Python 以列表为值的字典，使用原始键值的所有可能组合创建字典列表

Question

我有一个字典，其中所有值都是列表：

original_dict = {"key_00": [0, 1],
                    "key_01": [2, 3]
                    }

我想获得一个字典列表，其中每个键与原始对应值之一相关联，形成所有可能的组合......换句话说，像这样：

[
{"key_00": 0, "key_01": 2},
{"key_00": 0, "key_01": 3},
{"key_00": 1, "key_01": 2},
{"key_00": 1, "key_01": 3}
]

我尝试使用这样的东西：

res = []
for combs in product(*original_dict.values()):
    # zip used to perform cross keys combinations.
    res.append([{ele: cnt} for ele, cnt in zip(original_dict, combs)])

但这样我最终有

[{'key_00': 0}, {'key_01': 2}]
[{'key_00': 0}, {'key_01': 3}]
[{'key_00': 1}, {'key_01': 2}]
[{'key_00': 1}, {'key_01': 3}]    

Answer 1

您可以使用list comprehesntion 。

from itertools import product

original_dict = {"key_00": [0, 1],"key_01": [2, 3]}

res = [{ele: cnt for ele, cnt in zip(original_dict, combs)} 
       for combs in product(*original_dict.values())]

print(res)

您的代码可以修复如下：

res = []
for combs in product(*original_dict.values()):
    res.append({ele: cnt for ele, cnt in zip(original_dict, combs)})

---

[
    {'key_00': 0, 'key_01': 2}, 
    {'key_00': 0, 'key_01': 3}, 
    {'key_00': 1, 'key_01': 2}, 
    {'key_00': 1, 'key_01': 3}
]

Answer 2

您必须使用 itertools.product()：

import itertools
original_dict = {"key_00": [0, 1],
                 "key_01": [2, 3]}

keys, values = zip(*original_dict.items())
aim = [dict(zip(keys, v)) for v in itertools.product(*values)]
print(aim)

>>> [{'key_00': 0, 'key_01': 2},
     {'key_00': 0, 'key_01': 3},
     {'key_00': 1, 'key_01': 2},
     {'key_00': 1, 'key_01': 3}]

Answer 3

尝试这个

    res.append([{ele: cnt for ele, cnt in zip(original_dict, combs)}])