图元组列表之间的关系[重复]

Question

我有一个元组列表，比如说

tuplist = [('a','b'),('a','c'),('e','f'),('f','c'),('d','z'),('z','x')]

我正在尝试获取以下内容：

('a','b','c','e','f'),('d','z','x')

这是链接在一起的所有元素（就像图论中的树） 上述元组中的顺序（也可以是列表）无关紧要。

我已经设法获得单个元素的所有链接的字典，但我正在努力以一种干净有效的方式获得最终结果...... 到目前为止，这是我的代码：

ll=[('a','b'),('a','c'),('e','f'),('f','c'),('d','z'),('z','x')]
total = {}
total2={}
final=[]
for element in set([item for sublist in ll for item in sublist]):
    tp = []
    for tupl in ll:
        if element in tupl: 
            tp.append(tupl)

    tp = list(frozenset([item for sublist in tp for item in sublist]))
    total[element] = tp
print(total)

Answer 1

与this post 类似，您要查找的内容称为图的连通分量。一个简单的方法是用networkx建一个图，找到connected_components：

tuplist = [('a','b'),('a','c'),('e','f'),('f','c'),('d','z'),('z','x')]

import networkx as nx 

G=nx.Graph()
G.add_edges_from(tuplist)
list(nx.connected_components(G))
# [{'a', 'b', 'c', 'e', 'f'}, {'d', 'x', 'z'}]

Answer 2

可选的递归解决方案，虽然不如@yatu 的解决方案networkx 优雅：

tuplist = [('a','b'),('a','c'),('e','f'),('f','c'),('d','z'),('z','x')]
def get_graphs(start, c = [], seen = []):
  _r = [b for a, b in tuplist if a == start and b not in seen]+[a for a, b in tuplist if b == start and a not in seen]
  if _r:
     yield from [i for b in _r for i in get_graphs(b, c=c+[start, b, *_r], seen = seen+[start, b])]
  else:
     yield set(c)
     _r = [a for a, _ in tuplist if a not in seen]
     yield from ([] if not _r else get_graphs(_r[0], c = [], seen= seen))


result = list(get_graphs(tuplist[0][0]))
final_result = [a for i, a in enumerate(result) if all(all(k not in b for k in a) for b in result[i+1:])]
to_tuples = list(map(tuple, final_result))

输出：

[('f', 'e', 'a', 'c', 'b'), ('z', 'd', 'x')]