Question

我有一个这样的元组列表：

all_tuples=[(92, 242),(355, 403),(355, 436),(355, 489),(403, 436),(436, 489),(515, 517),(517, 859),(634, 775),(701, 859),(775, 859)]

，我需要取所有元组的交集并将它们合并。

The desired result = [{92, 242},{355, 403,436,489},{515, 517,859,701,775,634}]

那是相交的元组迭代合并。

我试图将元组转换为集合，然后采用交集，但没有用。任何想法？

Answer 1

这是网络问题，使用networkx

import networkx as nx 
G=nx.Graph()
all_tuples=[(92, 242),(355, 403),(355, 436),(355, 489),(403, 436),(436, 489),(515, 517),(517, 859),(634, 775),(701, 859),(775, 859)]
G.add_edges_from(all_tuples)
list(nx.connected_components(G))
Out[1216]: [{92, 242}, {355, 403, 436, 489}, {515, 517, 634, 701, 775, 859}]

Answer 2

此解决方案构建了一个等价类列表，其中在同一个元组中是我们的等价关系。对于每个元组，我们在列表中列出与该元组的某些元素匹配的所有集合。如果没有，我们将创建一组该元组并将其添加到列表中。如果存在，我们将更新该集合以包括元组的其他项。如果有多个，我们将它们从列表中删除，将它们和元组合并为一组，然后将该组添加到列表中。

res = []
for t in all_tuples:
    matched = []
    for s in res:
        if s.intersection(t):
            matched.append(s)
    if not matched:
        res.append(set(t))
    elif len(matched) == 1:
        matched[0].update(t)
    else:
        res = [subl for subl in res if subl not in matched]
        res.append(set.union(*matched, t))

print(res)
# [{242, 92}, {489, 436, 355, 403}, {515, 517, 775, 634, 859, 701}]

Answer 3

请说明原因，这是仅包含列表和元组的实现。

all_tuples=[(92, 242),(355, 403),(355, 436),(355, 489),(403, 436),(436, 489),(515, 517),(517, 859),(634, 775),(701, 859),(775, 859)]

curr_min, curr_max = all_tuples[0]
final = []
intermediate = [curr_min, curr_max]
for next_left, next_right in all_tuples[1:]:
    if curr_max >= next_left:
        intermediate.extend([next_left, next_right])
        curr_min, curr_max = min(curr_min, next_left), max(curr_max, next_right)
    else:
        final.append(set(intermediate))
        intermediate = []
        curr_min, curr_max = next_left, next_right
final.append(set(intermediate))

列表中元组的交集-python

3 个答案: