How to Get the Nth Element of a Python List or a Default If Not Available

How to get the nth element of a python list or a default if not available

l[index] if index < len(l) else default

To support negative indices we can use:

l[index] if -len(l) <= index < len(l) else default

How do I get the last element of a list?

some_list[-1] is the shortest and most Pythonic.

In fact, you can do much more with this syntax. The some_list[-n] syntax gets the nth-to-last element. So some_list[-1] gets the last element, some_list[-2] gets the second to last, etc, all the way down to some_list[-len(some_list)], which gives you the first element.

You can also set list elements in this way. For instance:

>>> some_list = [1, 2, 3]
>>> some_list[-1] = 5 # Set the last element
>>> some_list[-2] = 3 # Set the second to last element
>>> some_list
[1, 3, 5]

Note that getting a list item by index will raise an IndexError if the expected item doesn't exist. This means that some_list[-1] will raise an exception if some_list is empty, because an empty list can't have a last element.

Accessing nth element of list of lists python - index does not exist error

You could use a conditional list comprehension to check the length of the inner lists and index them only if they are above a certain length:

index = 1
[i[index] if len(i)>index else float('nan') for i in list_of_numbers]
# [10, 8, nan, 0, nan]

Get value at list/array index or None if out of range in Python

Try:

try:
    return the_list[i]
except IndexError:
    return None

Or, one liner:

l[i] if i < len(l) else None

Example:

>>> l=list(range(5))
>>> i=6
>>> print(l[i] if i < len(l) else None)
None
>>> i=2
>>> print(l[i] if i < len(l) else None)
2

Getting a default value on index out of range in Python

In the Python spirit of "ask for forgiveness, not permission", here's one way:

try:
    b = a[4]
except IndexError:
    b = 'sss'

Why does python return the n+1th list element rather than the nth?

Python uses something that is called zero based indexing, which means that the first element in a list is referred to element number 0 and not 1.

A quick way to return list without a specific element in Python

>>> suits = ["h","c", "d", "s"]
>>> noclubs = list(suits)
>>> noclubs.remove("c")
>>> noclubs
['h', 'd', 's']

If you don't need a seperate noclubs

>>> suits = ["h","c", "d", "s"]
>>> suits.remove("c")

How to get the nth element of a python list or a default if not available

l[index] if index < len(l) else default

To support negative indices we can use:

l[index] if -len(l) <= index < len(l) else default

Why doesn't list have safe get method like dictionary?

Ultimately it probably doesn't have a safe .get method because a dict is an associative collection (values are associated with names) where it is inefficient to check if a key is present (and return its value) without throwing an exception, while it is super trivial to avoid exceptions accessing list elements (as the len method is very fast). The .get method allows you to query the value associated with a name, not directly access the 37th item in the dictionary (which would be more like what you're asking of your list).

Of course, you can easily implement this yourself:

def safe_list_get (l, idx, default):
  try:
    return l[idx]
  except IndexError:
    return default

You could even monkeypatch it onto the __builtins__.list constructor in __main__, but that would be a less pervasive change since most code doesn't use it. If you just wanted to use this with lists created by your own code you could simply subclass list and add the get method.

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