在Python中组合两个字典列表中的值时，如何避免嵌套for循环？

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}}]}]}]

我想将清单1中的事件与清单2中各自的“几率”结合起来。 实际上，这两个列表中还有更多的元素，为了清晰起见，对示例进行了简化。 我当前的（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

期望输出：

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}

---

有什么方法可以使它更干净和/或更高效？

您可以编写一个小的助手函数。使用dicts可以使您从

O（n²）

中解脱出来

那么您的代码将是：

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'])

---

@长期的答案是完美的，但这里有另一种方法，你可能会觉得有趣：

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“dict of list of dict”转换为嵌套字典

然后我使用递归合并这些字典

我认为这种方法更好，因为你有一个“易于使用”的

结果

字典，这样做可以更容易地做其他事情，例如，你想要的确切的

打印

：

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内容。嵌套字典比字典列表更容易思考。。。等等。

但它是丑陋的。这些

list1

和

list2

来自哪里？我怀疑这是一个“XY问题”，您可以使用

zip

将嵌套“减半”：

对于sport_list1，sport_list2 in zip（list1，list2）：

等。它们来自收受赌注者的API。不确定zip是否有帮助，因为不能保证索引匹配。谢谢。行：对于league1，league2在zip_by_key（'id'，sport1['leagues']，sport2['leagues']）中给了我TypeError:字符串索引必须是整数哦，输入错误–返回应该是一个收益。固定的。

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'])