Python: How to check if two lists are not empty

Question

In Python, I know the pythonic way to check if a list is empty is

if not a:
 # do things with empty list

To check if a list is not empty then, we would do:

if a:
 # do things with my list

How would we check, simultaneously (as read), if two lists then are not empty?

if a and b:
 # do things with my two lists

The above does not seem to work, and I'm unsure what (a and b) actually means. For a = [2], b = [1,3], (a and b) = [1,3]. What is the and operator actually doing here? If I end up reducing b = [] at some point, (a and b) = [] even though a is not empty.

Edit: My use case goes something like

while (a and b are not empty):
 modify a
 modify b

I would have naively thought that since if a checks if a list is not empty, if a and b would check if neither were empty, which is not the case.

• vaultah: possible duplicate of Confusion found with and operator
• Wyrmwood: what do you mean when you say "does not .. work"?
• Padraic Cunningham: I am having trouble understanding,if a and b: will only be True if both are not empty
• drglove: I'll edit the question to be more concise and check out the duplicate. This one is hard to search for as 'and' is typically not indexed.
• Peter DeGlopper: if a and b does check whether or not both lists are empty. If your seeing things that seem otherwise, your problem is in your loop and not the conditional.

Answer 1

It's working fine. For a = [2] and b = [1, 3], a and b is returning [1, 3] which is truthy, exactly as you would expect, because True and True is True. When you change b to [] it returns [], which is falsy, again exactly as you would expect, because True and False is False. So if a and b does exactly what you want.

What is actually happening is that and is returning the value that decided the truth of the expression. and doesn't always evaluate both sub-expressions; when the first is falsy, the whole expression is falsy and the second doesn't need to be evaluated, and therefore isn't. This is called short-circuiting. or behaves similarly (skipping evaluating the second part if the first part is truthy). Wherever and or or is able to make its decision, it returns that value.

Another way of looking at it: bool(a) and bool(a) == bool(a and b) in Python.

• drglove: Thanks for explaining this, makes sense. Is there a better way to check if the two are not-empty than by checking their lengths?
• kindall: Yes. if a and b as you originally had it.

Answer 2

a and b is correct.

and returns the second argument if the first is true.

• Wyrmwood: the second comment, while true is only meaningful outside of an if (like an assignment)

Answer 3

You can do this

if len(a) and len(b):
 #do something

• drglove: I've thought about this, but am wondering if there's something closer to a and b. If not, I'll settle for it, but I'd still like to know what the operator is actually doing here.
• Kevin: Can you give an example where this would have different behavior from if a and b:?
• ForceBru: @Kevin, should there be such a situation?
• vaultah: @PeterDeGlopper this works because if a and b works. This is exactly the same as if a and b, only longer.
• Kevin: @ForceBru, if there's no such situation, then why use it over if a and b:?

Answer 4

I think what you want is

>>> a, b = list(), list()
>>> while not (a and b):
... a.append(1)
... b.append(2)
...
>>> a, b
([1], [2])
>>>

Answer 5

and operator is compare two boolean values

 bool([]) == False

so [] and b always return []; no matter what b is.

Answer 6

To build on kindall's answer - you could also concatenate the lists before doing the truthiness test:

a = []
b = []
c = [1]
bool(a + b) # False
bool(a + b + c) # True

Of course this would come with a performance penalty if you were doing something where that mattered.