python - 如何通过键列表导航字典？

Question

从这样的嵌套字典开始:

my_dict = {"North America" : { "USA" : { "Virginia" : ["Norfolk","Richmond","Charlottesville"], "New York": ["Albany"]}, "Canada" : {"Saskatchewan": ["Saskatoon"], "New Brunswick":["Moncton","Saint John"]}}}

print(my_dict)
North America
    USA
        Virginia
            ['Norfolk', 'Richmond', 'Charlottesville']
        New York
            ['Albany']
    Canada
        Saskatchewan
            ['Saskatoon']
        New Brunswick
            ['Moncton', 'Saint John']

当我不知道时，如何获取像 [key1, key2] 这样的字符串列表并以编程方式返回嵌套对象 my_dict[key1][key2]会有多少把 key ？示例:

keys = ['North America', 'USA']
print(my_dict.???)
    Virginia
        ['Norfolk', 'Richmond', 'Charlottesville']
    New York
        ['Albany']

keys = ['North America', 'Canada', 'Saskatchewan']
print(my_dict.???)
    ['Saskatoon']

keys = ['North America', 'Canada']
print(my_dict.???)
    Saskatchewan
        ['Saskatoon']
    New Brunswick
        ['Moncton', 'Saint John']

假设键的“路径”存在，但不要假设任何有关结构深度的信息。

Answer 1

你可以将functools.reduce扔到它上面。

>>> from functools import reduce
>>>
>>> my_dict = {"North America" : { "USA" : { "Virginia" : ["Norfolk","Richmond","Charlottesville"], "New York": ["Albany"]}, "Canada" : {"Saskatchewan": ["Saskatoon"], "New Brunswick":["Moncton","Saint John"]}}}
>>> keys = ['North America', 'Canada']
>>> reduce(dict.get, keys, my_dict)
>>> {'New Brunswick': ['Moncton', 'Saint John'], 'Saskatchewan': ['Saskatoon']}

...或者编写您自己的函数。

def nest_get(dic, keys):
    result = dic
    for k in keys:
        result = result[k]
    return result

演示:

>>> nest_get(my_dict, keys)
>>> {'New Brunswick': ['Moncton', 'Saint John'], 'Saskatchewan': ['Saskatoon']}