如何将字典组合成嵌套字典？

Question

我想将我的 3 个字典组合成 1 个嵌套字典。我使用 3 个嵌套的 for 循环编写了以下代码。但是对于同一件事是否有任何有效的方法或递归函数？

X = {"X1":["O","E","P"],"X2":["M"]}
Y = {"O":["a"],"E":["b","c"],"P":["d"],"M":["r"]}
Z = {"a":["1"],"b":["2","3"],"c":[],"d":["4","5"],"r":["6"]}

d1 = {}
for k in X:
    A = X[k]
    d2 = {}
    for v in A:
        B = Y[v]
        d3 = {}
        for i in B:
            C = Z[i]
            d3.update({i:C})
        d2.update({v:d3})
    d1.update({k:d2})

Answer 1

你可以使用简单的递归：

X = {"X1":["O","E","P"],"X2":["M"]}
Y = {"O":["a"],"E":["b","c"],"P":["d"],"M":["r"]}
Z = {"a":["1"],"b":["2","3"],"c":[],"d":["4","5"],"r":["6"]}
start = [X, Y, Z]
def group(d):
   return d if all(all(c not in i for i in start) for c in d) else \
           {i:group([c[i] for c in start if i in c][0]) for i in d}

r = {a:group(b) for a, b in X.items()}
print(r == d1) #d1 generated from OP's solution

输出：

{'X1': {'O': {'a': ['1']}, 'E': {'b': ['2', '3'], 'c': []}, 'P': {'d': ['4', '5']}}, 'X2': {'M': {'r': ['6']}}}
True

Answer 2

对 1 行的字典理解，与嵌套 for- 循环的过程基本相同：

{k: {v0:{v1: Z[v1] for v1 in Y[v0]} for v0 in v} for k, v in X.items()}

输出：

{'X1': {'O': {'a': ['1']},
  'E': {'b': ['2', '3'], 'c': []},
  'P': {'d': ['4', '5']}},
 'X2': {'M': {'r': ['6']}}}

解释： OP 的算法使用当前列表中的每个值作为键来查找下一个字典中的值列表，直到到达最后一个字典。 在伪代码中，嵌套看起来像：

# pseudo code
for key, values in X
    for valX in values:
        for valY in Y[valX]: # note Y[valX] is a list
            Z[valY]

把它翻译成一个理解，我们从最里面的循环开始，出去并添加必要的装饰

第 1 步：

{y:Z[y] for ys in Y.values() for y in ys}
# out: 
{'a': ['1'], 'b': ['2', '3'], 'c': [], 'd': ['4', '5'], 'r': ['6']}

第 2 步：现在我们直接查找 ys

{x:{y:Z[y] for y in Y[x]} for xs in X.values() for x in xs}
# out:
{'O': {'a': ['1']},
 'E': {'b': ['2', '3'], 'c': []},
 'P': {'d': ['4', '5']},
 'M': {'r': ['6']}}

第 3 步：现在我们将来自X 的键放入并添加另一层字典嵌套

{k:{x:{y:Z[y] for y in Y[x]} for x in xs} for k, xs in X.items()}

产生期望的结果

一般来说，当尝试将嵌套循环转换为推导式时，从最内层循环开始，然后向外工作。