If list index exists, do X

Could it be more useful for you to use the length of the list len(n) to inform your decision rather than checking n[i] for each possible length?

I need to code such that if a certain list index exists, then run a function.

This is the perfect use for a try block:

ar=[1,2,3]

try:
    t=ar[5]
except IndexError:
    print('sorry, no 5')   

# Note: this only is a valid test in this context 
# with absolute (ie, positive) index
# a relative index is only showing you that a value can be returned
# from that relative index from the end of the list...

However, by definition, all items in a Python list between 0 and len(the_list)-1 exist (i.e., there is no need for a try block if you know 0 <= index < len(the_list)).

You can use enumerate if you want the indexes between 0 and the last element:

names=['barney','fred','dino']

for i, name in enumerate(names):
    print(i + ' ' + name)
    if i in (3,4):
        # do your thing with the index 'i' or value 'name' for each item...

If you are looking for some defined 'index' though, I think you are asking the wrong question. Perhaps you should consider using a mapping container (such as a dict) versus a sequence container (such as a list). You could rewrite your code like this:

def do_something(name):
    print('some thing 1 done with ' + name)
        
def do_something_else(name):
    print('something 2 done with ' + name)        
    
def default(name):
    print('nothing done with ' + name)     
    
something_to_do={  
    3: do_something,        
    4: do_something_else
    }        
            
n = input ("Define number of actors: ")
count = 0
names = []

for count in range(n):
    print("Define name for actor {}:".format(count+1))
    name = raw_input ()
    names.append(name)
    
for name in names:
    try:
        something_to_do[len(name)](name)
    except KeyError:
        default(name)

Runs like this:

Define number of actors: 3
Define name for actor 1: bob
Define name for actor 2: tony
Define name for actor 3: alice
some thing 1 done with bob
something 2 done with tony
nothing done with alice

You can also use .get method rather than try/except for a shorter version:

>>> something_to_do.get(3, default)('bob')
some thing 1 done with bob
>>> something_to_do.get(22, default)('alice')
nothing done with alice

len(nams) should be equal to n in your code. All indexes 0 <= i < n "exist".

Using the length of the list would be the fastest solution to check if an index exists:

def index_exists(ls, i):
    return (0 <= i < len(ls)) or (-len(ls) <= i < 0)

This also tests for negative indices, and most sequence types (Like ranges and strs) that have a length.

If you need to access the item at that index afterwards anyways, it is easier to ask forgiveness than permission, and it is also faster and more Pythonic. Use try: except:.

try:
    item = ls[i]
    # Do something with item
except IndexError:
    # Do something without the item

This would be as opposed to:

if index_exists(ls, i):
    item = ls[i]
    # Do something with item
else:
    # Do something without the item