如何获取二维列表中元组的索引？

Question

我需要获取相互链接的元组位置，其中至少有一个元素。

 [('1','1','1'),

  ('X','1','X'),

  ('Z','Z','Z'),

  ('Y','Y','X')]

此处，在第一个元组中，值“1”出现在第二个元组中。并且，现在第二个元组值 'X' 出现在最后一个元组中。

因此，我需要将所有这些组合到一个列表中，而元组 3 值['Z'] 与任何其他元组都不匹配。因此，它存储在单独的列表中。

Expected : [[0,1,3],[2]]

我的做法：

df = [('1','1','1'),('X','1','X'),('Z','Z','Z'),('Y','Y','X')]
res = [list(sub) for sub in df]
res
count=0
list_A=[]
for i,j in enumerate(res): 
    for m,n in enumerate(j):
        if res[i][m] in res[i]:
            print(res.index(j))

Answer 1

试试这个代码：

df = [('1', '1', '1'),  # 0
      ('X', '1', 'X'),  # 1
      ('Z', 'Z', 'Z'),  # 2
      ('Y', 'Y', 'X'),  # 3
      ('Z', 'Z', 'Z'),  # 4
      ('2', '2', '2')]  # 5
# first gather all the sublist numbers that intersect
out = []
for x in df:
    out.append({n for n, v in enumerate(df) if set(x).intersection(v)})
# then remove intersected (duplicated) sublist number subsets
output = []
while out:
    buf = out.pop(0)
    for x in out:
        if buf & x:  # intersection not empty
            buf.update(x)
            out.remove(x)
    output.append(list(buf))
print(output)

输出：

[[0, 1, 3], [2, 4], [5]]

Answer 2

def find_linked_tuple_positions(tuple_list):
    tuple_list: list

    result_list = []
    checked_tuples = []
    for index, current_tuple in enumerate(tuple_list):
        if index in checked_tuples:
            continue
        linked_tuples = [index]
        result_list.append(linked_tuples)
        current_unique_items = set(current_tuple)
        for next_index, next_tuple in enumerate(tuple_list[index + 1:], index + 1):
            print(next_index, next_tuple)
            next_tuple_unique_items = set(next_tuple)
            match_items = [unique_item for unique_item in current_unique_items if
                           unique_item in next_tuple_unique_items]
            if match_items:  # tuples are linked
                current_unique_items = current_unique_items | next_tuple_unique_items  # updating set
                linked_tuples.append(next_index)  # updating the new link
                checked_tuples.append(next_index)
    return result_list


def main():
    tuple_list = [('1', '1', '1'), ('X', '1', 'X'), ('Z', 'Z', 'Z'), ('Y', 'Y', 'X')]
    result_list = find_linked_tuple_positions(tuple_list)
    print(result_list)

如何获取二维列表中元组的索引？

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