通过比较python中的示例JSON来填充丢失的键

Question

我想首先合并两个词典，这意味着第一个字典不会丢失其先前的值，并且只有丢失的键值被添加到第一个值中。

Original =  {key1: { key2:[{ key3: y1 , key5: z1 , key6 [{key7: n1}]}]}}

Example = {key1: { key2:[{ key3: None , key4: None ,key5: None , key6 [{key7: None,key8:None}]}]}}

def update2(final, first):
 for k, v in first.items():
    if isinstance(v, str):
        final[k] = v
    elif k in final:
        if isinstance(final[k], dict):
            final[k].update2(first[k])
        else:
            final[k] = first[k]
    else:
        final[k] = first[k]

    final_json = update2(Example['key1'], Original['key1'])
    print(final_json)

#returns only Original not appended output

    def update3(right,left):
        d = dict()
        for key,val in left.items():
         d[key]=[val]
        for key,val in right.items():
         if key not in d.keys():
          d[key].append(val)
         else:
          d[key]=[val]

    final_json = update3(Example['key1'], Original['key1'])
    print(final_json) #returns None

预期输出：

{key1：{key2：[{key3：y1，key4：none，key5：z1，ke1，key6 [{key7：n1，key8：none}]}}}}}}}}}}}}}}}}}}}}}}

}它必须在多个层面上迭代。

根据JSON Schema的设置默认值

” https://stackoverflow.com/questions/55694862/how-to-to-merge-two-dictionaries基于-n-key-beke-but-include-missing-missing-key-key-key-vorues-in-pyt"> in> wought"> wough>在键上，但在python中包括丢失的密钥值？

合并两个词典并坚持第一个字典的值

我的目标是在示例文件中添加丢失键的默认值。我是初学者到python。

Answer 1

尝试在以下步骤中递归处理。

1. 确定键是否存在，如果不是，则直接分配值，如果是的，请转到下一步
2. 确定值字典的类型

• ：递归呼叫
• 列表：通过内容迭代并为每个项目进行递归呼叫
• ：直接分配如果原始词典不是

这样的，那么

def update(orignal, addition):
    for k, v in addition.items():
        if k not in orignal:
            orignal[k] = v
        else:
            if isinstance(v, dict):
                update(orignal[k], v)
            elif isinstance(v, list):
                for i in range(len(v)):
                    update(orignal[k][i], v[i])
            else:
                if not orignal[k]:
                    orignal[k] = v




Original =  {
    "key1": {
        "key2":[
            { "key3": "y1" ,
              "key5": "z1" ,
              "key6": [
                {"key7": "n1"}
              ]
            }
        ]
    }
}

Example = {
    "key1": {
        "key2":[
            {
                "key3": None ,
                "key4": None ,
                "key5": None ,
                "key6": [
                    {"key7": None,"key8":None}
                ]
            }
        ]
    }
}
print("Original: ", Original)
print("Addition: ", Example)
update(Original, Example)
print("Merge: ", Original)

# Original:  {'key1': {'key2': [{'key3': 'y1', 'key5': 'z1', 'key6': [{'key7': 'n1'}]}]}}
# Addition:  {'key1': {'key2': [{'key3': None, 'key4': None, 'key5': None, 'key6': [{'key7': None, 'key8': None}]}]}}
# Merge:  {'key1': {'key2': [{'key3': 'y1', 'key5': 'z1', 'key6': [{'key7': 'n1', 'key8': None}], 'key4': None}]}}

Answer 2

我以为没有列表，并且不希望将键中的键添加到示例中。我不确定这是您要寻找的答案，但这是一个建议。

def f(base: dict, data: dict):
    keys = base.keys() & data.keys()
    for key in keys:
        if isinstance(base[key], dict):
            f(base[key], data[key])            
        elif base[key] is None:
            base[key] = data[key]

Python

>>> data = {"key1": {"key2": {"key3": "y1", "key5": "z1", "key6": {"key7": "n1"}}}}
>>> base = {"key1": {"key2": {"key3": "a1", "key4": None, "key5": None, "key6": {"key7": None, "key8": None}}}}
>>> f(base, data)
>>> print(base)
{'key1': {'key2': {'key3': 'a1', 'key4': None, 'key5': 'z1', 'key6': {'key7': 'n1', 'key8': None}}}}