python - 检查字典中是否存在键结构

Question

我想要一个函数，当给定的键列表导致字典中的现有结构时，该函数返回 True。每个key对应字典的深度级别

我遇到的困难是列表的长度(= 键的数量)和字典的深度都是动态的

#Example Code:

keys1 = ["K1", "K3", "K4"]
keys2 = ["K2", "K6"]
keys3 = ["K1", "K6", "K4"]

dict = {
    "K1": {
        "K3": {
            "K4": "a"
        }
    },
    "K2": {
        "K6": "b"
    }
}

result = function(keys1, dict) #result should be True    
result = function(keys2, dict) #result should be True    
result = function(keys3, dict) #result should be False

Answer 1

简单的递归方法:

def function(keys, dct):
    return not keys or (keys[0] in dct and function(keys[1:], dct[keys[0]]))

>>> function(keys1, dct)  # never shadow built-in names
True
>>> function(keys2, dct)
True
>>> function(keys3, dct)
False

这假设了一个非常统一的结构:所有中间值本身都是字典，深度始终至少是键的长度。否则，您将需要处理一些错误:

def function(keys, dct):
    try:
        return not keys or function(keys[1:], dct[keys[0]])
    except (TypeError, KeyError):  # this allows you to shorten the above
        return False