如何在Python中高效更新嵌套字典的元素数值

原字典 #

dic = {1:{"Warehouse1":100, "Warehouse2":100, "Warehouse3":100}, 
       2:{"Warehouse1":200, "Warehouse2":200, "Warehouse3":200}, 
       3:{"Warehouse1":300, "Warehouse2":300, "Warehouse3":300},
       4:{"Warehouse1":400, "Warehouse2":400, "Warehouse3":400}}

预期结果 #

dic = {1:{"Warehouse1":250, "Warehouse2":250, "Warehouse3":250}, 
       2:{"Warehouse1":230, "Warehouse2":230, "Warehouse3":230}, 
       3:{"Warehouse1":340, "Warehouse2":340, "Warehouse3":340},
       4:{"Warehouse1":500, "Warehouse2":500, "Warehouse3":500}}

最优实现方法 #

先明确每个外层键对应的增量：键1加150，键2加30，键3加40，键4加100。最高效的方式是原地修改原字典，避免额外内存开销，同时用增量映射表管理每个键的增加值，代码如下：

# 定义每个外层键对应的增量
increments = {1: 150, 2: 30, 3: 40, 4: 100}

# 嵌套循环原地更新
for key, warehouse_dict in dic.items():
    add = increments[key]
    for warehouse in warehouse_dict:
        warehouse_dict[warehouse] += add

方案优势 #

• 高效低开销：直接修改原字典的可变内层字典，无需创建新结构，内存占用极低。
• 清晰易维护：增量映射表单独定义，后续调整只需修改increments，逻辑一目了然。
• 时间复杂度最优：整体时间复杂度为O(n)（n为所有仓库值的总数），是理论上的最优效率。

运行代码后，原字典dic会直接更新为预期结果。

内容的提问来源于stack exchange，提问作者FZL