Sort list of dictionaries based on keys

Question

I want to sort a list of dictionaries based on the presence of keys. Let's say I have a list of keys [key2,key3,key1], I need to order the list in such a way the dictionary with key2 should come first, key3 should come second and with key1 last.

I saw this answer (Sort python list of dictionaries by key if key exists) but it refers to only one key

The sorting is not based on value of the 'key'. It depends on the presence of the key and that too with a predefined list of keys.

Answer 1

Just use sorted using a list like [key1 in dict, key2 in dict, ...] as the key to sort by. Remember to reverse the result, since True (i.e. key is in dict) is sorted after False.

>>> dicts = [{1:2, 3:4}, {3:4}, {5:6, 7:8}]
>>> keys = [5, 3, 1]
>>> sorted(dicts, key=lambda d: [k in d for k in keys], reverse=True)
[{5: 6, 7: 8}, {1: 2, 3: 4}, {3: 4}]

This is using all the keys to break ties, i.e. in above example, there are two dicts that have the key 3, but one also has the key 1, so this one is sorted second.

Answer 2

I'd do this with:

sorted_list = sorted(dict_list, key = lambda d: next((i for (i, k) in enumerate(key_list) if k in d), len(key_list) + 1))

That uses a generator expression to find the index in the key list of the first key that's in each dictionary, then use that value as the sort key, with dicts that contain none of the keys getting len(key_list) + 1 as their sort key so they get sorted to the end.