我有三个字典:
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
由卡车组成,键和值组成。我想更改我的packed_dict,使其以这种格式输出
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,我想将packed_items
中的键替换为trucks_dict
中的值,并将packed_items
中的值替换为items_dict
中的值。
答案 0 :(得分:1)
通过将列表键转换为元组,您可以执行以下操作:
result = {}
for k, v in packed_items.items():
for i in v:
result.setdefault(tuple(trucks_dict[k]), []).append(items_dict[i])
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)
{(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]]
}
答案 1 :(得分:1)
您不能将列表用作字典键,因为它们不能散列。
因为您要求输入字符串键,所以您可以执行以下操作:
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]]}
答案 2 :(得分:0)
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散列:
{tuple(trucks_dict[k]): [items_dict[i] for i in l] for k, l in packed_items.items()}
这将返回:
{(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]]}