Python: condition to check that none of list's element is an empty list

Question

I have a list of listA; each element of A is a list. I want a condition (to use for a if statement) that return True if all the element of A are non empty and False otherwise. How can I express this condition?

Example 1

A = [[], [5]]
if (condition):
    print("no empty list as elements of A")
else:
    print("at least an empty list inside A")

>>> at least an empty list inside A

Example 2

A = [[3,2], [5]]
if (condition):
    print("no empty list as elements of A")
else:
    print("at least an empty list inside A")

>>> no empty list as elements of A

I tried with the condition

if(not b for b in A):

but it seems not to work properly. What am I missing?

chepner · Accepted Answer · 2021-12-09 20:52:22Z

5

Since an non-empty list is considered truthy, you can use all:

if all(A):
    print("No empty list in A")
else:
    print("At least one empty list in A")

answered Dec 9, 2021 at 20:52

chepner

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2 Comments

user1172131 Over a year ago

Thank you, it's exactly what I was looking for! Do you know why the solution that I tried (end of the question) doesn't work?

chepner Over a year ago

(not b for b in A) is a generator expression, which is always true. You aren't actually iterating over it, and if you did, you'd have a list of boolean values, not a single one.

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