Python大量多个列表(list)合并(合并有相同元素的列表)
示例：
[[1,2],[3,4,5],[0,4]] 合并后 [[1,2],[0,3,4,5]
[[1],[1,2],[0,2]] 合并后 [[0,1,2]]
[[1, 2], [2, 3], [3, 4]] 合并后 [[1,2,3,4]]
1、使用itertools中combinations合并
from itertools import combinations
def 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 visited
def connected_components(G):
seen = set()
for v in G:
if v not in seen:
c = set(bfs(G, v))
yield c
seen.update©
def graph(edge_list):
result = {}
for source, target in edge_list:
result.setdefault(source, set()).add(target)
result.setdefault(target, set()).add(source)
return result
def 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 result
print(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、使用networkx合并
import networkx as nx
l = [[1,2],[3,4,5],[0,4]]
#l = [[1],[1,2],[0,2]]
#l = [[1, 2], [2, 3], [3, 4]]
G = nx.Graph()
#将节点添加到Graph
G.add_nodes_from(sum(l, []))
#从节点列表创建边
q = [[(s[i],s[i+1]) for i in range(len(s)-1)] for s in l]
for i in q:
#向Graph添加边
G.add_edges_from(i)
#查找每个组件的图形和列表节点中的所有连接组件
[list(i) for i in nx.connected_components(G)]
输出：
[[1, 2], [0, 3, 4, 5]]
#[[0, 1, 2]]
#[[1, 2, 3, 4]]
[[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 combinations
def 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 visited
def connected_components(G):
seen = set()
for v in G:
if v not in seen:
c = set(bfs(G, v))
yield c
seen.update©
def graph(edge_list):
result = {}
for source, target in edge_list:
result.setdefault(source, set()).add(target)
result.setdefault(target, set()).add(source)
return result
def 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 result
print(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]]