Python 创建嵌套dict的置换

我正在编写一个代码，它应该检查注册表行，并为（1）每个协议使用哪些行，以及（2）为了检查某些功能而断开哪些行，创建所有排列

reg_lines = {'primary': ['ETH', 'UDP', 'TCP'], 'secondary': ['ETH', 'TCP'], private': ['UDP', 'TCP']}

预期排列：

一,

二,

j

一,

i+1

i+2

n

---

首先，使用

itertools.combines

获取主要、次要和专用的所有组合，然后使用

itertools.product

获取这些组合的乘积。然后再次使用

itertools.compositions

获取

disc

字典的所有子集

from itertools import product, combinations, chain

reg_lines = {'primary': ['ETH', 'UDP', 'TCP'], 'secondary': ['ETH', 'TCP'], 'private': ['UDP', 'TCP']}

def all_combinations(lst):
    return chain(*[combinations(lst, i+1) for i in range(len(lst))])

for comb in all_combinations(reg_lines):
    for prod in product(*(reg_lines[k] for k in comb)):
        use = dict(zip(comb, prod))
        print("use", use)
        for comb2 in all_combinations(comb):
            disc = {k: use[k] for k in comb2}
            print("disc", disc)

输出：

use {'primary': 'ETH'}
disc {'primary': 'ETH'}
use {'primary': 'UDP'}
disc {'primary': 'UDP'}
... many many more ...
use {'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP'}
disc {'primary': 'TCP'}
disc {'secondary': 'TCP'}
disc {'private': 'TCP'}
disc {'primary': 'TCP', 'secondary': 'TCP'}
disc {'primary': 'TCP', 'private': 'TCP'}
disc {'secondary': 'TCP', 'private': 'TCP'}
disc {'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP'}

---

尝试使用itertools.compositions，但我发现了一些非常混乱的问题。对我来说，问题不清楚

priv_line

在哪里，因为给定的

reg_line

@SIslam，您是对的。我简化了问题。删除了priv_行。在您的示例中，您不会得到仅“secondary”或“secondary”+“private”（use=dict（zip）（“primary”，“secondary”，“private”），x）的排列@alnet是的，问题IMHO中的这部分有点不清楚。现在呢？

use = [{'secondary: 'TCP'}]
disc = [{'secondary: 'TCP'}]

use = [{'primary': 'ETH', 'secondary': 'ETH'}
disc = [{'primary': 'ETH'}]

use = [{'primary': 'ETH', 'secondary': 'ETH'}]
disc = [{'primary': 'ETH'}]

use = [{'primary': 'ETH', 'secondary': 'ETH'}]
disc = [{'primary': 'ETH', 'secondary': 'ETH'}]

use = [{'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP}]
disc = [{'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP}]

from itertools import product, combinations, chain

reg_lines = {'primary': ['ETH', 'UDP', 'TCP'], 'secondary': ['ETH', 'TCP'], 'private': ['UDP', 'TCP']}

def all_combinations(lst):
    return chain(*[combinations(lst, i+1) for i in range(len(lst))])

for comb in all_combinations(reg_lines):
    for prod in product(*(reg_lines[k] for k in comb)):
        use = dict(zip(comb, prod))
        print("use", use)
        for comb2 in all_combinations(comb):
            disc = {k: use[k] for k in comb2}
            print("disc", disc)

use {'primary': 'ETH'}
disc {'primary': 'ETH'}
use {'primary': 'UDP'}
disc {'primary': 'UDP'}
... many many more ...
use {'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP'}
disc {'primary': 'TCP'}
disc {'secondary': 'TCP'}
disc {'private': 'TCP'}
disc {'primary': 'TCP', 'secondary': 'TCP'}
disc {'primary': 'TCP', 'private': 'TCP'}
disc {'secondary': 'TCP', 'private': 'TCP'}
disc {'primary': 'TCP', 'secondary': 'TCP', 'private': 'TCP'}