从另一个字典的值替换字典键

Question

我有三本词典：

packed_items = {0: [0, 3],
                2: [1], 
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

packed_items包括卡车作为钥匙和价值作为项目清单。我想改变我的packed_dict，以便它以这种格式输出

packed_dict = {[9.5, 5.5, 5.5]:[[4.6, 4.3, 4.3],[8.75, 5.6, 6.6]]
               [16.0, 6.0, 7.0]:[[4.6, 4.3, 4.3]]
               [13.0, 5.5, 7.0]:[[6.0, 5.6, 9.0]]}

基本上我想用packed_items中的值替换trucks_dict中的键，使用packed_items中的值替换items_dict中的值。

Answer 1

通过将列表键转换为元组，您可以使用以下内容执行此操作：

Code:

result = {}
for k, v in packed_items.items():
    for i in v:
        result.setdefault(tuple(trucks_dict[k]), []).append(items_dict[i])

Test Code:

packed_items = {0: [0, 3],
                2: [1],
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

result = {}
for k, v in packed_items.items():
    for i in v:
        result.setdefault(tuple(trucks_dict[k]), []).append(items_dict[i])
print(result)

Results:

{(9.5, 5.5, 5.5): [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]], 
 (16.0, 6.0, 7.0): [[4.6, 4.3, 4.3]], 
 (13.0, 5.5, 7.0): [[6.0, 5.6, 9.0]]
}

Answer 2

您不能将列表作为字典键，因为它们不可用。

因为你要求输入字符串键，你可以这样做：

from collections import defaultdict

packed_items = {0: [0, 3],
                2: [1], 
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

d = defaultdict(list)

for k1, v1 in trucks_dict.items():
    for k2, v2 in items_dict.items():
        if k1 == k2 % 3:
            d[str(v1)].append(v2)

print(d)
# {'[9.5, 5.5, 5.5]': [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]], '[16.0, 6.0, 7.0]': [[4.6, 4.3, 4.3]], '[13.0, 5.5, 7.0]': [[6.0, 5.6, 9.0]]}

Answer 3

您可以使用dict理解将trucks_dict中的列表映射到items_dict中的项目。列表必须转换为元组，以便它们可以作为键使用：

{tuple(trucks_dict[k]): [items_dict[i] for i in l] for k, l in packed_items.items()}

返回：

{(9.5, 5.5, 5.5): [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]],
 (13.0, 5.5, 7.0): [[6.0, 5.6, 9.0]],
 (16.0, 6.0, 7.0): [[4.6, 4.3, 4.3]]}