英文:
List of all possible path of edges between two nodes, showing edges multiplicity
问题
I can help you with the translation. Here's the text you provided in Chinese:
我正在开发一个算法，将电网建模为无向多重图，以便使用Python分析电网上的故障。
我有一些函数，可以获取两个节点之间的所有最短路径（不包含循环），并需要获取考虑它们重复性的边缘。
假设我有源节点“LAG-138”和负载“SCA-24”，我的程序找到一条路径：
[“SCA-24”，“SCA-69”，“SCO-69”，“PAA-69”，“OCA-T4-69”，“OCA-138”，“LVR-138”，“LAG-138”]
还有许多其他路径，但为了简单起见，这里只有一条。
从这个输出中，我可以获取两个节点之间的所有边以及它们的重复性，如下所示：
[‘SCA-24 < - > SCA-69 (1)’, ‘SCA-24 < - > SCA-69 (2)’]
[‘SCO-69 < - > SCA-69 (1)’]
[‘SCO-69 < - > PAA-69 (1)’]
[‘PAA-69 < - > OCA-T4-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (2)’]
[‘OCA-T4-69 < - > OCA-138 (1)’]
[‘OCA-138 < - > LVR-138 (1)’]
[‘LVR-138 < - > LAG-138 (1)’, ‘LVR-138 < - > LAG-138 (2)’]
我从路径序列中的每一对节点获得这个输出。对于每一对节点，我都会获取它们之间的所有边。
现在，如何获取这些输出并获取所有路径？
我需要的输出将如下所示：
[
[‘SCA-24 < - > SCA-69 (1)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (1)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (1)’],
[‘SCA-24 < - > SCA-69 (1)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (1)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (2)’],
[‘SCA-24 < - > SCA-69 (1)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (2)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (1)’],
[‘SCA-24 < - > SCA-69 (1)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (2)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (2)’],
[‘SCA-24 < - > SCA-69 (2)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (1)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (1)’],
[‘SCA-24 < - > SCA-69 (2)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (1)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (2)’],
[‘SCA-24 < - > SCA-69 (2)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (2)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (1)’],
[‘SCA-24 < - > SCA-69 (2)’, ‘SCO-69 < - > SCA-69 (1)’, ‘SCO-69 < - > PAA-69 (1)’, ‘PAA-69 < - > OCA-T4-69 (2)’, ‘OCA-T4-69 < - > OCA-138 (1)’, ‘LVR-138 < - > LAG-138 (2)’],
]
这是所有路径的列表，按照边缘排列。这可能是一个简单的问题，但现在我想不出任何解决方法。
请帮忙。
If you have any further questions or need assistance with the problem described in this text, feel free to ask.
英文:
I'm working on a algorithm that models a electric grid as a Undirected Multigraph so I can analyse the faults on the grid using python.
I have the functions that get me all the minimum paths (paths that do not loop) between two nodes and need to get the edges accounting for their multiplicity.
Say I have the source "LAG-138", and the load "SCA-24", my program finds a path:
["SCA-24", "SCA-69", "SCO-69", "PAA-69", "OCA-T4-69", "OCA-138", "LVR-138", "LAG-138"]
And many others, but for the sake of simplicity, here's one.
From this output, I can get all the edges between two nodes, and their multiplicity like this:
['SCA-24 <-> SCA-69 (1)', 'SCA-24 <-> SCA-69 (2)']
['SCO-69 <-> SCA-69 (1)']
['SCO-69 <-> PAA-69 (1)']
['PAA-69 <-> OCA-T4-69 (1)', 'PAA-69 <-> OCA-T4-69 (2)']
['OCA-T4-69 <-> OCA-138 (1)']
['OCA-138 <-> LVR-138 (1)']
['LVR-138 <-> LAG-138 (1)', 'LVR-138 <-> LAG-138 (2)']
I get this output from every pair of nodes in the sequence of the path. For every pair, I get all the edges between them.
Now, how can I get these outputs and get all the paths?
The output I need from this would be like this:
[
['SCA-24 <-> SCA-69 (1)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (1)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (1)'],
['SCA-24 <-> SCA-69 (1)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (1)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (2)'],
['SCA-24 <-> SCA-69 (1)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (2)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (1)'],
['SCA-24 <-> SCA-69 (1)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (2)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (2)'],
['SCA-24 <-> SCA-69 (2)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (1)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (1)'],
['SCA-24 <-> SCA-69 (2)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (1)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (2)'],
['SCA-24 <-> SCA-69 (2)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (2)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (1)'],
['SCA-24 <-> SCA-69 (2)', 'SCO-69 <-> SCA-69 (1)', 'SCO-69 <-> PAA-69 (1)', 'PAA-69 <-> OCA-T4-69 (2)', 'OCA-T4-69 <-> OCA-138 (1)', 'LVR-138 <-> LAG-138 (2)'],
]
A list of all the paths, arrenged among the edges. It's probably a simple problem, but right now I can't think of anything
Please Help.
答案1
得分: 2
这听起来你想要使用itertools.product。
for path in itertools.product(a, b, c, d, ....):
print(path)
将为你提供每一种可能性，从a中取一个，从b中取一个，依此类推。
如果像我怀疑的那样，你实际上有一个列表，比如paths_list，其中paths_list[0]是从第一个节点到第二个节点的路径，path_lists[1]是从第二个节点到第三个节点的路径，依此类推，
for path in itertools.product(*paths_list):
...
这两种方法都会返回一个迭代器。如果你只想要整个列表，那么
list(itertools.product(*paths_list))
英文:
It sounds like you want itertools.product.
for path in itertools.product(a, b, c, d, ....):
print(path)
will give you every possibility, taking one from a, one from b, etc.
If, as I suspect, you actually have a list, say paths_list where paths_list[0] are the paths from the first node to the second, path_lists[1] are the paths from the second node to the third, etc., then
for path in itertools.product(*paths_list):
...
Both of these return an iterator. If you just want the whole list, then
list(itertools.product(*paths_list))
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