Python 将大量多个list中有重复的元素的合成一个list

本文主要介绍Python中，将大量多个的list，其中有重复元素，或者交集的元素的list合成一个，没有的不变。

示例效果：

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

[[1],[1,2],[0,2]] 会成为 [[0,1,2]]

[[1, 2], [2, 3], [3, 4]] 会成为 [[1,2,3,4]]

1、使用itertools实现

from itertools import combinationsdef bfs(graph, start):    visited, queue = set(), [start]    while queue:        vertex = queue.pop(0)        if vertex not in visited:            visited.add(vertex)            queue.extend(graph[vertex] - visited)    return visiteddef connected_components(G):    seen = set()    for v in G:        if v not in seen:            c = set(bfs(G, v))            yield c            seen.update(c)def graph(edge_list):    result = {}    for source, target in edge_list:        result.setdefault(source, set()).add(target)        result.setdefault(target, set()).add(source)    return resultdef concat(l):    edges = []    s = list(map(set, l))    for i, j in combinations(range(len(s)), r=2):        if s[i].intersection(s[j]):            edges.append((i, j))    G = graph(edges)    result = []    unassigned = list(range(len(s)))    for component in connected_components(G):        union = set().union(*(s[i] for i in component))        result.append(sorted(union))        unassigned = [i for i in unassigned if i not in component]    result.extend(map(sorted, (s[i] for i in unassigned)))    return resultprint(concat([[1, 2], [3, 4, 5], [0, 4]]))print(concat([[1], [1, 2], [0, 2]]))print(concat([[1, 2], [2, 3], [3, 4]]))

输出:

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

2、迭代方法实现

original_list = [[1,2],[3,4,5],[0,4]]mapping = {}rev_mapping = {}for i, candidate in enumerate(original_list):    sentinel = -1    for item in candidate:        if mapping.get(item, -1) != -1:            merge_pos = mapping[item]            #update previous list with all new candidates            for item in candidate:                mapping[item] = merge_pos            rev_mapping[merge_pos].extend(candidate)            break    else:        for item in candidate:            mapping[item] = i        rev_mapping.setdefault(i, []).extend(candidate)result = [list(set(item)) for item in rev_mapping.values()]print(result)

输出：

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