使用同一字典中的键扩展字典值

Question

我尝试使用同一字典中的键来扩展字典的值。换句话说，我尝试替换 dic 值（现有集合与另一个/可能扩展的集合）。当值包含字典中的键时会发生扩展。

输入：

dic = {
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'05', '10'},
    '21': {'06', '11'},
    '30': {'07', '20'},
    '40': {'08', '21', '30'},
    '50': {'09', '40'}
}

预期输出：

{
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'05', '01', '02'},
    '21': {'06', '03', '04'},
    '30': {'07', '05', '01', '02'},
    '40': {'08', '06', '03', '04', '07', '05', '01', '02'},
    '50': {'09', '08', '06', '03', '04', '07', '05', '01', '02'}
}

我正在尝试创建一个递归函数...

def transform_dic(d):
    def func(k):
        v = d.get(k, k)
        if v != k:
            for e in v:
                v = func(v)
        return v
    d2 = {}
    for k, v in d.items():
        d2[k] = {func(i) for i in v}
    return d2

print(transform_dic(dic))

TypeError: unhashable type: 'set'

Answer 1

您可以迭代输出字典并检查集合中的值是否是结果字典的成员：

from typing import Dict, Set


dic = {
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'05', '10'},
    '21': {'06', '11'},
    '30': {'07', '20'},
    '40': {'08', '21', '30'},
    '50': {'09', '40'}
}

output = {
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'05', '01', '02'},
    '21': {'06', '03', '04'},
    '30': {'07', '05', '01', '02'},
    '40': {'08', '06', '03', '04', '07', '05', '01', '02'},
    '50': {'09', '08', '06', '03', '04', '07', '05', '01', '02'}
}

def conv(d: Dict[str, set[str]]) -> Dict[str, set[str]]:        
    result: Dict[str, set[str]] = {}
    for k,vs in d.items():
        result[k] = set()
        for v in vs:
            if v in result:
                for t in result[v]:
                    result[k].add(t)
            else:
                result[k].add(v)
    return result

print(conv(dic))
print(conv(dic)==output)

输出将是

{'10': {'02', '01'}, '11': {'03', '04'}, '20': {'02', '05', '01'}, '21': {'06', '03', '04'}, '30': {'02', '05', '07', '01'}, '40': {'05', '03', '02', '01', '06', '07', '08', '04'}, '50': {'05', '03', '09', '02', '01', '06', '07', '08', '04'}}
True

Answer 2

你要递归展开的是字典中的集合，而不是字典本身。你永远只有一个字典。

出于这个原因，我认为编写一个递归函数来扩展一个集合，然后使用这个函数来扩展字典会更容易。

def expanded_dict(d):
    return {k: expanded_set(v, d) for k,v in d.items()}

def expanded_set(s, d):
    return {
        y
        for x in s
            for y in (expanded_set(d[x], d) if x in d
                      else (x,))
    }

注意 expanded_set returns 一个新的集，没有修改输入集和输入字典；和 expanded_dict returns 一个不修改输入字典的新字典。

最好明确说明您的函数 returns 是新对象还是修改输入对象。例如，sorted returns 一个新列表而不修改其输入，而 list.sort 修改一个列表而不返回任何内容。

这是一个通过修改字典而不返回任何内容来扩展字典的函数：

def expand_dict(d):
    stable = False
    while not stable:
        stable = True
        for k in d:
            for x in list(d[k]):
                if x in d:
                    d[k].remove(x)
                    d[k].update(d[x])
                    stable = False

正在使用您的示例进行测试：

dic = {
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'05', '10'},
    '21': {'06', '11'},
    '30': {'07', '20'},
    '40': {'08', '21', '30'},
    '50': {'09', '40'}
}

print(expanded_dict(dic))
# {'10': {'01', '02'}, '11': {'03', '04'}, '20': {'01', '05', '02'}, '21': {'03', '06', '04'}, '30': {'07', '01', '02', '05'}, '40': {'04', '06', '08', '05', '02', '01', '07', '03'}, '50': {'04', '06', '08', '02', '05', '01', '07', '09', '03'}}

print(dic)
# {'10': {'01', '02'}, '11': {'03', '04'}, '20': {'10', '05'}, '21': {'11', '06'}, '30': {'07', '20'}, '40': {'08', '30', '21'}, '50': {'40', '09'}}

expand_dict(dic)
# no return value

print(dic)
# {'10': {'01', '02'}, '11': {'03', '04'}, '20': {'01', '02', '05'}, '21': {'03', '06', '04'}, '30': {'01', '05', '07', '02'}, '40': {'04', '06', '08', '05', '02', '01', '07', '03'}, '50': {'04', '06', '08', '05', '02', '01', '07', '09', '03'}}

Answer 3

一定要递归吗？

def transform_dic(d):
    for k, v in d.items():
        for n in set(v):
            if n in d:
                d[k].remove(n)
                d[k].update(d[n])

transform_dic(dic)
print(dic)

如果您不想修改原始词典，请使用deepcopy

def transform_dic(d):
    d = deepcopy(d)
    for k, v in d.items():
        for n in set(v):
            if n in d:
                d[k].remove(n)
                d[k].update(d[n])
    return d

print(transform_dic(dic))

输出：

{
    '10': {'01', '02'},
    '11': {'03', '04'},
    '20': {'01', '05', '02'},
    '21': {'03', '04', '06'},
    '30': {'01', '07', '05', '02'},
    '40': {'01', '08', '07', '04', '06', '03', '05', '02'},
    '50': {'01', '08', '07', '04', '06', '03', '05', '02', '09'}
}

使用同一字典中的键扩展字典值

Expanding dict value using keys in the same dictionary

python

recursion

dictionary

set