Question

我希望生成许多不同的JSON结构排列作为同一数据集的表示，最好不必对实现进行硬编码。例如，给定以下JSON：

{"name": "smith", "occupation": "agent", "enemy": "humanity", "nemesis": "neo"}`

应该产生许多不同的排列，例如：

更改名称：{"name":"smith"}- > {"last_name":"smith"}
按顺序更改：{"name":"...","occupation":"..."} -> {"occupation":"...", "name":"..."}
改变安排：{"name":"...","occupation":"..."} -> "smith":{"occupation":"..."}
更改模板：{"name":"...","occupation":"..."} -> "status": 200, "data":{"name":"...","occupation":"..."}
等

目前，实施情况如下：

我正在使用itertools.permutations和OrderedDict（）来查看可能的键和各自的值组合以及它们返回的顺序。

key_permutations = SchemaLike(...).permutate()

all_simulacrums = []
for key_permutation in key_permutations:
   simulacrums = OrderedDict(key_permutation)
   all_simulacrums.append(simulacrums)
for x in itertools.permutations(all_simulacrums.items()):
    test_data = json.dumps(OrderedDict(p))
    print(test_data)
    assert json.loads(test_data) == data, 'Oops! {} != {}'.format(test_data, data)

当我尝试实现排列和模板的排列时，我的问题就出现了。我不知道如何最好地实现这个功能，有什么建议吗？

Answer 1

如需订购，只需使用有序的dicts：

>>> data = OrderedDict(foo='bar', bacon='eggs', bar='foo', eggs='bacon')
>>> for p in itertools.permutations(data.items()):
...     test_data = json.dumps(OrderedDict(p))
...     print(test_data)
...     assert json.loads(test_data) == data, 'Oops! {} != {}'.format(test_data, data)

{"foo": "bar", "bacon": "eggs", "bar": "foo", "eggs": "bacon"}
{"foo": "bar", "bacon": "eggs", "eggs": "bacon", "bar": "foo"}
{"foo": "bar", "bar": "foo", "bacon": "eggs", "eggs": "bacon"}
{"foo": "bar", "bar": "foo", "eggs": "bacon", "bacon": "eggs"}
{"foo": "bar", "eggs": "bacon", "bacon": "eggs", "bar": "foo"}
{"foo": "bar", "eggs": "bacon", "bar": "foo", "bacon": "eggs"}
{"bacon": "eggs", "foo": "bar", "bar": "foo", "eggs": "bacon"}
{"bacon": "eggs", "foo": "bar", "eggs": "bacon", "bar": "foo"}
{"bacon": "eggs", "bar": "foo", "foo": "bar", "eggs": "bacon"}
{"bacon": "eggs", "bar": "foo", "eggs": "bacon", "foo": "bar"}
{"bacon": "eggs", "eggs": "bacon", "foo": "bar", "bar": "foo"}
{"bacon": "eggs", "eggs": "bacon", "bar": "foo", "foo": "bar"}
{"bar": "foo", "foo": "bar", "bacon": "eggs", "eggs": "bacon"}
{"bar": "foo", "foo": "bar", "eggs": "bacon", "bacon": "eggs"}
{"bar": "foo", "bacon": "eggs", "foo": "bar", "eggs": "bacon"}
{"bar": "foo", "bacon": "eggs", "eggs": "bacon", "foo": "bar"}
{"bar": "foo", "eggs": "bacon", "foo": "bar", "bacon": "eggs"}
{"bar": "foo", "eggs": "bacon", "bacon": "eggs", "foo": "bar"}
{"eggs": "bacon", "foo": "bar", "bacon": "eggs", "bar": "foo"}
{"eggs": "bacon", "foo": "bar", "bar": "foo", "bacon": "eggs"}
{"eggs": "bacon", "bacon": "eggs", "foo": "bar", "bar": "foo"}
{"eggs": "bacon", "bacon": "eggs", "bar": "foo", "foo": "bar"}
{"eggs": "bacon", "bar": "foo", "foo": "bar", "bacon": "eggs"}
{"eggs": "bacon", "bar": "foo", "bacon": "eggs", "foo": "bar"}

