如果一个元素在2个或更多不同的列表中相同，我正在尝试创建一个将多个列表串联的函数。
示例：
[[1,2],[3,4,5],[0,4]]将成为[[1,2],[0,3,4,5]
[[1],[1,2],[0,2]]将成为[[0,1,2]]
[[1, 2], [2, 3], [3, 4]]将成为[[1,2,3,4]]
实际上，只要它们具有一个公共元素，我们就将列表重新分组，然后删除两个元素之一。决赛名单必须具有独特的元素。
我尝试执行以下功能。它可以工作，但是当使用大列表（大约100或200个列表列表）时，出现以下递归错误：
RecursionError: maximum recursion depth exceeded while getting the repr of an object
def concat(L):
break_cond = False
print(L)
for L1 in L:
for L2 in L:
if (bool(set(L1) & set(L2)) and L1 != L2):
break_cond = True
if (break_cond):
i, j = 0, 0
while i < len(L):
while j < len(L):
if (bool(set(L[i]) & set(L[j])) and i != j):
L[i] = sorted(L[i] + list(set(L[j]) - set(L[i])))
L.pop(j)
j += 1
i += 1
return concat(L)
此外，我想只使用基本的python而不用那么多的库。任何想法 ？谢谢
出现错误的列表示例：
[[0, 64], [1, 120, 172], [2, 130], [3, 81, 102], [5, 126], [6, 176], [7, 21, 94], [8, 111, 167], [9, 53, 60, 138], [10, 102, 179], [11, 45, 72], [12, 53, 129], [14, 35, 40, 58, 188], [15, 86], [18, 70, 94], [19, 28], [20, 152], [21, 24], [22, 143, 154], [23, 110, 171], [24, 102, 144], [25, 73, 106, 187], [26, 189], [28, 114, 137], [29, 148], [30, 39], [31, 159], [33, 44, 132, 139], [34, 81, 100, 136, 185], [35, 53], [37, 61, 138], [38, 144, 147, 165], [41, 42, 174], [42, 74, 107, 162], [43, 99, 123], [44, 71, 122, 126], [45, 74, 144], [47, 94, 151], [48, 114, 133], [49, 130, 144], [50, 51], [51, 187], [52, 124, 142, 146, 167, 184], [54, 97], [55, 94], [56, 88, 128, 166], [57, 63, 80], [59, 89], [60, 106, 134, 142], [61, 128, 145], [62, 70], [63, 73, 76, 101, 106], [64, 80, 176], [65, 187, 198], [66, 111, 131, 150], [67, 97, 128, 159], [68, 85, 128], [69, 85, 169], [70, 182], [71, 123], [72, 85, 94], [73, 112, 161], [74, 93, 124, 151, 191], [75, 163], [76, 99, 106, 129, 138, 152, 179], [77, 89, 92], [78, 146, 156], [79, 182], [82, 87, 130, 179], [83, 148], [84, 110, 146], [85, 98, 137, 177], [86, 198], [87, 101], [88, 134, 149], [89, 99, 107, 130, 193], [93, 147], [95, 193], [96, 98, 109], [104, 105], [106, 115, 154, 167, 190], [107, 185, 193], [111, 144, 153], [112, 128, 188], [114, 136], [115, 146], [118, 195], [119, 152], [121, 182], [124, 129, 177], [125, 156], [126, 194], [127, 198], [128, 149], [129, 153], [130, 164, 196], [132, 140], [133, 181], [135, 165, 170, 171], [136, 145], [141, 162], [142, 170, 187], [147, 171], [148, 173], [150, 180], [153, 191], [154, 196], [156, 165], [157, 177], [158, 159], [159, 172], [161, 166], [162, 192], [164, 184, 197], [172, 199], [186, 197], [187, 192]]
答案 0 :(得分：7)
如@ScottBoston所述，这是一个图形问题，称为connected components，我建议您按照@ScottBoston的指示使用networkx，以防万一您无法使用没有networkx的版本：
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(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 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]]
答案 1 :(得分：6)
您可以使用networkx库，因为存在graph theory和connected components问题：
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()
#Add nodes to Graph
G.add_nodes_from(sum(l, []))
#Create edges from list of nodes
q = [[(s[i],s[i+1]) for i in range(len(s)-1)] for s in l]
for i in q:
#Add edges to Graph
G.add_edges_from(i)
#Find all connnected components in graph and list nodes for each component
