展开 python 嵌套字典答案

【问题标题】：Unwrap python nested dictionary展开 python 嵌套字典
【发布时间】：2018-04-17 19:21:23
【问题描述】：

我有一个这个序列的嵌套字典：

data_dict = {0: [{1: 2}, {2: 3}, {3: 6}, {7: 8}], 1: [{0: 2}, {4: 7}, {2: 5}], 2: [{0: 3}, {1: 5}, {5: 4}, {11: 5}], 
3: [{0: 6}, {6: 2}, {11: 4}], 4: [{1: 7}], 5: [{2: 4}, {8: 3}], 6: [{3: 2}, {9: 3}], 7: [{0: 8}], 8: [{5: 3}], 9: [{6: 3}, {10: 6}], 
10: [{9: 6}], 11: [{2: 5}, {3: 4}]}

我想把字典解包成一个列表：

确切地说，我想从嵌套字典中创建一个networkX 图。请帮忙。

【问题讨论】：

问题不清楚，尤其是最后一部分。什么是“我的算法”？ adj_list 是从哪里来的？为什么它与graph 形状相同，但包含不同的数据？什么是内部字典值，它们是边缘权重吗？
另外，将一对值表示为单元素字典也没有意义。使用包含 2 个元素的元组。
在你的最终列表中，哪些是你的节点？

标签： python dictionary nested networkx

【解决方案1】：

嵌套`dict`

{0: [{1: 2}, {2: 3}, {3: 6}, {7: 8}],
 1: [{0: 2}, {4: 7}, {2: 5}],
 2: [{0: 3}, {1: 5}, {5: 4}, {11: 5}],
 3: [{0: 6}, {6: 2}, {11: 4}],
 4: [{1: 7}],
 5: [{2: 4}, {8: 3}],
 6: [{3: 2}, {9: 3}],
 7: [{0: 8}],
 8: [{5: 3}],
 9: [{6: 3}, {10: 6}],
 10: [{9: 6}],
 11: [{2: 5}, {3: 4}]}

解开

results = []
for v,list_dicts in data_dict.items():
    for d in list_dicts:
        results.append([v,list(d.keys())[0],list(d.values())[0]])
results

[出]

[[0, 1, 2],
 [0, 2, 3],
 [0, 3, 6],
 [0, 7, 8],
 [1, 0, 2],
 [1, 4, 7],
 [1, 2, 5],
 [2, 0, 3],
 [2, 1, 5],
 [2, 5, 4],
 [2, 11, 5],
 [3, 0, 6],
 [3, 6, 2],
 [3, 11, 4],
 [4, 1, 7],
 [5, 2, 4],
 [5, 8, 3],
 [6, 3, 2],
 [6, 9, 3],
 [7, 0, 8],
 [8, 5, 3],
 [9, 6, 3],
 [9, 10, 6],
 [10, 9, 6],
 [11, 2, 5],
 [11, 3, 4]]

【讨论】：

【解决方案2】：

这是使用 itertools 和理解的一种解决方案。（虽然我承认它可能不是最容易一见钟情的。）

它还可以处理每个字典的一对以上。

如果您希望将这对作为输出中的元组，请删除 *pair 中的解包 *。

from itertools import chain
from pprint import pprint

data_dict = {
    0: [{1: 2}, {2: 3}, {3: 6}, {7: 8}],
    1: [{0: 2}, {4: 7}, {2: 5}],
    2: [{0: 3}, {1: 5}, {5: 4}, {11: 5}],
    3: [{0: 6}, {6: 2}, {11: 4}],
    4: [{1: 7}],
    5: [{2: 4}, {8: 3}],
    6: [{3: 2}, {9: 3}],
    7: [{0: 8}],
    8: [{5: 3}],
    9: [{6: 3}, {10: 6}],
    10: [{9: 6}],
    11: [{2: 5}, {3: 4}],
}

data_list = list(chain(*(
    [(key, *pair) for pair in chain(*(d.items() for d in dicts))]
    for (key, dicts)
    in data_dict.items()
)))

pprint(data_list)

输出：

[(0, 1, 2),
 (0, 2, 3),
 (0, 3, 6),
 (0, 7, 8),
 (1, 0, 2),
 (1, 4, 7),
 (1, 2, 5),
 (2, 0, 3),
 (2, 1, 5),
 (2, 5, 4),
 (2, 11, 5),
 (3, 0, 6),
 (3, 6, 2),
 (3, 11, 4),
 (4, 1, 7),
 (5, 2, 4),
 (5, 8, 3),
 (6, 3, 2),
 (6, 9, 3),
 (7, 0, 8),
 (8, 5, 3),
 (9, 6, 3),
 (9, 10, 6),
 (10, 9, 6),
 (11, 2, 5),
 (11, 3, 4)]

【讨论】：

【解决方案3】：

    ls = []
rows = 0
for key in data_dict:
    for tempValue in data_dict[key]:
        # print(tempValue)
        ls.append([])
        for (k, v) in tempValue.items():
            ls[rows].append(key)
            ls[rows].append(k)
            ls[rows].append(v)
        rows+=1
print(ls)

这应该可以解决您的问题

【讨论】：

嵌套dict

解开

嵌套`dict`