If list index exists, do X

Question

In my program, user inputs number n, and then inputs n number of strings, which get stored in a list.

I need to code such that if a certain

Answer 1

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?

Answer 2

You can try something like this

list = ["a", "b", "C", "d", "e", "f", "r"]

for i in range(0, len(list), 2):
    print list[i]
    if len(list) % 2 == 1 and  i == len(list)-1:
        break
    print list[i+1];

Answer 3

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

You already know how to test for this and in fact are already performing such tests in your code.

The valid indices for a list of length n are 0 through n-1 inclusive.

Thus, a list has an index i if and only if the length of the list is at least i + 1.

Answer 4

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

Answer 5

It can be done simply using the following code:

if index < len(my_list):
    print(index, 'exists in the list')
else:
    print(index, "doesn't exist in the list")

Answer 6

A lot of answers, not the simple one.

To check if a index 'id' exists at dictionary dict:

dic = {}
dic['name'] = "joao"
dic['age']  = "39"

if 'age' in dic

returns true if 'age' exists.