[list(i) for i in nx.connected_components(G)]
输出：
[[1, 2], [0, 3, 4, 5]]
如果取消注释第2行和注释第1行，则输出
[[0, 1, 2]]
与第3行类似，
[[1, 2, 3, 4]]
答案 2 :(得分：1)
这是一种迭代方法，应该与纯python一样高效。一件事是必须花费额外的通行证才能消除重复项。
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)
[[1, 2], [0, 3, 4, 5]]
答案 3 :(得分：0)
您可以使用广度优先搜索的递归版本，而无需导入：
def group_vals(d, current, _groups, _seen, _master_seen):
if not any(set(current)&set(i) for i in d if i not in _seen):
yield list({i for b in _groups for i in b})
for i in d:
if i not in _master_seen:
yield from group_vals(d, i, [i], [i], _master_seen+[i])
else:
for i in d:
if i not in _seen and set(current)&set(i):
yield from group_vals(d, i, _groups+[i], _seen+[i], _master_seen+[i])
def join_data(_data):
_final_result = list(group_vals(_data, _data[0], [_data[0]], [_data[0]], []))
return [a for i, a in enumerate(_final_result) if a not in _final_result[:i]]
c = [[[1,2],[3,4,5],[0,4]], [[1],[1,2],[0,2]], [[1, 2], [2, 3], [3, 4]]]
print(list(map(join_data, c)))
输出：
[
[[1, 2], [0, 3, 4, 5]],
[[0, 1, 2]],
[[1, 2, 3, 4]]
]
答案 4 :(得分：0)
如果您希望以简单的形式在此处找到解决方法：
def concate(l):
len_l = len(l)
i = 0
while i < (len_l - 1):
for j in range(i + 1, len_l):
# i,j iterate over all pairs of l's elements including new
# elements from merged pairs. We use len_l because len(l)
# may change as we iterate
i_set = set(l[i])
j_set = set(l[j])
if len(i_set.intersection(j_set)) > 0:
# Remove these two from list
l.pop(j)
l.pop(i)
# Merge them and append to the orig. list
ij_union = list(i_set.union(j_set))
l.append(ij_union)
# len(l) has changed
len_l -= 1
# adjust 'i' because elements shifted
i -= 1
# abort inner loop, continue with next l[i]
break
i += 1
return l
答案 5 :(得分：0)
如果您想了解算法的工作原理，可以使用以下使用连通性矩阵的脚本：
import numpy
def Concatenate(L):
result = []
Ls_length = len(L)
conn_mat = numpy.zeros( [Ls_length, Ls_length] ) # you can use a list of lists instead of a numpy array
check_vector = numpy.zeros( Ls_length ) # you can use a list instead of a numpy array
idx1 = 0
while idx1 < Ls_length:
idx2 = idx1 + 1
conn_mat[idx1,idx1] = 1 # the diaginal is always 1 since every set intersects itself.
while idx2 < Ls_length:
if bool(set(L[idx1]) & set(L[idx2]) ): # 1 if the sets idx1 idx2 intersect, and 0 if they don't.
conn_mat[idx1,idx2] = 1 # this is clearly a symetric matrix.
conn_mat[idx2,idx1] = 1
idx2 += 1
idx1 += 1
print (conn_mat)
idx = 0
while idx < Ls_length:
if check_vector[idx] == 1: # check if we already concatenate the idx element of L.
idx += 1
continue
connected = GetAllPositiveIntersections(idx, conn_mat, Ls_length)
r = set()
for idx_ in connected:
r = r.union(set(L[idx_]))
check_vector[idx_] = 1
result.append(list(r))
return result
def GetAllPositiveIntersections(idx, conn_mat, Ls_length):
# the elements that intersect idx are coded with 1s in the ids' row (or column, since it's a symetric matrix) of conn_mat.
connected = [idx]
i = 0
idx_ = idx
while i < len(connected):
j = 0
while j < Ls_length:
if bool(conn_mat[idx_][j]):
if j not in connected: connected.append(j)
j += 1
i += 1
if i < len(connected): idx_ = connected[i]
return list(set(connected))
那你就是：
L = [[1,2],[3,4,5],[0,4]]
r = Concatenate(L)
print(r)