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
使用reduce()遍历词典:
reduce()
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():
getFromDict
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。
mapList
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():
Python PEP8
snake_case
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