Recipe4.3.Returning an Element of a List If It Exists

Recipe 4.3. Returning an Element of a List If It Exists

Credit: Nestor Nissen, A. Bass

Problem

You have a list L and an index i, and you want to get L[i] when i is a valid index into L; otherwise, you want to get a default value v. If L were a dictionary, you'd use L.get(i, v), but lists don't have a get method.

Solution

Clearly, we need to code a function, and, in this case, the simplest and most direct approach is the best one:

def list_get(L, i, v=None):     if -len(L) <= i < len(L): return L[i]     else: return v

Discussion

The function in this recipe just checks whether i is a valid index by applying Python's indexing rule: valid indices are negative ones down to -len(L) inclusive, and non-negative ones up to len(L) exclusive. If almost all calls to list_get pass a valid index value for i, you might prefer an alternative approach:

def list_get_egfp(L, i, v=None):     try: return L[i]     except IndexError: return v

However, unless a vast majority of the calls pass a valid index, this alternative (as some time-measurements show) can be up to four times slower than the list_get function shown in the solution. Therefore, this "easier to get forgiveness than permission" (EGFP) approach, although it is often preferable in Python, cannot be recommended for this specific case.

I've also tried quite a few fancy, intricate and obscure approaches, but, besides being hard to explain and to understand, they all end up slower than the plain, simple function list_get. General principle: when you write Python code, prefer clarity and readability to compactness and tersenesschoose simplicity over subtlety. You often will be rewarded with code that runs faster, and invariably, you will end up with code that is less prone to bugs and is easier to maintain, which is far more important than minor speed differences in 99.9% of the cases you encounter in the real world.

Recipe4.3.Returning an Element of a List If It Exists