Optional chaining in Python

Question

In JavaScript, if I'm not sure whether every element of the chain exists/is not undefined, I can do foo?.bar, and if bar does not exist on foo, the interpreter will silently short circuit it and not throw an error.

Is there anything similar in Python? For now, I've been doing it like this:

if foo and foo.bar and foo.bar.baz:
    # do something

My intuition tells me that this isn't the best way to check whether every element of the chain exists. Is there a more elegant/Pythonic way to do this?

What you want is PEP 505 https://www.python.org/dev/peps/pep-0505 — brandonscript, Apr 21 '21 at 15:17
That's exactly what I'm looking for but the maybe-dot and maybe-subscript operators haven't been added to Python (yet?) — Xeoth, Apr 22 '21 at 16:09
Maybe? I'm not certain because I'm not privileged enough to use Py 3.8 yet, but it appears to have been ratified? — brandonscript, Apr 23 '21 at 00:12
It's not in Python 3.8. PEP 505 is marked as Deferred which means no progress is being made. See discussion at https://discuss.python.org/t/pep-505-status/4612. — John Mellor, Jun 28 '21 at 22:19

score 14 · Answer 1 · edited Oct 09 '20 at 18:33

14

Most pythonic way is:

try:
    # do something
    ...
except (NameError, AttributeError) as e:
    # do something else
    ...

edited Oct 09 '20 at 18:33

theEpsilon

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answered Oct 09 '20 at 18:25

soumya-kole

641
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score 13 · Answer 2 · answered Oct 09 '20 at 18:11

13

You can use getattr:

getattr(getattr(foo, 'bar', None), 'baz', None)

answered Oct 09 '20 at 18:11

theEpsilon

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2

This is't more readable than what OP has `if foo and foo.bar and foo.bar.baz:` – Kuldeep Jain Mar 30 '22 at 23:51

score 12 · Answer 3 · answered Mar 04 '21 at 11:38

12

If it's a dictionary you can use get(keyname, value)

{'foo': {'bar': 'baz'}}.get('foo', {}).get('bar')

answered Mar 04 '21 at 11:38

Aliaksandr Sushkevich

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Sérgio Rafael Siqueira · Answer 4 · 2022-01-20T17:44:45.960

5

You can use the Glom.

from glom import glom

target = {'a': {'b': {'c': 'd'}}}
glom(target, 'a.b.c', default=None)  # returns 'd'

edited Jan 20 '22 at 17:44

answered Jun 02 '21 at 16:31

Sérgio Rafael Siqueira

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This is not optional chaining, this only works when the path is present. With optional chaining you want it to simply return None instead of failing. – svenema Jan 18 '22 at 14:04
You can set a default value too. `glom(target, 'a.b.d', default=None)` – Sérgio Rafael Siqueira Jan 19 '22 at 15:11
1

Thanks for clarifying, this then makes it a good and correct answer. I cannot upvote until you update your answer, by the way. – svenema Jan 19 '22 at 16:47

Hielke Walinga · Answer 5 · 2021-11-09T16:01:29.113

Combining a few things I see here.

from functools import reduce


def optional_chain(obj, keys):
    try:
        reduce(getattr, keys.split('.'), root)
    except AttributeError:
        return None

optional_chain(foo, 'bar.baz')

Or instead extend getattr so you can also use it as a drop-in replacement for getattr

from functools import reduce


def rgetattr(obj, attr, *args):
    def _getattr(obj, attr):
        return getattr(obj, attr, *args)
    return reduce(_getattr, attr.split('.'), obj)

With rgetattr it can still raise an AttributeError if the path does not exist, and you can specify your own default instead of None.

Bernard Swart · Answer 6 · 2021-09-22T09:58:26.657

Combining some of the other answers into a function gives us something that's easily readable and something that can be used with objects and dictionaries.

def optional_chain(root, *keys):
    result = root
    for k in keys:
        if isinstance(result, dict):
            result = result.get(k, None)
        else:
            result = getattr(result, k, None)
        if result is None:
            break
    return result

Using this function you'd just add the keys/attributes after the first argument.

obj = {'a': {'b': {'c': {'d': 1}}}}
print(optional_chain(obj, 'a', 'b'), optional_chain(obj, 'a', 'z'))

Gives us:

{'c': {'d': 1}} None

score 4 · Answer 7 · answered Nov 18 '21 at 21:23

4

I like modern languages like Kotlin which allow this:

foo?.bar?.baz

Recently I had fun trying to implement something similar in python: https://gist.github.com/karbachinsky/cc5164b77b09170edce7e67e57f1636c

Unfortunately, the question mark is not a valid symbol in attribute names in python, thus I used a similar mark from Unicode :)

answered Nov 18 '21 at 21:23

Igor Karbachinsky

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It should also be noted that this only works in python 3 because python 2 variables names do not allow unicode – smac89 Dec 08 '21 at 17:00

Optional chaining in Python

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