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"),"J")]
使用python我想有效地将此列表拆分为：
L1 = [("A","D")]
L2 = [("E","F")]
L3 = [("G","J"),"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.
解决方法
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',('C','D',0)],[('E','F',1)],[('G','H',2),('H','I',('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'),'C'),'D')]
L2
[('E','F')]
L3
[('G','H'),'I'),'J')]
