组合 Python 中两个字典列表中的值时，如何避免嵌套 for 循环？

Question

假设我有两个列表：

list1 = [{"sport": 'Soccer',
          "leagues": [{"id": 1001,
                       "name": "League1",
                       "events": [{"id": 100,
                                   "home": "team1",
                                   "away": "team2"},
                                  {"id": 101,
                                   "home": "team3",
                                   "away": "team4"}]}]},
         {"sport": 'Basketball',
          "leagues": [{"id": 1002,
                       "name": "League2",
                       "events": [{"id": 200,
                                  "home": "team5",
                                  "away": "team6"},
                                 {"id": 201,
                                  "home": "team7",
                                  "away": "team8"}]},
                      {"id": 1003,
                       "name": "League3",
                       "events": [{"id": 300,
                                   "home": "team9",
                                   "away": "team10"},
                                  {"id": 301,
                                   "home": "team11",
                                   "away": "team12"}]}],
      }
     ]

list2 = [{"sport": 'Soccer',
          "leagues": [{"id": 1001,
                       "events": [{"id": 100,
                                   "odds": {"home": 1.862, "away": 1.847}},
                                  {"id": 101,
                                   "odds": {"home": 1.70, "away": 2.10}}]}]},
         {"sport": 'Basketball',
          "leagues": [{"id": 1002,
                       "events": [{"id": 200,
                                   "odds": {"home": 1.952, "away": 1.952}},
                                  {"id": 201,
                                   "odds": {"home": 1.90, "away": 2.05}}]},
                      {"id": 1003,
                       "events": [{"id": 300,
                                   "odds": {"home": 1.5, "away": 2.7}},
                                  {"id": 301,
                                   "odds": {"home": 1.75, "away": 2.09}}]}]}]

我想将 list1 中的事件与 list2 中它们各自的“赔率”结合起来。 实际上，这两个列表中还有更多元素，为了清楚起见，示例被简化了。 我目前的（丑陋的）解决方案：

for sport_list1 in list1:
    for sport_list2 in list2:
        if sport_list1['sport'] == sport_list2['sport']:
            for league_list1 in sport_list1['leagues']:
                for league_list2 in sport_list2['leagues']:
                    if league_list1['id'] == league_list2['id']:
                        for event_list1 in league_list1['events']:
                            for event_list2 in league_list2['events']:
                                if event_list1['id'] == event_list2['id']:
                                    print(sport_list1['sport'], league_list1['name'], event_list1['home'], event_list1['away'], event_list2['odds'])
                                    break

所需的 output：

Soccer League1 team1 team2 {'home': 1.862, 'away': 1.847}
Soccer League1 team3 team4 {'home': 1.7, 'away': 2.1}
Basketball League2 team5 team6 {'home': 1.952, 'away': 1.952}
Basketball League2 team7 team8 {'home': 1.9, 'away': 2.05}
Basketball League3 team9 team10 {'home': 1.5, 'away': 2.7}
Basketball League3 team11 team12 {'home': 1.75, 'away': 2.09}

有什么方法可以让这个更清洁和/或更高效？

Answer 1

@Chronial 答案是完美的，但这是您可能会觉得有趣的另一种方法：

def convert_to_dicts(x):
    if type(x) == list:
        id_field = {"sport", "id"}.intersection(set(x[0].keys())).pop()
        return {y.pop(id_field): convert_to_dicts(y) for y in x}
    elif type(x) == dict:
        return{z: convert_to_dicts(y) for z, y in x.items()}
    return x

def recursive_dict_merge(x, y):
    new_dict = {}
    for key in set(x.keys()).union(set(y.keys())):
        x_val = x.get(key, None)
        y_val = y.get(key, None)
        if type(x_val) == dict and type(y_val) == dict:
            new_dict[key] = recursive_dict_merge(x_val, y_val)
        else:
            new_dict[key] = x_val or y_val
    return new_dict


result = recursive_dict_merge(convert_to_dicts(list1), convert_to_dicts(list2))

我首先将“字典列表的字典列表”转换为嵌套字典。

然后我使用递归来合并这些字典。

我认为这种方法更好，因为您有一个“易于使用”的result字典，然后可以更轻松地执行其他操作，例如您想要的确切print ：

for sport, leagues in result.items():
    for league in leagues["leagues"].values():
        for event in league["events"].values():
            print(sport, league['name'], event['home'], event['away'], event['odds'])

一般来说，对于这样的问题，我发现最好的第一步通常是将输入重塑为更易于管理的东西。 嵌套字典比...的字典列表的字典列表更容易考虑。

Answer 2

可以写一个小帮手function。 使用 dicts 可以让你摆脱O(n²) 。

def zip_by_key(key, list1, list2):
    map1 = {x[key]: x for x in list1}
    map2 = {x[key]: x for x in list2}
    for k in map1.keys() & map2.keys():
        yield (map1[k], map2[k])

那么你的代码将是：

for sport1, sport2 in zip_by_key('sport', list1, list2):
    for league1, league2 in zip_by_key('id', sport1['leagues'], sport2['leagues']):
        for event1, event2 in zip_by_key('id', league1['events'], league2['events']):
            print(sport1['sport'], league1['name'], event1['home'], event1['away'], event2['odds'])