使用“路径”访问不同级别的嵌套词典

Question

我想定义一个小包装函数，它采用某种“路径”来访问不同级别的嵌套字典：

D = {'key1': valueA,
     'key2': {'key21': valueB,
             {'key22': valueC}

在这个简单的例子中，我想写一个函数，作为参数，它接受一个元组，比如

dict_path = ('key2', 'key22')

>>>nested_getter(dict_path)
valueC

Answer 1

def nested_getter(d, keys):
    return reduce(dict.get, keys, d)

Answer 2

D = {'key1': valueA,
     'key2': {'key21': valueB,
              'key22': valueC}}


def nested_getter(root, path):
    for elem in path:
        root = root[elem]
    return root

有了它，您可以：

>>> nested_getter(D, ('key2', 'key22'))
3