Check if list is empty without using the `not` command

Question

How can I find out if a list is empty without using the not command?
Here is what I tried:

if list3[0] == []: 
 print("No matches found") 
else: 
 print(list3)

I am very much a beginner so excuse me if I do dumb mistakes.

• halex: What's wrong with not?
• user2240288: Its for a school task, and I'm not supposed to know not right now. I only know about it because I searched for the answer before.
• Grijesh Chauhan: possible duplicate of Best way to check if a list is empty
• mike3996: Top tip: not is an operator, not a command

Answer 1

In order of preference:

# Good
if not list3:
# Okay
if len(list3) == 0:
# Ugly
if list3 == []:
# Silly
try:
 next(iter(list3))
 # list has elements
except StopIteration:
 # list is empty

If you have both an if and an else you might also re-order the cases:

if list3:
 # list has elements
else:
 # list is empty

• Martijn Pieters: Indeed, empty lists are False in a boolean context. No need to explicitly test for len().
• Silas Ray: It should be noted that under the hood, 1 and 2 are really doing the same thing most of the time.
• Aya: @sr2222 Sorry. Deleted comment due to syntax error in Python 2.x. Could do None if list3 else print("No matches found") in Python 3.x, though.
• Silas Ray: @Aya You could also do try: iter(l).next() except StopIteration: #do stuff. Of course, 'can' and 'should' are different things... Edit: Doh, John beat me to it... though try: l[0] except IndexError: #do stuff is equally silly...
• Aya: @sr2222 Indeed. Although it'd be interesting to see how many crazy ways people can come up with. list3 or print("No matches found") might also do the trick. ;-)

Answer 2

You find out if a list is empty by testing the 'truth' of it:

>>> bool([])
False
>>> bool([0]) 
True

While in the second case 0 is False, but the list [0] is True because it contains something. (If you want to test a list for containing all falsey things, use all or any: any(e for e in li) is True if any item in li is truthy.)

This results in this idiom:

if li:
 # li has something in it
else:
 # optional else -- li does not have something 
if not li:
 # react to li being empty
# optional else...

According to PEP 8, this is the proper way:

• For sequences, (strings, lists, tuples), use the fact that empty sequences are false.

Yes: if not seq:
 if seq:
No: if len(seq)
 if not len(seq)

You test if a list has a specific index existing by using try:

>>> try:
... li[3]=6
... except IndexError:
... print 'no bueno'
... 
no bueno

So you may want to reverse the order of your code to this:

if list3: 
 print list3 
else: 
 print "No matches found"

Answer 3

Check its length.

l = []
print len(l) == 0

Answer 4

this is how you would do that

if len(list3) == 0:
 print("No matches found") 

Answer 5

Python provides an inbuilt any() function to check whether an iterable is empty or not:

>>>list=[]
>>>any(list)
False

The function returns True if the iterable contains a 'True' value, and False otherwise.

However, note that the list [0] also returns False with any().