一尘不染

Python-通过键列表访问嵌套字典项?

python

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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2020-02-10

共1个答案

一尘不染

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
2020-02-10