alist):
    try:
        for a in alist:
            if not isEmpty(a):
                return False
    except:
        # we will reach here if alist is not a iterator/list
        return False

    return True

alist = []
blist = [alist]               # [[]]
clist = [alist, alist, alist] # [[], [], []]
dlist = [blist]               # [[[]]]
elist = [1, isEmpty, dlist]

if isEmpty(alist): 
    print "alist is empty"

if isEmpty(dlist): 
    print "dlist is empty"

if not isEmpty(elist): 
    print "elist is not empty"

You can further improve it to check for recursive list or no list objects, or may be empty dicts etc.

Answer 2

I don't think there is an obvious way to do it in Python. My best guess would be to use a recursive function like this one :

def empty(li):
    if li == []:
        return True
    else:
        return all((isinstance(sli, list) and empty(sli)) for sli in li)

Note that all only comes with Python >= 2.5, and that it will not handle infinitely recursive lists (for example, a = []; a.append(a)).

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

def isEmpty(a):
    return all([isEmpty(b) for b in a]) if isinstance(a, list) else False

Simply.