如何在 Python 中规范化包含列表（应保存为列表）的 json 文件熊猫？

Question

我正在尝试使用 json_normalize 函数将 json 文件转换为数据帧。

源 JSON

• json 是一个字典列表，看起来像这样：

  {
        "sport_key": "basketball_ncaab",
        "sport_nice": "NCAAB",
        "teams": [
            "Bryant Bulldogs",
            "Wagner Seahawks"
        ],
        "commence_time": 1608152400,
        "home_team": "Bryant Bulldogs",
        "sites": [
            {
                "site_key": "marathonbet",
                "site_nice": "Marathon Bet",
                "last_update": 1608156452,
                "odds": {
                    "h2h": [
                        1.28,
                        3.54
                    ]
                }
            },
            {
                "site_key": "sport888",
                "site_nice": "888sport",
                "last_update": 1608156452,
                "odds": {
                    "h2h": [
                        1.13,
                        5.8
                    ]
                }
            },
            {
                "site_key": "unibet",
                "site_nice": "Unibet",
                "last_update": 1608156434,
                "odds": {
                    "h2h": [
                        1.13,
                        5.8
                    ]
                }
            }
        ],
        "sites_count": 3
    }
  

问题是未来的列之一包含一个列表（应该是这种情况），但是在 json_normalize 函数的元部分中包含此列会引发以下错误：

ValueError: operands could not be broadcast together with shape (22,) (11,)

当我尝试在以下代码的列表中添加“团队”时出现错误：

pd.json_normalize(data, 'sites', ['sport_key', 'sport_nice', 'home_team', 'teams'])

Answer 1

假设data 是一个字典列表，您仍然可以使用json_normalize，但您必须为data 中的每个对应字典分别分配teams 列：

def normalize(d):
    return pd.json_normalize(d, 'sites', ['sport_key', 'sport_nice', 'home_team'])\
           .assign(teams=[d['teams']]*len(d['sites']))


df = pd.concat([normalize(d) for d in data], ignore_index=True)

你也可以试试：

data = [{**d, 'teams': ','.join(d['teams'])} for d in data]
df = pd.json_normalize(data, 'sites', ['sport_key', 'sport_nice', 'home_team', 'teams'])
df['teams'] = df['teams'].str.split(',')

结果：

      site_key     site_nice  last_update      odds.h2h         sport_key sport_nice        home_team                               teams
0  marathonbet  Marathon Bet   1608156452  [1.28, 3.54]  basketball_ncaab      NCAAB  Bryant Bulldogs  [Bryant Bulldogs, Wagner Seahawks]
1     sport888      888sport   1608156452   [1.13, 5.8]  basketball_ncaab      NCAAB  Bryant Bulldogs  [Bryant Bulldogs, Wagner Seahawks]
2       unibet        Unibet   1608156434   [1.13, 5.8]  basketball_ncaab      NCAAB  Bryant Bulldogs  [Bryant Bulldogs, Wagner Seahawks]