相同的原则可以应用于键/值排列：

>>> for p in itertools.permutations(data.keys()):
...:     test_data = json.dumps(OrderedDict(zip(p, data.values())))
...:     print(test_data)
...:     
{"foo": "bar", "bacon": "eggs", "bar": "foo", "eggs": "bacon"}
{"foo": "bar", "bacon": "eggs", "eggs": "foo", "bar": "bacon"}
{"foo": "bar", "bar": "eggs", "bacon": "foo", "eggs": "bacon"}
{"foo": "bar", "bar": "eggs", "eggs": "foo", "bacon": "bacon"}
{"foo": "bar", "eggs": "eggs", "bacon": "foo", "bar": "bacon"}
{"foo": "bar", "eggs": "eggs", "bar": "foo", "bacon": "bacon"}
{"bacon": "bar", "foo": "eggs", "bar": "foo", "eggs": "bacon"}
{"bacon": "bar", "foo": "eggs", "eggs": "foo", "bar": "bacon"}
{"bacon": "bar", "bar": "eggs", "foo": "foo", "eggs": "bacon"}
{"bacon": "bar", "bar": "eggs", "eggs": "foo", "foo": "bacon"}
{"bacon": "bar", "eggs": "eggs", "foo": "foo", "bar": "bacon"}
{"bacon": "bar", "eggs": "eggs", "bar": "foo", "foo": "bacon"}
{"bar": "bar", "foo": "eggs", "bacon": "foo", "eggs": "bacon"}
{"bar": "bar", "foo": "eggs", "eggs": "foo", "bacon": "bacon"}
{"bar": "bar", "bacon": "eggs", "foo": "foo", "eggs": "bacon"}
{"bar": "bar", "bacon": "eggs", "eggs": "foo", "foo": "bacon"}
{"bar": "bar", "eggs": "eggs", "foo": "foo", "bacon": "bacon"}
{"bar": "bar", "eggs": "eggs", "bacon": "foo", "foo": "bacon"}
{"eggs": "bar", "foo": "eggs", "bacon": "foo", "bar": "bacon"}
{"eggs": "bar", "foo": "eggs", "bar": "foo", "bacon": "bacon"}
{"eggs": "bar", "bacon": "eggs", "foo": "foo", "bar": "bacon"}
{"eggs": "bar", "bacon": "eggs", "bar": "foo", "foo": "bacon"}
{"eggs": "bar", "bar": "eggs", "foo": "foo", "bacon": "bacon"}
{"eggs": "bar", "bar": "eggs", "bacon": "foo", "foo": "bacon"}

依此类推......如果您不需要所有组合，您可以使用一组预定义的键/值。您还可以使用for循环random.choice来翻转硬币以跳过某些组合，或使用random.shuffle冒着重复组合的风险。

对于模板，我猜你必须创建一个不同模板的列表（或列表列表，如果你想要嵌套结构），然后迭代它以创建数据。为了给出更好的建议，我们需要对您想要的内容进行更严格的规范。

请注意，有几个库在Python中生成测试数据：

>>> from faker import Faker
>>> faker = Faker()
>>> faker.credit_card_full().strip().split('\n')
['VISA 13 digit', 'Jerry Gutierrez', '4885274641760 04/24', 'CVC: 583']

Faker有几个模式，很容易创建自己的自定义虚假数据提供程序。

Answer 2

由于已经回答了dict顺序的shuffle，我将跳过它。

当我想到新事物时，我会加入这个答案。

from random import randint
from collections import OrderedDict

#Randomly shuffles the key-value pairs of a dictionary
def random_dict_items(input_dict):
    items = input_dict.items()
    new_dict = OrderedDict()
    for i in items:
        rand = randint(0, 1)
        if rand == 0:
            new_dict[i[0]] = i[1]
        else:
            new_dict[i[1]] = i[0]
    return new_dict

为数据集生成随机JSON结构排列

2 个答案: