python - 通过键列表访问嵌套字典项？

Question

我有一个复杂的字典结构，我想通过一个键列表来访问它以解决正确的项目。

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

maplist = ["a", "r"]

或

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

我已经编写了以下有效的代码，但如果有人有想法，我相信有更好、更有效的方法来做到这一点。

# 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

Answer 1

使用reduce()遍历字典:

from functools import reduce  # forward compatibility for Python 3
import operator

def getFromDict(dataDict, mapList):
    return reduce(operator.getitem, mapList, dataDict)

并重用 getFromDict 来找到存储 setInDict() 值的位置:

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

mapList 中除了最后一个元素之外的所有元素都需要找到“父”字典来添加值，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 风格指南 prescribes snake_case names for functions .以上方法同样适用于列表或字典和列表的混合，因此名称应该是 get_by_path() 和 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

为了完成，一个删除键的函数:

def del_by_path(root, items):
    """Delete a key-value in a nested object in root by item sequence."""
    del get_by_path(root, items[:-1])[items[-1]]