keep highest value of duplicate keys in dicts

问题

For school i am writing a small program for a rankinglist for a game. I am using dicts for this, with the name of the player as keyname, and the score as keyvalue. there will be 10 games, and each game will have an automatic ranking system which i print to file. ive already managed to code the ranking system, but now im facing a bigger challange which i cannot solve:

I have to make an overall ranking, which means someplayername can be in several contests with several scores, but i need to only keep the highest score of a duplicate.

In short: I need some help with keeping the duplicate key with the highest value:

like this:

dict1 = {"a": 6, "b": 4, "c": 2, "g": 1}
dict2 = {"a": 3, "f": 4, "g": 5, "d": 2}
dictcombined = {'a': 6, 'b': 4, 'c': 2, 'g': 5, 'f': 4, 'd': 2}

the normal merge option just takes the second dict and thus that value.

thnx in advance

回答1:

You need to have a function that will keep track of the highest scores for each player. It will add a player to the total if not already there, otherwise adding it if it's higher. Something like this:

def addScores(scores, total):
    for player in scores:
        if player not in total or total[player] < scores[player]:
            total[player] = scores[player]

回答2:

This works like a charm:

dict1 = {"a": 6, "z": 4, "g": 1, "hh": 50, "ggg": 1}

dict2 = {"a": 3, "g": 5, "d": 2, "hh": 50}

for key in dict1:

    if key not in dict2 or dict1[key] > dict2[key]:
        dict2[key] = dict1[key]

print (dict1)
print (dict2)

dict3 = {**dict1, **dict2}

print (dict3)

Now I can compare dict3 with other dicts and so on.

回答3:

Here is my generalized solution to your question. It's a function that can combine an arbitrary number of dictionaries and has an option for other comparison functions should you want to say, keep track of the minimum values instead.

import collections

def combine_dicts(func, *dicts):
    default = collections.defaultdict(set)
    for d in dicts:
        for k, v in d.items():
            default[k].add(v)
    return {k: func(v) for k, v in default.items()}

It uses a defaultdict with set as its default_factory to keep track of repetitions of keys with different values. Then it returns a dictionary comprehension to filter out the desired values.

dict1 = {"a": 6, "b": 4, "c": 2, "g": 1}
dict2 = {"a": 3, "d": 2, "f": 4, "g": 5}
dict_comb = combine_dicts(max, dict1, dict2)
print(dict_comb)  # -> {'a': 6, 'b': 4, 'c': 2, 'd': 2, 'f': 4, 'g': 5}

回答4:

Here's a variation on Matt Eding's answer that compares each value individually instead of creating sets of values. As a plus, it doesn't need any imports.

def combine_dicts(func, *dicts):
    d0 = {}
    for d in dicts:
        for k, v in d.items():
            if k not in d0:
                d0[k] = v
            else:
                d0[k] = func(v, d0[k])
    return d0

来源：https://stackoverflow.com/questions/47082410/keep-highest-value-of-duplicate-keys-in-dicts