在python中查找未出现元组元素的最近索引

Question

R=[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2 , 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]

从上面的 LIST R 中，我需要为其未出现的任何元素的每个元组找到最近的索引。我的意思是没有元组的值是相同的。而且这个元组的顺序也是固定的。

最后，当我们组合三个元组时，我们应该拥有所有源元素 (1,2,3,4,5,6)

预期输出：

对于 0: (0 , 9 , 14): (1,2) (3,4) (5,6) 对于 1: (1 , 6 , 14): (1,3) (2,4) (5,6) . . . . 对于 4:( 4 , 2 , 12): (1,2) (3,4) (5,6) . . .等等

请帮助...谢谢。

我尝试过的东西太大了，我自己对此并不满意

for j in range(i+1,i+10):
    b=set(Results[j])
    if (len(a&b)==0):

        for k in range(i+10, i+200):
            c=set(Results[k])
            if ( (len(a&c)==0) and (len(b&c)==0) ):

                for l in range(i+200, i+600):
                    d=set(Results[l])
                    if ( (len(a&d)==0) and (len(b&d)==0)  and (len(c&d)==0) ):

Answer 1

这消除了您的嵌套循环之一。我不知道您从哪里获得 i 值，也不知道您为什么在内部循环中使用如此大的范围。

R = [(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]
# keep track of the indices where tuple elements are disjoint
matches = {}
# loop over R
for index in range(len(R)):
  # track the elements seen so far
  elements = set(R[index])
  # track the index where elements were seen
  matches[index] = [index]
  # loop over subsequent items
  for _index in range(index, len(R)):
    # if the tuple has all items not already seen, record it
    if elements.isdisjoint(set(R[_index])):
      elements = elements.union(R[_index])
      matches[index] += [_index]
# discard anything that does not have all six items
final_matches = {k: v for k, v in matches.viewitems() if len(v) == 3}