CODEWITHSUNDEEP
pythondictionary
blackmamba
blackmamba

Reputation: 2002

Sort list of dictionaries based on keys

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.

Upvotes: 3

Views: 13233

Answers (3)

Dr K
Dr K

Reputation: 426

How about something like this

def sort_key(dict_item, sort_list):
    key_idx = [sort_list.index(key) for key in dict_item.iterkeys() if key in sort_list]
    if not key_idx:
        return len(sort_list)
    return min(key_idx)

dict_list.sort(key=lambda x: sort_key(x, sort_list))

If the a given dictionary in the list contains more than one of the keys in the sorting list, it will use the one with the lowest index. If none of the keys are present in the sorting list, the dictionary is sent to the end of the list.

Dictionaries that contain the same "best" key (i.e. lowest index) are considered equal in terms of order. If this is a problem, it wouldn't be too hard to have the sort_key function consider all the keys rather than just the best. To do that, simply return the whole key_idx instead of min(key_idx) and instead of len(sort_list) return [len(sort_list)]

Upvotes: 1

tobias_k
tobias_k

Reputation: 82949

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.

Upvotes: 4

Peter DeGlopper
Peter DeGlopper

Reputation: 37364

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.

Upvotes: 1

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