Question

我有这个片段

dict_a = {'key_1_lv_1': {'key_1': 'value_1'},
      'key_2_lv_1': {'key_1_lv_2': 'value_1', 'key_2_lv_2': 'value_2', 'key_3_lv_2': 'value_3'},
      }

dict_b = {'key_1_lv_1': {'key_1': 'value_1'},
      'key_2_lv_1': {'key_1_lv_2': 'new_value'},
      }

当我执行这一行时：

dict_a.update(dict_b)

并打印dict_a['key_2_lv_1']，它给了我：

{'key_1_lv_2': 'new_value'}, only. The other 2 key-value pairs are missing,, but i need them.

如果字典没有嵌套，我会得到其他 2 个键值对。那么是否有任何巧妙的解决方案来获取嵌套字典中缺失的键值？

Answer 1

代码：

dict_a = {'key_1_lv_1': {'key_1': 'value_1'},
      'key_2_lv_1': {'key_1_lv_2': 'value_1', 'key_2_lv_2': 'value_2', 'key_3_lv_2': 'value_3'},
      }
dict_b = {'key_1_lv_1': {'key_1': 'value_1'},
      'key_2_lv_1': {'key_1_lv_2': 'new_value'},
      }
for k,v in dict_b.items():
    dict_a[k] = {**dict_a.get(k,{}),**v}
print(dict_a)

结果：

{'key_1_lv_1': {'key_1': 'value_1'}, 'key_2_lv_1': {'key_1_lv_2': 'new_value', 'key_2_lv_2': 'value_2', 'key_3_lv_2': 'value_3'}}

Answer 2

您可以递归更新：


def recursive_update(d1, d2):
    for key in d2:
        # Check if the key exists on both dicts and merge them if they do
        if key in d1 and isinstance(d1[key], dict) and isinstance(d2[key], dict):
            # Recursively call the same function
            recursive_update(d1[key], d2[key])
        # You can add other conditions for merging lists or other data types here too if you want
        else:
            d1[key] = d2[key]
    # Note this function works in place, so there is no return

用法：

recursive_update(dict_a, dict_b)

如果您确定您的数据结构将是一个具有 1 级嵌套的 dicts dict，@leaf_yakitori 的答案也将起作用并且更加简洁。如果某些值不是字典或嵌套超过 2 级，则此函数起作用。

嵌套字典更新

2 个答案: