如何修改嵌套Json的键

Question

我正在尝试更新如下所示的 JSON 对象的键：

results = 
 {  
   'Game':12345,
   'stats':[  
      {  
         'detail':[  
            {  
               'goals':4,
               'refs':{  
                  'number':0

我目前正在手动更新每个键，如下所示

##update Game to newValue
results['newValue'] = results['Game']
    del results['Game']

## update nested key "goals" to "goals_against"
results['stats'][0]['detail'][0]['goals_against'] = results['stats'][0]['detail'][0]['goals']
    del results['stats'][0]['detail'][0]['goals']

必须有更好的方法，因为我发现自己必须更新结果中的多个键。例如，我还想将“number”键更新为“assis_ref”。

如果 json 文件是“简单”的，我知道如何更新密钥：即如果我可以这样做：

result['stats']['details']['refs']

然而，'stats' 和 'details' 需要 [0] 在它旁边，我假设它是我接下来要尝试的元素的索引。

Answer 1

我编写了一个递归函数来处理 json 对象中的转换键。对于深度嵌套的 json 尤其有用！

def walk_json(obj, key_transform):
    """
    Recurse over a json object modifying the keys according to the `key_transform` function.
    Returns a new json object with the modified keys.
    """
    assert isinstance(obj, dict), "walk_json expects obj to be of type dict"

    def _walk_json(obj, new):

        if isinstance(obj, dict):

            if isinstance(new, dict):
                for key, value in obj.items():

                    new_key = key_transform(key)

                    if isinstance(value, dict):
                        new[new_key] = {}
                        _walk_json(value, new=new[new_key])

                    elif isinstance(value, list):
                        new[new_key] = []
                        for item in value:
                            _walk_json(item, new=new[new_key])

                    else:  # take value as is
                        new[new_key] = value

            elif isinstance(new, list):
                new.append(_walk_json(obj, new={}))

        else:  # take object as is
            new.append(obj)

        return new

    return _walk_json(obj, new={})

这是一个过于简单的 json 对象的用法：

def my_key_transform(x):
    return x.upper()

my_obj = {"a": 1, "b": 2, "c": 3}

result = walk_json(my_obj, key_transform=my_key_transform)

result
{"A": 1, "B": 2, "C": 3}

Answer 2

导航和修改从 JSON 对象派生的深层嵌套对象可能会很痛苦。在Functions that help to understand json(dict) structure 中，我发布了允许您导航此类对象的代码。请阅读该答案中的解释。在这个答案中，我将展示如何使用该代码来修改此类对象中的字典键。

简而言之，find_key 是一个递归生成器，它将找到具有给定名称的所有键。您可以使用next 函数来获取第一个（或唯一一个）匹配的名称。如果您需要使用多个具有相同名称的键，或者在 for 循环中调用 find_key。

find_key 产生的每个值都是dict 键的列表，列表索引需要到达所需的键。

from json import dumps

def find_key(obj, key):
    if isinstance(obj, dict):
        yield from iter_dict(obj, key, [])
    elif isinstance(obj, list):
        yield from iter_list(obj, key, [])

def iter_dict(d, key, indices):
    for k, v in d.items():
        if k == key:
            yield indices + [k], v
        if isinstance(v, dict):
            yield from iter_dict(v, key, indices + [k])
        elif isinstance(v, list):
            yield from iter_list(v, key, indices + [k])

def iter_list(seq, key, indices):
    for k, v in enumerate(seq):
        if isinstance(v, dict):
            yield from iter_dict(v, key, indices + [k])
        elif isinstance(v, list):
            yield from iter_list(v, key, indices + [k])


results = {
    "Game": 12345,
    "stats": [
        {
            "detail": [
                {
                    "goals": 4,
                    "refs": {
                        "number": 0
                    }
                }
            ]
        }
    ]
}

# Change oldkey to newkey      
oldkey, newkey = 'goals', 'goals_against'
# Find the first occurrence of the oldkey
seq, val = next(find_key(results, oldkey))
print('seq:', seq, 'val:', val)

# Get the object that contains the oldkey 
obj = results
for k in seq[:-1]:
    obj = obj[k]

# Change the key
obj[newkey] = obj.pop(oldkey)

print(dumps(results, indent=4))

输出

seq: ['stats', 0, 'detail', 0, 'goals'] val: 4
{
    "Game": 12345,
    "stats": [
        {
            "detail": [
                {
                    "refs": {
                        "number": 0
                    },
                    "goals_against": 4
                }
            ]
        }
    ]
}