使用列表中的项目更改嵌套字典的dict的值？

Question

您将如何基于列表的值来修改/创建嵌套dict的dict中的键/值，其中列表的最后一项是dict的值，而其余项会返回到dict中的键？这将是列表：

list_adddress = [ "key1", "key1.2", "key1.2.1", "value" ]

仅在诸如解析命令行参数之类的情况下，这才是问题。显然，在脚本中修改/创建此值将非常容易dict_nested["key1"]["key1.2"]["key1.2.1"]["value"]。

这将是字典的嵌套字典：

dict_nested = { 
    "key1": {
                "key1.1": { 
                            "...": "...",
                },
                "key1.2": { 
                            "key1.2.1": "change_this",
                },
            },

    "key2": {
                "...": "..."
            },
}

我猜在这种情况下，将需要诸如递归函数或列表理解之类的东西。

def ValueModify(list_address, dict_nested):
    ...
    ...
    ValueModify(..., ...)

另外，如果中的项目list_address会重新引用不存在的字典中的键，则应创建它们。

Answer 1

单线：

keys, (newkey, newvalue) = list_address[:-2], list_address[-2:]
reduce(dict.__getitem__, keys, dict_nested)[newkey] = newvalue

注：dict.get与operator.getitem将在这里产生错误的异常。

乔尔·科内特（Joel
Cornett）的答案中
明确的for循环可能更具可读性。

如果要创建不存在的中间词典：

reduce(lambda d,k: d.setdefault(k, {}), keys, dict_nested)[newkey] = newvalue

如果要覆盖不是字典的现有中间值，例如字符串，整数：

from collections import MutableMapping

def set_value(d, keys, newkey, newvalue, default_factory=dict):
    """
    Equivalent to `reduce(dict.get, keys, d)[newkey] = newvalue`
    if all `keys` exists and corresponding values are of correct type
    """
    for key in keys:
        try:
            val = d[key]
        except KeyError:
            val = d[key] = default_factory()
        else:
            if not isinstance(val, MutableMapping):
                val = d[key] = default_factory()
        d = val
    d[newkey] = newvalue

例

list_address = ["key1", "key1.2", "key1.2.1", "key1.2.1.1", "value"]
dict_nested = {
    "key1": {
                "key1.1": {
                            "...": "...",
                },
                "key1.2": {
                            "key1.2.1": "change_this",
                },
            },

    "key2": {
                "...": "..."
            },
}

set_value(dict_nested, list_address[:-2], *list_address[-2:])
assert reduce(dict.get, list_address[:-1], dict_nested) == list_address[-1]

测验

>>> from collections import OrderedDict
>>> d = OrderedDict()
>>> set_value(d, [], 'a', 1, OrderedDict) # non-existent key
>>> d.items()
[('a', 1)]
>>> set_value(d, 'b', 'a', 2) # non-existent intermediate key
>>> d.items()
[('a', 1), ('b', {'a': 2})]
>>> set_value(d, 'a', 'b', 3) # wrong intermediate type
>>> d.items()
[('a', {'b': 3}), ('b', {'a': 2})]
>>> d = {}
>>> set_value(d, 'abc', 'd', 4)
>>> reduce(dict.get, 'abcd', d) == d['a']['b']['c']['d'] == 4
True
>>> from collections import defaultdict
>>> autovivify = lambda: defaultdict(autovivify)
>>> d = autovivify()
>>> set_value(d, 'abc', 'd', 4)
>>> reduce(dict.get, 'abcd', d) == d['a']['b']['c']['d'] == 4
True
>>> set_value(1, 'abc', 'd', 4) #doctest:+IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
TypeError:
>>> set_value([], 'abc', 'd', 4) #doctest:+IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
TypeError:
>>> L = [10]
>>> set_value(L, [0], 2, 3)
>>> L
[{2: 3}]