Sorting list of dicts by value of a key (or default-value, if key is missing)

Sorting list of dicts by value of a key (or default-value, if key is missing)

Imagine that you have to sort a list of dicts, by the value of a particular key. Note that the key might be missing from some of the dicts, in which case you default to the value of that key to being 0.

Imagine that you have to sort a list of dicts, by the value of a particular key. Note that the key might be missing from some of the dicts, in which case you default to the value of that key to being 0.

sample input

  • input = [{'a': 1, 'b': 2}, {'a': 10, 'b': 3}, {'b': 5}]

sample output (sorted by value of key 'a')

  • [{'b': 5}, {'a': 1, 'b': 2}, {'a': 10, 'b': 3}]

note that {'b': 5} is first in the sort-order because it has the lowest value for 'a' (0)

I would've used input.sort(key=operator.itemgetter('a')), if all the dicts were guaranteed to have the key 'a'. Or I could convert the input dicts to collections.defaultdict and then sort.

Is there a way to do this in-place without having to creating new dicts or updating the existing dicts? Can operator.itemgetter handle missing keys?

python

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