python - 字典列表 : aggregate value by grouping by inner dictionary key

Question

我有这个签名:

def aggregate_by_player_id(input, playerid, fields):

我所说的“字段”是指在“输入”中按“playerID”进行分组的字段。

我这样调用该函数:

aggregate_by_player_id(input, 'player', ['stat1','stat3'])

输入如下所示:

[{'player': '1', 'stat1': '3', 'stat2': '4', 'stat3': '5'},
{'player': '1', 'stat1': '1', 'stat2': '4', 'stat3': '1'},
{'player': '2', 'stat1': '1', 'stat2': '2', 'stat3': '3'},
{'player': '2', 'stat1': '1', 'stat2': '2', 'stat3': '1'},
{'player': '3', 'stat1': '4', 'stat2': '1', 'stat3': '6'}]

我的输出结构是:

nested_dic = {value_of_playerid1: {'playerid': value_of_playerid1, 'stat1': value_of_stat1, 'stat2': value_of_stat2}, 
              value_of_playerid2: {'playerid': value_of_playerid2, 'stat2': value_of_stat2, 'stat2': value_of_stat2}, 
              value_of_playerid3: {'playerid': value_of_playerid3, 'stat3': value_of_stat3, 'stat3': value_of_stat3}}

因此输出应如下所示:

{'1': {'player': '1', 'stat1': 4, 'stat3': 6}, 
 '2': {'player': '2', 'stat1': 2, 'stat3': 4}, 
 '3': {'player': '3', 'stat1': 4, 'stat3': 6}} 

Answer 1

我们可以使用itertools.groupby以便根据 playerid 进行分组，然后对各个字段的值求和。

from itertools import groupby
from operator import itemgetter

def aggregate_by_player_id(input_, playerid, fields):
    player = itemgetter(playerid)
    output = {}

    for k, v in groupby(input_, key=player):
        data = list(v)
        stats = {playerid: k}

        for field in fields:
            stats[field] = sum(int(d.get(field, 0)) for d in data)
        output[k] = stats
    return output

data.sort(key=player) # data must be pre-sorted on grouping key
results = aggregate_by_player_id(data, 'player', ['stat1', 'stat3'])

{'1': {'player': '1', 'stat1': 4, 'stat3': 6},
 '2': {'player': '2', 'stat1': 2, 'stat3': 4},
 '3': {'player': '3', 'stat1': 4, 'stat3': 6}}