I like to eliminate duplicate components from double list, and then combine the double lists-CodePudding

I need your help to figure out this problem.

when I use this code, I can get this result.

input=[[1],[1,2],[5,7]] output=[[1,2],[5,7]]

Tb2 = Tb[:1] 
for t in Tb[1:]: 
    if set(t).isdisjoint(Tb2[-1]):
        Tb2.append(t)
    else:
        Tb2[-1] = sorted({*t,*Tb2[-1]})

But I can't solve the problem when another second list have same number in input.

input=[[2,3],[1,2],[5,7],[5,8],[7,8,9],[1]] expected output=[[1,2,3],[5,7,8,9]]

Would you give me advice or help?

CodePudding user response：

This seems like connected components problem from graph theory, so you could use networkx for it:

import networkx as nx
from itertools import combinations

# input
lst = [[2, 3], [1, 2], [5, 7], [5, 8], [7, 8, 9], [1]]

# build graph
g = nx.Graph()
g.add_edges_from([edge for ls in lst for edge in combinations(ls, 2)])

# compute components
components = nx.algorithms.components.connected_components(g)

res = list(components)
print(res)

Output

[{1, 2, 3}, {8, 9, 5, 7}]

The idea is to build an edge between each pair of elements of the same list, this is achieved with the following list comprehension:

[edge for ls in lst for edge in combinations(ls, 2)]
# [(2, 3), (1, 2), (5, 7), (5, 8), (7, 8), (7, 9), (8, 9)]

Once that is done simply run the connected_components algorithm on the graph.

CodePudding user response：

for t in Tb:

    for j in range(len(Tb2)):

        if set(t).isdisjoint(Tb2[j]):
            Tb2.append(t)
        else:
            Tb2[j] = sorted({*t,*Tb2[j]})

Output= [[1, 2, 3], [5, 7, 8, 9], [5, 7, 8, 9], [7, 8, 9], [1], [1], [1]]

I can't eliminate same and small size of double lists.