Can you embed a try in a Python list comprehension?

Question

Is there anyway to do the equivalent of this?

my_list = [try: my_dict["some_key"] except KeyError: 0 for my_dict in my_list]

Since dictionaries throw KeyErrors I want to catch the error if the element in the list does not have a "some_key" property. I know I could create a defaultdict by importing collections and sidestepping the exception, but I want to know if this is possible with out of the box dictionaries.

Answer 1

No, you can't. You can put only for loops, if's and else's.

What you can do, though, is use .get(), which never throws a KeyError:

my_list = [my_dict.get("some_key", 0) for my_dict in my_list]

The second argument is the default value in case the key does not exist. If you don't specify a default, the default default is None.

Answer 2

No, you can not do this directly in a list comprehension. You must factor the try/except logic out into a separate function, and then call the function.

There is an easy alternative for the use case shown in your question, though, using dict.get:

my_list = [my_dict.get('some_key', 0) for my_dict in my_list]

Answer 3

As other answers mentioned you can not use an try-except inside a list comprehension. But as a tricky approach you can use collections.defaultdict() and override the __missing__ attribute in order to catch the exceptions. Here is an example:

from collections import defaultdict


class Mydefaultdict(defaultdict):
    def __init__(self, *args, **kwargs):
        super(Mydefaultdict, self).__init__(*args, **kwargs)

    def __getitem__(self, key):
        try:
            return defaultdict.__getitem__(self, key)
        except KeyError:
            return self.__missing__(key)

    def __missing__(self, key):
        # Instead of printing you can catch your exceptions in any way you like
        print("{} doesn't exist".format(key))
        if self.default_factory:
            return self.default_factory()  # You can rturn whatever you want here

Demo:

d = Mydefaultdict(None, {4: 'b', 1: 'a'})
print([d[i] for i in [1, 2]])

2 doesn't exist
['a', None]