Given a list of dictionaries, how can I eliminate duplicates of one key, and sort by another

Question

I\'m working with a list of dict objects that looks like this (the order of the objects differs):

[
    {\'name\': \'Foo\', \'score         


        

Answer 1

One way to do that is:

data = collections.defaultdict(list)
for i in my_list:
    data[i['name']].append(i['score'])
output = [{'name': i, 'score': max(j)} for i,j in data.items()]

so output will be:

[{'score': 2, 'name': 'Baz'},
 {'score': 3, 'name': 'Foo'},
 {'score': 3, 'name': 'Bar'}]

Answer 2

Just for fun, here is a purely functional approach:

>>> map(dict, dict(sorted(map(sorted, map(dict.items, s)))).items())
[{'score': 3, 'name': 'Bar'}, {'score': 2, 'name': 'Baz'}, {'score': 3, 'name': 'Foo'}]

Answer 3

This is the simplest way I can think of:

names = set(d['name'] for d in my_dicts)
new_dicts = []
for name in names:
    d = dict(name=name)
    d['score'] = max(d['score'] for d in my_dicts if d['name']==name)
    new_dicts.append(d)

#new_dicts
[{'score': 2, 'name': 'Baz'},
 {'score': 3, 'name': 'Foo'},
 {'score': 3, 'name': 'Bar'}]

Personally, I prefer not to import modules when the problem is too small.

Answer 4

There's no need for defaultdicts or sets here. You can just use dirt simple dicts and lists.

Summarize the best running score in a dictionary and convert the result back into a list:

>>> s = [
    {'name': 'Foo', 'score': 1},
    {'name': 'Bar', 'score': 2},
    {'name': 'Foo', 'score': 3},
    {'name': 'Bar', 'score': 3},
    {'name': 'Foo', 'score': 2},
    {'name': 'Baz', 'score': 2},
    {'name': 'Baz', 'score': 1},
    {'name': 'Bar', 'score': 1}
]
>>> d = {}
>>> for entry in s:
        name, score = entry['name'], entry['score']
        d[name] = max(d.get(name, 0), score)

>>> [{'name': name, 'score': score} for name, score in d.items()]
[{'score': 2, 'name': 'Baz'}, {'score': 3, 'name': 'Foo'}, {'score': 3, 'name': 'Bar'}]

Answer 5

I think I can come up with an one-liner here:

result = dict((x['name'],x) for x in sorted(data,key=lambda x: x['score'])).values()

Answer 6

In case you haven't heard of group by, this is nice use of it:

from itertools import groupby

data=[
    {'name': 'Foo', 'score': 1},
    {'name': 'Bar', 'score': 2},
    {'name': 'Foo', 'score': 3},
    {'name': 'Bar', 'score': 3},
    {'name': 'Foo', 'score': 2},
    {'name': 'Baz', 'score': 2},
    {'name': 'Baz', 'score': 1},
    {'name': 'Bar', 'score': 1}
]

keyfunc=lambda d:d['name']
data.sort(key=keyfunc)

ans=[]
for k, g in groupby(data, keyfunc):
    ans.append({k:max((d['score'] for d in g))})
print ans

>>>
[{'Bar': 3}, {'Baz': 2}, {'Foo': 3}]