Python-通过键列表访问嵌套字典项?

发布于 2021-02-02 23:22:52

我有一个复杂的字典结构,我想通过一个键列表来访问该字典以解决正确的项。

dataDict = {
    "a":{
        "r": 1,
        "s": 2,
        "t": 3
        },
    "b":{
        "u": 1,
        "v": {
            "x": 1,
            "y": 2,
            "z": 3
        },
        "w": 3
        }
}    

maplist = ["a", "r"]

要么

maplist = ["b", "v", "y"]

我已经制作了下面的代码,但是可以肯定的是,如果有人有想法,那么我可以找到一种更好,更有效的方法。

# Get a given data from a dictionary with position provided as a list
def getFromDict(dataDict, mapList):    
    for k in mapList: dataDict = dataDict[k]
    return dataDict

# Set a given data in a dictionary with position provided as a list
def setInDict(dataDict, mapList, value): 
    for k in mapList[:-1]: dataDict = dataDict[k]
    dataDict[mapList[-1]] = value
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1 个回答
  • 面试哥
    面试哥 2021-02-02
    为面试而生,有面试问题,就找面试哥。

    使用reduce()遍历词典:

    from functools import reduce  # forward compatibility for Python 3
    import operator
    
    def getFromDict(dataDict, mapList):
        return reduce(operator.getitem, mapList, dataDict)
    

    并重getFromDict用以查找用于存储值的位置setInDict()

    def setInDict(dataDict, mapList, value):
        getFromDict(dataDict, mapList[:-1])[mapList[-1]] = value
    

    除了最后一个元素外,所有元素mapList都需要查找“父”字典以将值添加到其中,然后使用最后一个元素将值设置为右键。

    演示:

    >>> getFromDict(dataDict, ["a", "r"])
    1
    >>> getFromDict(dataDict, ["b", "v", "y"])
    2
    >>> setInDict(dataDict, ["b", "v", "w"], 4)
    >>> import pprint
    >>> pprint.pprint(dataDict)
    {'a': {'r': 1, 's': 2, 't': 3},
     'b': {'u': 1, 'v': {'w': 4, 'x': 1, 'y': 2, 'z': 3}, 'w': 3}}
    

    请注意,Python PEP8样式指南规定了函数的snake_case名称。上面的方法同样适用于列表或字典和列表的混合,因此名称应为get_by_path()and set_by_path()

    from functools import reduce  # forward compatibility for Python 3
    import operator
    
    def get_by_path(root, items):
        """Access a nested object in root by item sequence."""
        return reduce(operator.getitem, items, root)
    
    def set_by_path(root, items, value):
        """Set a value in a nested object in root by item sequence."""
        get_by_path(root, items[:-1])[items[-1]] = value
    


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