我有一个成对(元组)列表，为简化起见，像这样:
L = [("A","B"), ("B","C"), ("C","D"), ("E","F"), ("G","H"), ("H","I"), ("G","I"), ("G","J")]
我想使用python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")]
如何有效地将列表分成几对，对于该组中的对，必须始终至少有一对与其他项共享一个项目？ 如答案之一所述，这实际上是网络问题。目标是有效地将网络分成断开(隔离)的网络部分。
类型列表，元组(集合)可以更改以实现更高的效率。
最佳答案
这更像是网络问题，因此我们使用networks
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')]
关于python - 如何根据共享项目有效地对配对？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55232459/