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.

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

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"