您可以使用该
reduce()函数遍历一系列嵌套字典：
def get_nested(d, path): return reduce(dict.__getitem__, path, d)
演示：
>>> def get_nested(d, path):... return reduce(dict.__getitem__, path, d)... >>> my_dict = {'key1': {'key2': {'foo': 'bar', 'key3': {'key4': {'key5': 'blah'}}}}}>>> get_nested(my_dict, ('key1', 'key2', 'key3', 'key4', 'key5'))'blah'当密钥不存在时，此版本引发异常：
>>> get_nested(my_dict, ('key1', 'nonesuch'))Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 2, in get_nestedKeyError: 'nonesuch'但您可以替换
dict.__getitem__为
lambda d, k: d.setdefault(k, {})，使其创建空字典：def get_nested_default(d, path): return reduce(lambda d, k: d.setdefault(k, {}), path, d)演示：
>>> def get_nested_default(d, path):... return reduce(lambda d, k: d.setdefault(k, {}), path, d)... >>> get_nested_default(my_dict, ('key1', 'nonesuch')){}>>> my_dict{'key1': {'key2': {'key3': {'key4': {'key5': 'blah'}}, 'foo': 'bar'}, 'nonesuch': {}}}要在给定路径上 设置 值，请遍历除最后一个键以外的所有键，然后在常规词典分配中使用最后一个键：
def set_nested(d, path, value): get_nested_default(d, path[:-1])[path[-1]] = value
这使用该
get_nested_default()函数根据需要添加空字典：
>>> def set_nested(d, path, value):... get_nested_default(d, path[:-1])[path[-1]] = value... >>> my_dict = {'key1': {'key2': {'foo': 'bar'}}}>>> set_nested(my_dict, ('key1', 'key2', 'key3', 'key4', 'key5'), 'blah')>>> my_dict{'key1': {'key2': {'key3': {'key4': {'key5': 'blah'}}, 'foo': 'bar'}}}
