How to check if end of list was reached?

Question

If have a list, say a=[1,2,3], and I want to see if a[4] is null, is there a way to do that, without using an exception or assertion?

Answer 1

len will tell you the length of the list. To quote the docs:

len(s)
    Return the length (the number of items) of an object. The argument may be a sequence
    (string, tuple or list) or a mapping (dictionary).

Of course, if you want to get the final element in a list, tuple, or string, since indexes are 0 based, and the length of an item is the element count, a[len(a)-1] will be the last item.

---

As an aside, generally, the proper way to access the last element in an object which allows numeric indexing (str, list, tuple, etc) is using a[-1]. Obviously, that does not involve len though.

Answer 2

Here is an approach which I applied in one of the Arcade Challenges from Code Fights.

Basically, the end of a list is defined by:

• list length - current index (iteration) == 1

#!/usr/bin/python3

numbers = [1, 3, 5, 8, 10, 13, 16]

list_len = len(numbers)

for n in numbers:
    current_idx = numbers.index(n)
    print("Current Number:", numbers[current_idx])
    list_end = list_len - current_idx
    if list_end != 1:
        next_idx = current_idx + 1
        print("Next Number:   ", numbers[next_idx])
    else:
        print("End Of List!")

Answer 3

with a = [1,2,3] :

a[2:3] is [3]

a[3:4] is [ ]

So a[i:i+1] != [ ] tells if is an index of a

a[i:] does the same, but a[i:] creates another list, possible very long, while a[i:i+1] is 1 element if not empty

Answer 4

Use len

if len(a) <= index:
   ...

Note: Your question asks how you would find out "if a[4] is null". a[4] isn't anything, which is why you get an IndexError when you try to check it.

Answer 5

a[4] in this case will throw a IndexError exception, which isn't the same as comparing the value of a at index 4 to None. You can have values of None in a list, and if you were to compare values of a, then when you encounter a None, it doesn't mean that the index is not found in the list. For example:

>>> a=[1,None,2]
>>> a[1]==None
True
>>> a[3]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

Since lists are contiguous and indexed sequentially, the correct way to check if an index is in a list is to compare it to the len() of a list, but depending on the application, there are other ways around it, like catching an IndexError, or iteration.

>>> for index, value in enumerate(a):
...     print index, value
... 
0 1
1 None
2 2

Answer 6

You could write a function which behaves kind of like dict.get() does for dictionaries:

def listget(list_, index, default=None):
    """Return the item for index if index is in the range of the list_,
    else default. If default is not given, it defaults to None, so that
    this method never raises an IndexError."""
    if index >= len(list_) or index < -len(list_):
        return default
    else:
        return list_[index]

Example usage:

>>> names = ["Mark","Frank","James"]
>>> listget(names, 2)
'James'
>>> listget(names,-3)
'Mark'
>>> listget(names,3) # returns None
>>> listget(names,4,0)
0

So it will always return a value and you get no exceptions.

Answer 7

Here is the logic statement I use to check whether the end of your list has been reached:

arr = [1,3,2,4,5]
#counter for the array
arr_counter = 0

for ele in array:
    # check if end of list has been reached
    if (arr_counter+1) != len(arr):
        #put all your code here
        pass
    # increment the array counter
    arr_counter += 1

Hope this helps ! :)

Answer 8

You're not providing a specific use-case, but generally for a list your would use len to see how many elements are in the list.

if len(a) > 3:
    # Do something

Answer 9

The general way to check if you're currently looking at the element at the end of a list (in any language) is to compare the current index you're looking at with the length of the list minus one (since indexes start at 0).

a[4] isn't really anything, because it doesn't exist - some languages may implement that as being null (or undefined) but many will simply throw an exception if you try to access it instead.

Answer 10

look here: https://www.geeksforgeeks.org/python-how-to-get-the-last-element-of-list/

test_list = [1, 4, 5, 6, 3, 5] 
# printing original list  
print ("The original list is : " + str(test_list)) 
# First naive method 
# using loop method to print last element  
for i in range(0, len(test_list)): 
    if i == (len(test_list)-1): 
        print ("The last element of list using loop : "+  str(test_list[i])) 
# Second naive method         
# using reverse method to print last element 
test_list.reverse() `enter code here`
print("The last element of list using reverse : "+  str(test_list[0]))