Optional Chaining in Python

JavaScript’s addition of the optional chaining (?.) has been useful since its beginning. The optional chaining helps us to make our codes more concise by obviating the needs for membership checks. For example,

let phone = { "name": "iPhone 13 Pro", "dimension": 15, "screen_size": { "width": 390, "height": 450 } };
console.log(phone.name); // "iPhone 13 Pro"
console.log(phone.screen.width); // error! because you are getting the "width" of undefined
console.log(phone.screen?.width); // undefined

This further helps us when we need a default value for an undefined attribute. For instance,

let phone_width = phone.screen?.width || 350;
console.log(phone_width); // 350
phone_width = phone.screen_size.width || 350;
console.log(phone_width); // 390

However, this is so difficult in Python! To do the same thing in Python, you should have multiple lines (particularly Python requires “indent” to be perefect)

phone = { "name": "iPhone 13 Pro", "dimension": 15, "screen_size": { "width": 390, "height": 450 } }
if "screen_size" in phone:
    if "width" in phone["screen_size"]:
        phone_width = phone["screen_size"]["width"]

One possible workaround is using a recursive function.

phone_width = getDictValue(phone, "screen_size.width", 350)

where the first argument is the dict variable, the second is a key chain, and the third is an alternative value to provide in case the intended value is unavailable.

To implement, we could do the following:

def getDictValue(d, k, alt=None):
    # @d: a dictionary
    # @k: a key chain (str ("a.b.c") or list ["a", "b", "c"])
    # @alt: an alternative value to return when the key is not in the dict
    #      (default: None)

    # if the key chain @k is str, convert to a list
    if type(k) == str:
        k = k.split(".")
    # if the first key in @k is not in the dict @d, return @alt
    if d.get(k[0]) is None:
        return alt
    # else if the key chain @k's length is 1, return its value
    # implicitly, it assumes that the value is found
    elif len(k) == 1:
        return d[k[0]]
    # else (recursion)
    else:
        return getDictValue(d[k[0]], k[1:], alt)

The first if statement is to normalize the key chain to a list. The first two conditions in the other if-else statement are the stop conditions for the recursion, and the rest does the recursion.