如何基于共享项有效地分组对？

2019年8月4日 108次阅读

我有一个对(元组)列表,为简化这样的事情：

L = [("A","B"), ("B","C"), ("C","D"), ("E","F"), ("G","H"), ("H","I"), ("G","I"), ("G","J")]

使用python我想有效地将​​此列表拆分为：

L1 = [("A","B"), ("B","C"), ("C","D")]
L2 = [("E","F")]
L3 = [("G","H"), ("G","I"), ("G","J"), ("H","I")]

如何有效地将列表拆分为成对的组,对于组中的对,必须始终至少有一对与其他对共享一个项目？如其中一个答案中所述,这实际上是网络问题.目标是有效地将网络分成不连续(隔离)的网络部分.

可以更改类型列表,元组(集合)以实现更高的效率.

最佳答案 这更像是一个网络问题,所以我们使用网络

import networkx as nx
G=nx.from_edgelist(L)

l=list(nx.connected_components(G))
# after that we create the map dict , for get the unique id for each nodes
mapdict={z:x for x, y in enumerate(l) for z in y }
# then append the id back to original data for groupby 
newlist=[ x+(mapdict[x[0]],)for  x in L]
import itertools
#using groupby make the same id into one sublist
newlist=sorted(newlist,key=lambda x : x[2])
yourlist=[list(y) for x , y in itertools.groupby(newlist,key=lambda x : x[2])]
yourlist
[[('A', 'B', 0), ('B', 'C', 0), ('C', 'D', 0)], [('E', 'F', 1)], [('G', 'H', 2), ('H', 'I', 2), ('G', 'I', 2), ('G', 'J', 2)]]

匹配您的输出

L1,L2,L3=[[y[:2]for y in x] for x in yourlist]
L1
[('A', 'B'), ('B', 'C'), ('C', 'D')]
L2
[('E', 'F')]
L3
[('G', 'H'), ('H', 'I'), ('G', 'I'), ('G', 'J')]
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