如何使用Python逐步移除JSON对象（字典）中的属性

首先你提供的这段JSON存在语法错误，正确的JSON结构应该用冒号分隔键值，且引号需配对，修正后的结构如下：

{
  "name": "John",
  "address": "123 bb",
  "age": "40",
  "children": [
    {
      "name": "smith",
      "address": "458 ff",
      "age": "25",
      "grand_children": [
        {"name": "smith", "address": "458 ff", "age": "25"},
        {"name": "smith", "address": "458 ff", "age": "25"}
      ]
    },
    {
      "name": "smith",
      "address": "458 ff",
      "age": "25",
      "grand_children": [
        {"name": "smith", "address": "458 ff", "age": "25"},
        {"name": "smith", "address": "458 ff", "age": "25"}
      ]
    }
  ]
}

针对这种多层嵌套的字典/列表结构，要批量移除指定属性（比如name），最简便的方式是写一个递归函数，自动遍历所有层级的结构，遇到字典就删除目标键，遇到列表就递归处理每个元素。

示例代码（Python） #

直接修改原数据

def remove_key(obj, key):
    if isinstance(obj, dict):
        # 删除当前字典的目标键，键不存在时不报错
        obj.pop(key, None)
        # 递归处理字典内的所有值
        for value in obj.values():
            remove_key(value, key)
    elif isinstance(obj, list):
        # 递归处理列表中的每个元素
        for item in obj:
            remove_key(item, key)
    return obj

# 假设data是修正后的字典
data = {
  "name": "John",
  "address": "123 bb",
  "age": "40",
  "children": [
    {
      "name": "smith",
      "address": "458 ff",
      "age": "25",
      "grand_children": [
        {"name": "smith", "address": "458 ff", "age": "25"},
        {"name": "smith", "address": "458 ff", "age": "25"}
      ]
    },
    {
      "name": "smith",
      "address": "458 ff",
      "age": "25",
      "grand_children": [
        {"name": "smith", "address": "458 ff", "age": "25"},
        {"name": "smith", "address": "458 ff", "age": "25"}
      ]
    }
  ]
}

# 移除所有层级的name属性
result = remove_key(data, "name")
print(result)

不修改原数据（返回新对象）

如果不想改动原始数据，可以在函数中创建副本处理：

def remove_key(obj, key):
    if isinstance(obj, dict):
        new_obj = obj.copy()
        new_obj.pop(key, None)
        # 递归处理副本中的每个值
        for k, v in new_obj.items():
            new_obj[k] = remove_key(v, key)
        return new_obj
    elif isinstance(obj, list):
        # 递归处理列表中的每个元素并生成新列表
        return [remove_key(item, key) for item in obj]
    else:
        return obj

# 调用后原data不受影响
result = remove_key(data, "name")

这种递归方法通用性极强，不管嵌套多少层（比如children、grand_children层级），都能一次性清理所有指定属性，无需手动逐层遍历处理。

内容的提问来源于stack exchange，提问作者Budd