根据单独的值匹配查找字典中的最小值

Question

I have a dict of dicts. I want to find top-level keys based on uniqueness of one value or minimum of another.

Example:

d = {'first': {'key_to_match': 'a', 'key_to_compare': 50}, 
'second': {'key_to_match': 'b', 'key_to_compare': 50}, 
'third': {'key_to_match': 'a', 'key_to_compare': 10}}

I want the keys of each dict with the lowest key_to_compare for each key_to_match. In this case, ['second', 'third'].

I have tried turning them into a list of tuples and comparing that way but that seems needlessly convoluted.

Answer 1

Here is an example.

d = {
    'first': {'key_to_match': 'a', 'key_to_compare': 50}, 
    'second': {'key_to_match': 'b', 'key_to_compare': 50}, 
    'third': {'key_to_match': 'a', 'key_to_compare': 10}
}

# Output format: {'a': {'key': 'first', 'value': 50}}
lowest_keys = {}
for k, v in d.items():
    s_key = v['key_to_match']
    s_value = v['key_to_compare']
    if not lowest_keys.get(s_key) or lowest_keys[s_key]['value'] > s_value:
        lowest_keys[s_key] = {'key': k, 'value': s_value}

print(lowest_keys)
print([v['key'] for v in lowest_keys.values()])

I tried to make it readable. Processed values are saved in lowest_keys dictionary as the loop progresses. It keeps in memory the top level key ('first', 'second' etc.), sub key group ('key_to_match') and the lowest found value so far ('key_to_compare'). If the loop encounters a lower value from the same group, it saves the value and top key in a dict. The last print shows how you can access this data.